Definite Integration

$\begin{aligned} & \int_\limits0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_\limits0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\ & \frac{1}{2}\left[\int_\limits0^1 \sqrt{3+x} d x-\int_\limits0^1(\sqrt{1+x}) d x\right] \end{aligned}$

$\begin{aligned} & \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 \\ & \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] \\ & \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] \\ & =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} \\ & a=3, b=-\frac{2}{3}, c=-1 \\ & 2 a+3 b-4 c=6-2+4=8 \end{aligned}$

Equation of CE

$\begin{aligned} & y-1=-(x-3) \\ & x+y=4 \end{aligned}$

orthocentre lies on the line $x+y=4$

so, $a+b=4$

$I_1=\int_\limits a^b x \sin (x(4-x)) d x\quad$ ..... (i)

Using king rule

$I_1=\int_\limits a^b(4-x) \sin (x(4-x)) d x\quad$ .... (ii)

$\begin{aligned} & \text { (i) }+ \text { (ii) } \\ & 2 \mathrm{I}_1=\int_\limits{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \\ & 2 \mathrm{I}_1=4 \mathrm{I}_2 \\ & \mathrm{I}_1=2 \mathrm{I}_2 \\ & \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \\ & \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72 \end{aligned}$

1. The given integral is:

$ I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)} $ .............(i)

1. Perform a substitution, $x \rightarrow 1-x$. This gives:

$ I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)} $

1. Simplify the expression inside the integral:

$ I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)} $ ............(ii)

1. Add the original integral (i) and the integral after substitution (ii):

$ 2I=\int_0^1 \frac{dx}{5+2x-2x^2} $

1. Factor the quadratic expression in the denominator:

$ 2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)} $

1. To solve this integral, we can perform a change of variables using the substitution
   
   $x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$, then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$:

$ I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du $

1. Using the identity $\sec^2{u}=1+\tan^2{u}$:

$ I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C) $

1. Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:

$ I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right] $

1. Using the properties of the arctangent function, we can rewrite the integral as:

$ I=\frac{1}{\sqrt{11}} \ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right) $

1. From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:

$ \alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21 $

Thus, $\alpha^4 - \beta^4 = 21$.

We're given the expression:

$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$

Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$, where $n$ is an integer. Let's denote this integral as $I(n)$:

$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$

We can then rewrite the original expression in terms of $I(n)$:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$

Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:

$\int u dv = uv - \int v du$

We'll choose:

$u = \tan^n x, \quad dv = e^{-x} dx$

Then we get:

$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$

Applying integration by parts, we have:

$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$

Since $\tan(\pi / 4) = 1$, the first term evaluates to:

$-e^{-\pi / 4}$

The second term becomes:

$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$

This is equal to:

$n(I(n-1) + I(n+1))$

So we have:

$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$

Now we can substitute $n = 50$ into this equation:

$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$

So the original expression becomes:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$

$ \begin{aligned} & S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\\\ & \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\\\ & S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)^{15} \\\\ & =\int_0^1 x^{15} d x=\frac{1}{16} \\\\ & \therefore \text { Both } S_1 \text { and } S_2 \text { are correct. } \end{aligned} $

$ \begin{aligned} & \mathrm{l}=\int_0^{\infty} \frac{6}{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+2\right)\left(\mathrm{e}^{\mathrm{x}}+3\right)} \mathrm{dx} \\\\ & =6 \int_0^{\infty}\left(\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+1}+\frac{-1}{\mathrm{e}^{\mathrm{x}}+2}+\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+3}\right) \mathrm{dx} \\\\ & =3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}-6 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}} \mathrm{dx}}{1+2 \mathrm{e}^{-\mathrm{x}}}+3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+3 \mathrm{e}^{-\mathrm{x}}} \mathrm{dx} \\\\ & =3\left[-\ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty}+6 \frac{1}{2}\left[\ln \left(1+2 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} -\frac{3}{3}\left[\ln \left(1+3 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} \\\\ & =3 \ln 2-3 \ln 3+\ln 4 \\\\ & =3 \ln \frac{2}{3}+\ln 4 \\\\ & =\ln \frac{32}{27} \end{aligned} $

The integral equation is given by :

$\int\limits_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$

Step 1 : Break the first integral into two parts :

$I = \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \sin x \, dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx$

3. Apply the King's property, $\int\limits_a^b f(x) dx = \int\limits_a^b f(a+b-x) dx$, to the first integral and substitute $x - \frac{\pi}{4} = t$ in the second integral.

This gives :

$ \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin(\frac{\pi}{4}-x) \, dx + \int\limits_0^{\frac{\pi}{4}} f(\cos 2t) \sin(\frac{\pi}{4}+t) \, dt + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0 $

$ \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) [2 \sin \frac{\pi}{4} \cos x + \alpha \cos x] \, dx = 0 $

Then, noticing that $2 \sin \frac{\pi}{4} = \sqrt{2}$, you can factor out the term $\cos x$:

$ = (\alpha + \sqrt{2}) \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0 $

In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero. Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:

$ \alpha + \sqrt{2} = 0 \Rightarrow \alpha = -\sqrt{2} $

So, the correct answer to the original problem is $\alpha = -\sqrt{2}$, which corresponds to Option C.

$ \begin{aligned} \operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\ {\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2) \end{aligned} $

$ \therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}\right. $

$ \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} $

$ =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2 $

$ \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned} $

$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$

Let $e^x=t \Rightarrow e^x d x=d t$

When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$

When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$

$ I=\int_\limits{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}\right)\right] d t $ ...........(i)

On applying integration by part method in Eq. (i), we get

$ \begin{aligned} & I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\\\ & =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t \end{aligned} $

$ =\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t $ .............(ii)

Let $\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t$

Let $1+t^2=w$

$2 t d t=d w$

When, $t \rightarrow \frac{1}{2}$, then $w=\frac{5}{4}$

When, $t \rightarrow 2$, then $w=5$

$ \begin{aligned} I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\\\ & =[2 \sqrt{w}]_{5 / 4}^5 \\\\ & =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5} \end{aligned} $

On substitute value of $I_1$ in Eq. (ii), we get

$ I=\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2} $

Given that

$ \int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0 $

On differentiating using Leibnitz rule, we get

$ \begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned} $

On substituting $\frac{\pi}{2}$ for $t$, we get

$ f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right) $

Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)

$ \begin{aligned} & =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\ & =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii) \end{aligned} $

On adding Equations (i) and (ii), we get

$ \begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned} $

Let $L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)\right\}$

Here, $\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right)$ $ \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

$ \Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0$

$\therefore L = 0$

We have,

$ 5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0 $ ..........(i)

On replacing $x$ by $\frac{1}{x}$ in (i), we get

$ 5 f\left(\frac{1}{x}\right)+4 f(x)=x+3 $ ..........(ii)

Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get

$ \begin{aligned} & 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\ &\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\ &\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii) \end{aligned} $

$ \begin{aligned} \therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\ = & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x \end{aligned} $

$ \begin{aligned} & =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\ & =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\ & =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\ & =10 \log _e 2-6 \end{aligned} $

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x $

Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$ I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x $

$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi d x}{2(2-\cos 2 x)} \\\\ & \Rightarrow I=\frac{2 \pi}{4} \int_0^{\frac{\pi}{4}}\left(\frac{d x}{\left.\frac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\\\ & \Rightarrow I=\frac{\pi}{2} \int_0^{\frac{\pi}{4}}\left(\frac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned}$

Now,

$ \begin{aligned} & \tan x=t \\\\ & =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\ & =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\ & =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned} $

$ \begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\ & =\int_0^1 \frac{d x}{x+1} \\\\ & =\left.\log _e(1+\mathrm{x})\right|_0 ^1 \\\\ & =\log _e^2 \end{aligned} $

$I=\int\limits_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha>0$

$ \begin{aligned} & =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x \end{aligned} $

$\eqalign{ & = {1 \over \alpha }\int\limits_0^\alpha {\left( {x\sqrt {x + \alpha } + x\sqrt x } \right)} dx \cr & = {1 \over \alpha }\int\limits_0^\alpha {\left[ {\left( {x + \alpha - \alpha } \right)\sqrt {x + \alpha } + {x^{{3 \over 2}}}} \right]} dx \cr} $

$ \begin{aligned} & =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\ & =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\\\ & =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\\\ & =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\\\ & =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10) \end{aligned} $

Now, $\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}=\frac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}$

$ \Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2} $

$\Rightarrow \alpha=2 $

$\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)$

At, $x=\frac{\pi}{4}$

$\phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{-1}{2\left(\frac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\frac{4}{\pi}}\left(4 \sqrt{2} \times \frac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)\left[1+\frac{6}{\sqrt{\pi}}\right]=\frac{2}{\sqrt{\pi}} \times 4$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt{x}}$

$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$

$ \begin{aligned} & = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\ & =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\ & =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t \end{aligned} $

$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$

$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$

$=-\mathrm{P}_{50}(\alpha)-\beta$

Let I = $\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x $

$ \begin{aligned} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x\left(1-\cos ^2 x\right)} & d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x \end{aligned} $

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x $

$ \begin{array}{r} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x \end{array} $

$ =2 I_1-2 I_2+\frac{3}{2} I_3 $

Now,

$ \begin{aligned} I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\\\ & =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x \end{aligned} $

Put $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$

When, $x=\frac{\pi}{3} \Rightarrow t=\frac{1}{\sqrt{3}} \text { and } x=\frac{\pi}{2} \Rightarrow t=0$

$ \begin{aligned} \therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\\\ = & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\\\ = & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\\\ = & \frac{1}{3}+\frac{1}{2} \log \sqrt{3} \end{aligned} $

$ I_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x $

Put $\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t$

When, $x=\frac{\pi}{3}$ $\Rightarrow t=\frac{1}{\sqrt{3}}$ and $x=\frac{\pi}{2} \Rightarrow t=0$

$ \therefore \begin{aligned} I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\\\ & =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6} \end{aligned} $

$ \begin{aligned} & I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\\\ = & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} & \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\\\ & =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right) \end{aligned} $

$ \begin{aligned} & =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\\\ & =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3} \end{aligned} $

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $

$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $

$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$

$ = {3^3} - {2^3} = 19$

$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $

$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $ ($\because$ $[x] = 1$ when $x \in (12)$)

$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $

Let ${x^3} = t$

$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $

$ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})$

$ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}$

$ = {e^8} - e$

$\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt} $

$ = \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} } $

$ = \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} } $

$ = \left. {{{\tan }^{ - 1}}(2t + \sqrt 3 ) + {{\tan }^{ - 1}}(2t - \sqrt 3 )} \right|_1^2 - \left. {{2 \over 3}{{\tan }^{ - 1}}({t^3})} \right|_1^2$

$ = {\tan ^{ - 1}}(4 + \sqrt 3 ) + {\tan ^{ - 1}}(4 - \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2\sqrt 3 ) - {2 \over 3}({\tan ^{ - 1}}8 - {\tan ^{ - 1}}1)$

$ = {\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$

$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $ ..... (i)

$x \to {1 \over x}$

$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $ ..... (ii)

$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx} $

$ = \left. {{\pi \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2$

$ \Rightarrow I = {\pi \over 2}\ln 2$

$f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$

$f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$

On comparing with

$f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then

$\Rightarrow \frac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)$

$\Rightarrow \frac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)$

Add (2) and (3)

$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)$

$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y\quad....(5)$

Add (4) and (5)

$\frac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y$

$=\pi+\frac{a+b}{\pi^{2}-4}\left(\frac{\pi}{2}+1\right)$

$(a+b)=-2 \pi(\pi+2)$

$I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$

$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$

$=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$

Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$

$=\frac{11}{6}-\ln 4$

$ f(x)=\int_0^2 e^{|x-t|} d t $

For $x>2$

$ f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right) $

For $x<0$

$ f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right) $

For $x \in[0,2]$

$ \begin{aligned} & f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\ & =e^{2-x}+e^x-2 \end{aligned} $

For $x>2$

$ \left.f(x)\right|_{\min =e^2-1} $

For $\mathrm{x}<0$

$ \left.f(x)\right|_{\min =e^2-1} $

For $x \in[0,2]$

$ \left.f(x)\right|_{\min }=2(e-1) $

$ \int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x $

We have $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$

Hence $\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}$

$=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right]$ $=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right]$

$=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$

$=24 \times \frac{\pi}{12}=2 \pi$

$\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx} $

$3{x^2} - 5x + 2 = 0$

$ \Rightarrow 3{x^2} - 3x - 2x + 2 = 0$

$ \Rightarrow 3x(x - 1) - 2(x - 1) = 0$

$ \Rightarrow (x - 1)(3x - 2) = 0$

$\therefore$ In 0 to ${2 \over 3},\,|3{x^2} - 5x + 2| = 3{x^2} - 5x + 2$

And in ${2 \over 3}$ to 1, $|3{x^2} - 5x + 2| = - (3{x^2} - 5x + 2)$

$\therefore$ $\int_0^{{2 \over 3}} {\left[ {2x - 3{x^2} + 5x - 2 + 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {2x + 3{x^2} - 5x + 2 + 1} \right]dx} } $

$ = \int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} } $

Now, let $f(x) = - 3{x^2} + 7x - 1$

$ \Rightarrow f'(x) = - 6x + 7$

$ = - 6\left( {x - {7 \over 6}} \right)$

$f(0) = - 1$

$f\left( {{2 \over 3}} \right) = - 3 \times {4 \over 9} + {{14} \over 3} - 1$

$ = {7 \over 3} = 2.33$

$\therefore$ Range of $f(x) = \left[ { - 1,{7 \over 3}} \right]$

Integer value between $ - 1$ to ${7 \over 3} = - 1,0,1,2,{7 \over 3}$

When $f(x) = - 1$

$ \Rightarrow - 3{x^2} + 7x - 1 = - 1$

$ \Rightarrow 3{x^2} - 7x = 0$

$ \Rightarrow x(3x - 7) = 0$

$ \Rightarrow x = 0,{7 \over 3}$ (not possible as ${7 \over 3} \notin (0,2)$)

When $f(x) = 0$

$ \Rightarrow - 3{x^2} + 7x - 1 = 0$

$ \Rightarrow 3{x^2} - 7x + 1 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 12} } \over 6}$

$ = {{7\, \pm \sqrt {37} } \over 6}$

$\therefore$ $x = {{7 - \sqrt {37} } \over 6}$

When $f(x) = 1$

$ \Rightarrow - 3{x^2} + 7x - 1 = 1$

$ \Rightarrow 3{x^2} - 7x + 2 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 24} } \over 6}$

$ = {{7\, \pm \,\sqrt {25} } \over 6}$

$ = {{7\, \pm \,5} \over 6}$

$ \Rightarrow x = 2,\,{1 \over 3}$

When $f(x) = 2$

$ \Rightarrow - 3{x^2} + 7x - 1 = 2$

$ \Rightarrow 3{x^2} - 7x + 3 = 0$

$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 36} } \over 6}$

$ = {{7\, \pm \,\sqrt {13} } \over 6}$

$\therefore$ $x = {{7\, - \sqrt {13} } \over 6}$

$\therefore$ Possible values of $x = 0,\,{{7\, - \,\sqrt {37} } \over 6},{1 \over 3},{{7\, - \,\sqrt {13} } \over 6},{2 \over 3}$

$\therefore$ $\int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx} $

$ = \int_0^{{{7\, - \,\sqrt {37} } \over 6}} {( - 1)dx + \int_{{{7\, - \,\sqrt {37} } \over 6}}^{{1 \over 3}} {0\,dx + \int_{{1 \over 3}}^{{{7\, - \,\sqrt {13} } \over 6}} {(1)\,dx + \int_{{{7\, - \,\sqrt {13} } \over 6}}^{{2 \over 3}} {2\,dx} } } } $

$ = {{ - 7\, + \,\sqrt {37} } \over 6} + 0 + {{7\, - \,\sqrt {13} } \over 6} - {1 \over 3} + 2\left[ {{2 \over 3} - {{7\, - \,\sqrt {13} } \over 6}} \right]$

$ = {{ - 7 + \sqrt {37} + 7 - \sqrt {13} - 2 + 8 - 14 + 2\sqrt {13} } \over 6}$

$ = {{\sqrt {37} + \sqrt {13} - 8} \over 6}$

Now, $\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $

Let $f(x) = 3{x^2} - 3x + 3$

$f'(x) = 6x - 3 = 3(2x - 1)$

$f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = {1 \over 2}$

$f\left( {{2 \over 3}} \right) = 3\left( {{4 \over 9} - {2 \over 3} + 1} \right)$

$ = {7 \over 3} = 2.3$

$f(1) = 3(1 - 1 + 1)$

$ = 3$

No integer values present in between ${7 \over 3}$ and 3.

$\therefore$ Possible value of $x = {2 \over 3},\,1$

So, integration don't break anywhere.

$\therefore$ $\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $

$ = \int_{{2 \over 3}}^1 {\left[ {2.33} \right]\,dx} $

$ = \int_{{2 \over 3}}^1 {2\,dx = 2\left( {1 - {2 \over 3}} \right) = {2 \over 3}} $

$\therefore$ Value of Integration

$ = {{\sqrt {37} + \sqrt {13} - 1} \over 6} + {2 \over 3}$

$ = {{\sqrt {37} + \sqrt {13} - 8 + 4} \over 6}$

$ = {{\sqrt {37} + \sqrt {13} - 4} \over 6}$

$I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} $

$ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} $

Let $\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt$

$I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}} $

$ = \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} } $

$ = \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}$

$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt} $ ...... (i)

$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $

Substituting $t \to {1 \over p}$

$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $

$ = \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $ ....... (ii)

By (i) + (ii)

$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $

$ = {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$

$\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$

${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt} $

$ = \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt} $

$ = \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt} $

$ = {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt} $

${I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)$

$10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)$

For $n = 5$

$50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)$

$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $

$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $

${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $

Differentiate on both side

${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1$

$f(x) = {e^x}(2x - 1)$

$f'(x) = {e^x}(2) + {e^x}(2x - 1)$

$ = {e^x}(2x + 1)$

$x = - {1 \over 2}$

$f''(x) = {e^x}(2) + (2x + 1){e^x}$

$ = {e^x}(2x + 3)$

For $x = - {1 \over 2}\,\,f''(x) > 0$

$\Rightarrow$ Maxima

$\therefore$ Max. $ = {e^{ - {1 \over 2}}}( - 1 - 1)$

$\therefore$ $ - {2 \over {\sqrt e }}$

$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$

$\therefore$ $f(x) = \left\{ {\matrix{ { - x} & , & {x < - 1} \cr {x + 2} & , & { - 1 \le x < 0} \cr {3x + 2} & , & {0 \le x < 1} \cr {x + 4} & , & {x \ge 1} \cr } } \right.$

$\therefore$ $f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$

and $\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $

$ = \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2$

$ = {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$

$\therefore$ Only (S2) is correct

$\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} } $

$ = \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } } $

$ = \left. {\left( {{{3{x^2}} \over 2} - {{2{x^3}} \over 3}} \right)} \right|_0^{3/2} + \left. {\left( {{{2{x^3}} \over 3} - {{3{x^2}} \over 2}} \right)} \right|_{3/2}^2 - {1 \over 2} + {1 \over 2}$

$ = \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right)$

$ = {{19} \over {12}}$

$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$

Now,

$\because$ $\mathop {\lim }\limits_{x \to - 1} f(x)$ exist

$\therefore$ $\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)$

$ \Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2$

$ \Rightarrow - a = 1 \Rightarrow a = - 1$

Now, $\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} } $

$ = \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } } $

$ = 1 - 1 - 1 - 1 = - 2$

I comes out around 1.536 which is not satisfied by any given options.

$\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } $

${\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} $

${{\sin x} \over x}$ is decreasing in $\left( {{\pi \over 4},{\pi \over 3}} \right)$ so it attains maximum at $x = {\pi \over 4}$

$I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} $

$I > \sqrt 3 - {\pi \over 6}$

$\because$ f(x) is continuous at x = 4

$ \Rightarrow f({4^ - }) = f({4^ + })$

$ \Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt} $

$ = \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} } $

$ = \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4$

$ = 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)$

$16 + 4b = 15$

$ \Rightarrow b = {{ - 1} \over 4}$

$ \Rightarrow f(x) = \left\{ {\matrix{ {\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \cr {{x^2} - {x \over 4}} & {x \le 4} \cr } } \right.$

$ \Rightarrow f'(x) = \left\{ {\matrix{ {5 - |x - 3|} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$

$ \Rightarrow f'(x) = \left\{ {\matrix{ {8 - x} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$

$f'(x) < 0 = x \in \left( { - \infty ,{1 \over 8}} \right) \cup (8,\infty )$

$f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}$

$f'(x) = 0 \Rightarrow x = {1 \over 8}$ have local minima

$\therefore$ (C) is only incorrect option.

$I = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx} $

$ = 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx} $

$ = 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx} $

$ = \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}$

$ = 40\left( {{\pi \over 2} + {1 \over 2} + {1 \over 2}} \right) = 20(\pi + 2)$

$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $

$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}} $

$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\int_0^1 { = {\pi \over 2}} } $

$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} ,\,x \in (0,1)$

$f(x) = {{1 - \cos x} \over {\sin x}} = \cos ec\,x - \cot x$

$f'(x) = \cos e{c^2}x - \cos ec\,x\cot x$

$\left. {\matrix{ {f\left( {{a \over 2}} \right) = f\left( {{\pi \over 4}} \right) = \sqrt 2 - 1} \cr {f'\left( {{a \over 2}} \right) = f'\left( {{\pi \over 4}} \right) = 2 - \sqrt 2 } \cr } } \right\}f'\left( {{a \over 2}} \right) = \sqrt 2 \,.\,f\left( {{a \over 2}} \right)$

$I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$

Let ${2^n} = t$ and if $n \to \infty $ then $t \to \infty $

$I = \mathop {\lim }\limits_{n \to \infty } {1 \over t}\left( {\sum\limits_{r = 1}^{t - 1} {{1 \over {\sqrt {1 - {r \over t}} }}} } \right)$

$I = \int\limits_0^1 {{{dx} \over {\sqrt {1 - x} }} = \int\limits_0^1 {{{dx} \over {\sqrt x }}\,\,\,\,\,\,\,\,\,\left( {\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } } \right.} } $

$ = \left[ {2{x^{{1 \over 2}}}} \right]_0^1 = 2$

$I = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx} $

$[\sin \pi x]$ is periodic with period 2 and ${{e^{[\cos (2\pi x)]}}}$ is periodic with period 1.

So,

$I = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx} $

$ = 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}$

$ = {{52} \over e}$

Case 1 :

Let $0 \le x < 1$

then $\left[ x \right] = 0$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 0 - x$

$ = 1 - x$

Case 2 :

Let $1 \le x < 2$

then $\left[ x \right] = 1$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 1$

Case 3 :

Let $2 \le x < 3$

then $\left[ x \right] = 2$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 2 - x$

$ = 3 - x$

Case 4 :

Let $3 \le x < 4$

then $\left[ x \right] = 3$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 3$

$\therefore$ $f(x) = \left\{ {\matrix{ {1 - x} & ; & {0 \le x < 1} \cr {x - 1} & ; & {1 \le x < 2} \cr {3 - x} & ; & {2 \le x < 3} \cr {x - 3} & ; & {3 \le x < 4} \cr } } \right.$

$\therefore$ $f(x)$ is periodic and period of $f(x) = 2$

And period of $\cos \pi x = {{2\pi } \over \pi } = 2$

$\therefore$ Period of $f(x)\cos \pi x = 2$

Now,

$I = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx} $

$ = {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx} $

$\therefore$ $I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]$

$ = {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]$

$ = {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]$

$ = {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]$

$\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]$

$ = {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]$

$ = {\pi ^2}\left[ { - \left\{ {\left( { - {1 \over {{\pi ^2}}}} \right) - \left( {{1 \over {{\pi ^2}}}} \right)} \right\} + \left( {\left( { + {1 \over {{\pi ^2}}}} \right) - \left( { - {1 \over {{\pi ^2}}}} \right)} \right.} \right]$

$ = {\pi ^2}\left[ { - \left( { - {2 \over {{\pi ^2}}}} \right) + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2} \times {4 \over {{\pi ^2}}}$

$ = 4$

$\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $

$ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}} $

$ = \int_a^b {f(x)dx} $

$a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0$

$b = \mathop {\lim }\limits_{n \to \alpha } \left( {{n \over n}} \right) = 1$

and ${r \over n} \to x$

$ = \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx} $

$ = \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx} $

$ = \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx} $

$ = \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx} $

$ = \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx} $

$ = \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1$

$ = - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]$

$ = - {3 \over 2}\log \left( {{1 \over 3}} \right) + 2\log \left( {{1 \over 2}} \right)$

$ = + {3 \over 2}\log 3 - 2\log 2$

$ = \log \sqrt {{3^3}} - \log 4$

$ = \log {{\sqrt {{3^2} \times 3} } \over 4}$

$ = \log {{3\sqrt 3 } \over 4}$

Given,

$f(x) = x + \int_0^1 {(x - t)f(t)dt} $

$ = x + x\int_0^1 {f(t)dt - \int_0^1 {tf(t)dt} } $

$ = x\left( {1 + \int_0^1 {f(t)dt} } \right) - \int_0^1 {tf(t)dt} $

Now,

let $A = 1 + \int_0^1 {f(t)dt} $

and $B = \int_0^1 {tf(t)dt} $

$\therefore$ $f(x) = Ax - B$

$ \Rightarrow f(t) = At - B$

So, $A = 1 + \int_0^1 {f(t)dt} $

$ = 1 + \int_0^1 {(At - B)dt} $

$ = 1 + \left[ {{{A{t^2}} \over 2} - Bt} \right]_0^1$

$ = 1 + {A \over 2} - B$

$ \Rightarrow {A \over 2} = 1 - B$ ...... (1)

$B = \int_0^1 {tf(t)dt} $

$ = \int_0^1 {t(At - B)dt} $

$ = \int_0^1 {(A{t^2} - Bt)dt} $

$ = \left[ {{{A{t^3}} \over 3} - {{B{t^2}} \over 2}} \right]_0^1$

$ = {A \over 3} - {B \over 2}$

$ \Rightarrow {{3B} \over 2} = {A \over 3}$ ....... (2)

Solving equation (1) and (2) we get,

$A = {{18} \over {13}}$ and $B = {4 \over {13}}$

$\therefore$ $f(x) = {{18} \over {13}}x - {4 \over {13}}$

By checking all the options you can see when x = 6 we get

$y = f(x) = {{18} \over {13}} \times 6 - {4 \over {13}}$

$ = {{108 - 4} \over {13}} = 8$

$\therefore$ Point (6, 8) lies on the curve.

Given,

$\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $

Now,

$L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx} $

$ = \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} } $

$ = \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx} $

$ = \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx} $

[We know,

$\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C$]

$ = {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2$

$ = {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]$

$ = {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)$

$ = {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}$

$ = {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}$

$ = {8 \over 3} - {{2\pi } \over 4}$

$ = {8 \over 3} - {\pi \over 2}$

Now in R.H.S.,

$\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy} $

$ = \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } } $

$ = \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1$

$ = 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}$

$ = 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}$

$ = 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}$

$ = {5 \over 6} - {\pi \over 4}$

Now,

$\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy} $

$ = \left[ {2y - {{{y^3}} \over 6}} \right]_1^2$

$ = \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]$

$ = 4 - {8 \over 6} - 2 + {1 \over 6}$

$ = 2 - {7 \over 6}$

$ = {5 \over 6}$

$\therefore$ $R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I$

As $L.H.S. = R.H.S.$

$ \Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I$

$ \Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}$

$ \Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}$

$ \Rightarrow I = 1 - {\pi \over 4}$

From option (c)

$\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy} $

$ = \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy} $

$ = 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1$

$ = 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]$

$ = 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]$

$ = 1 - {\pi \over 4} = I$

$\therefore$ Option (c) is the right answer.

Note :

There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.

$ \left\{\begin{array}{l} f(x)=x^3-3 x, x \leq-1 \\\\ 2,-1 < x < 2 \\\\ x^2+2 x-6,2 < x < 3 \\\\ 9,3 \leq x < 4 \\\\ 10,4 \leq x < 5 \\\\ 11, x=5 \\\\ 2 x+1, x>5 \end{array}\right. $

Clearly $\mathrm{f}(\mathrm{x})$ is not differentiable at

$ \begin{aligned} & x=2,3,4,5 \Rightarrow m=4 \\\\ & I=\int_{-2}^{-1}\left(x^3-3 x\right) d x+\int_{-1}^2 2 \cdot d x=\frac{27}{4} \end{aligned} $

We know,

$\left[ {{x \over 2}} \right]$ is discontinuous at 1, 2, 3, 4 ........

$\therefore$ [ x ] is discontinuous at 2, 4, 6, 8 .....

In between 0 to 5 it is discontinuous at 2 and 4.

Break the integration into 3 parts

(1) 0 to 2

(2) 2 to 4

(3) 4 to 5

$\therefore$ $\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $

$ = \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } } $

$ = \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } } $

$ = \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } } $

$ = \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5$

$ = 0 - 0 + 0$

$ = 0$

Given : $f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0$ and $f^{\prime}\left(\frac{\pi}{2}\right)=1$

$ \begin{aligned} &g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t \\\\ &=[\sec t+f(t)]_{x}^{\frac{\pi}{4}}=2-\sec x f(x) \end{aligned} $

Now, $\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\frac{\pi}{2}-h\right)$

$ =\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right) $

$=\lim \limits_{h \rightarrow 0}\left[2-\frac{f\left(\frac{\pi}{2}-h\right)}{\sin h}\right]$

$=\lim \limits_{h \rightarrow 0}\left[2+\frac{f^{\prime}\left(\frac{\pi}{2}-h\right)}{\cos h}\right]$

$ =3 $

$f: R \rightarrow R$ and $f(x)+f(x+k)=n \quad \forall x \in R$

$ \begin{aligned} &x \rightarrow x+k \\\\ &f(x+k)+f(x+2 k)=n \\\\ &\therefore \quad f(x+2 k)=f(x) \end{aligned} $

So, period of $f(x)$ is $2 k$

$ \begin{aligned} &\text { Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\\\ &x=t+k \Rightarrow d x=d t \text { (in second integral) } \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\\\ &=2 n^{2} k \end{aligned} $

Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$

$ I_{2}=2(n k) $

$\therefore \quad l_{1}+n l_{2}=4 n^{2} k$

$\int_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$


$=\int_{0}^{\frac{1}{4}}(-1) d x+\int_{\frac{1}{4}}^{\frac{3}{4}} 0 d x+\int_{\frac{1}{2}}^{\frac{3}{4}}-1 d x+\int_{\frac{3}{4}}^{8}-2 d x+\int_{\frac{3+\sqrt{17}}{8}}^{1}-3 d x$

$=-\frac{1}{4}-\frac{1}{4}-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$

$=\frac{\sqrt{17}-13}{8}$