Definite Integration

$ I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{12(3+[x])}{3+[\sin x]+[\cos x]}\right) d x $

Breakdown the Interval Since $\frac{\pi}{2} \approx 1.57$ and $-\frac{\pi}{2} \approx-1.57$, we break the integral at the integer points $x=-1,0,1$.

Notes on values :

• 1. For $x \in(0,1), 0<\cos x<1$, so $[\cos x]=0$. (At $x=0,[\cos x]=1$, but a single point doesn't change the integral).
  
  2. For $x \in\left[1, \frac{\pi}{2}\right), 0<\sin x<1$, so $[\sin x]=0$.

To find the total value, we integrate the constant values found above over their respective subintervals:

$ I=\int\limits_{-\pi / 2}^{-1} 6 d x+\int\limits_{-1}^1 12 d x+\int\limits_1^{\pi / 2} 16 d x $

$\Rightarrow $ $ I=6[x]_{-\pi / 2}^{-1}+12[x]_{-1}^1+16[x]_1^{\pi / 2} $

$\Rightarrow $ $ I=6\left(-1-\left(-\frac{\pi}{2}\right)\right)+12(1-(-1))+16\left(\frac{\pi}{2}-1\right) $

$\Rightarrow $ $ I=-6+3 \pi+24+8 \pi-16 \\ I=11 \pi+2 $

We are given the functional equation:

$ f\left(x^2+1\right)=x^4+5 x^2+2 $

Let $t=x^2+1$. This implies that $x^2=t-1$. Now, substitute this into the right side of the equation:

$ \begin{gathered} f(t)=(t-1)^2+5(t-1)+2 \\ f(t)=\left(t^2-2 t+1\right)+5 t-5+2 \\ f(t)=t^2+3 t-2 \end{gathered} $

Replacing $t$ back with $x$, the polynomial function is:

$ f(x)=x^2+3 x-2 $

Now, we calculate the definite integral from 0 to 3 :

$ \begin{gathered} \quad \int_0^3 f(x) d x=\int_0^3\left(x^2+3 x-2\right) d x \\ {\left[\frac{x^3}{3}+\frac{3 x^2}{2}-2 x\right]_0^3} \\ =\left(\frac{3^3}{3}+\frac{3\left(3^2\right)}{2}-2(3)\right)-(0) \\ =\left(\frac{27}{3}+\frac{27}{2}-6\right)=9+\frac{27}{2}-6=3+\frac{27}{2} \\ =\frac{6+27}{2}=\frac{33}{2} \end{gathered} $

$ \begin{aligned} &\begin{aligned} I= & \int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{1}{1+(\tan 2 x)^{1 / 3}} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\\ \Rightarrow & I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{1}{1+\tan \left(2\left(\frac{5 \pi}{24}+\frac{\pi}{24}-4\right)\right)^{1 / 3}} d x \\ & I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{1}{1+\tan \left(\frac{\pi}{2}-2 x\right)^{1 / 3}} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{aligned}\\ &\text { From equation (1) and (2) } \end{aligned} $

$ \begin{aligned} & 2 I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{(\tan (2 x))^{1 / 3}+1}{1+(\tan 2 x)^{1 / 3}} d x \\ & =\left.(x)\right|_{\frac{\pi}{24}} ^{\frac{5 \pi}{24}} \Rightarrow I=\frac{1}{2} \times \frac{4 \pi}{24}=\frac{\pi}{12} \end{aligned} $

$ \begin{aligned} & \int_{-\frac{\pi}{2}}^{-1} \frac{1}{2} d x+\int_{-1}^0 \frac{1}{3} d x+\int_0^1 \frac{1}{4} d x+\int_1^{\frac{\pi}{2}} \frac{1}{5} d x \\ & =\frac{21 \pi-7}{60}=\frac{7}{60}(3 \pi-1) \end{aligned} $

$ \begin{aligned} &6 \int_1^x f(t) d t=3 x f(x)+x^2-4\\ &\text { By using leibnz rule }\\ &\begin{aligned} & 6 f(x)=3 f(x)+3 x f^{\prime}(x)+3 x^2 \\ & \frac{f(x)-x f^{\prime}(x)}{x^2}=1 \\ & \frac{d}{d x}\left(\frac{f(x)}{x}\right)=-1 f(x)=-x^2+k x \\ & 1=-1+k(\text { since } f(1)=1) \\ & f(x)=-x^2+2 x \\ & f(2)-f(3)=3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \int_0^{\frac{\pi}{6}} \frac{\pi+4 x^{11}}{1-\sin \left(|x|+\frac{\pi}{6}\right)}+\frac{\pi-4 x^{11}}{1-\sin \left(|x|+\frac{\pi}{6}\right)} d x \\ & =2 \pi \int_0^{\frac{\pi}{6}} \frac{1}{1-\sin \left(x+\frac{\pi}{6}\right)} d x \\ & =2 \pi \int_0^{\frac{\pi}{6}} \frac{1+\sin \left(x+\frac{\pi}{6}\right)}{\cos ^2\left(x+\frac{\pi}{6}\right)} d x \\ & =2 \pi \int_0^{\frac{\pi}{6}} \sec ^2\left(x+\frac{\pi}{6}\right)+\tan \left(x+\frac{\pi}{6}\right) \sec \left(x+\frac{\pi}{6}\right) d x \\ & \Rightarrow 2 \pi\left[\tan \left(x+\frac{\pi}{6}\right)+\sec \left(x+\frac{\pi}{6}\right)\right]_0^{\frac{\pi}{6}} \\ & \Rightarrow 4 \pi \end{aligned} $

$\begin{aligned} &\text { Let, } \mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}\\ &\begin{aligned} & =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \\ & =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \end{aligned} \end{aligned}$

$\begin{aligned} &\text { Consider }\\ &\begin{aligned} & \int \mathrm{x} \sin \pi \mathrm{xdx} \\ & -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \\ & =--\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2} \\ & \mathrm{I}=\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \\ & =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \\ & =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \\ & =3 \pi+1 \end{aligned} \end{aligned}$

$\begin{aligned} & I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x \\ & \Rightarrow 2 x=t \Rightarrow 2 d x=d t \quad \Rightarrow I_1=\frac{1}{2} \int_{-1}^2 t f(t(1-t)) d t \\ & \Rightarrow 2 I_1=\int_{-1}^2(1-t) f(1-t)(1-(1-t)) d t \end{aligned}$

$ \Rightarrow 2{I_1} = \int\limits_{ - 1}^2 {f(t(1 - t)dt - \int\limits_{ - 1}^2 {tf(t(1 - t)dt} } $

$\begin{aligned} & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \\ & \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \\ & \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4 \end{aligned}$

$\begin{aligned} & I=\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (i)}\\ & =\int_0^\pi \frac{(\pi-x+3) \sin (\pi-x)}{1+3 \cos ^2(\pi-x)} d x \\ & =\int_0^\pi \frac{(\pi+3) \sin x-x \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (ii)} \end{aligned}$

Add (i) & (ii)

$2 I=\int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos ^2 x} d x$

Let $\cos x=t \Rightarrow-\sin x d x=d t$

$(\pi+6) \int_1^{-1} \frac{-d t}{\left(1+3 t^2\right)}=\left(\frac{\pi+6}{3}\right) \int_{-1}^1 \frac{d t}{t^2+\left(\frac{1}{\sqrt{3}}\right)^2}$

$\Rightarrow \quad 2 I=\left.\left(\frac{\pi+6}{3}\right) \cdot \frac{1}{\left(\frac{1}{\sqrt{3}}\right)} \cdot \tan ^{-1} \frac{t}{\frac{1}{\sqrt{3}}}\right|_{-1} ^1$

$\begin{aligned} & \Rightarrow \quad 2 I=\left(\frac{\pi+6}{\sqrt{3}}\right)\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(-\sqrt{3})\right] \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot \frac{2 \pi}{3} \\ & \Rightarrow \quad I=\frac{\pi(\pi+6)}{3 \sqrt{3}} \end{aligned}$

$\begin{aligned} & f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \\ & 2 f\left(\frac{1}{x}\right)+4 f(x)=2\left(\frac{1}{x^2}+5\right) \\ & 3 f(x)=\frac{2}{x^2}-x^2+5 \\ & f(x)=\frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) \\ & 2 g(x)-3 g\left(\frac{1}{x}\right)=x \\ & 2 g\left(\frac{1}{x}\right)-3 g(x)=\frac{1}{x} \\ & \text { Or } 4 g(x)-6 g\left(\frac{1}{x}\right)=2 x \\ & 6 g\left(\frac{1}{x}\right)-9 g(x)=\frac{3}{x} \\ & -5 g(x)=2 x+\frac{3}{x} \end{aligned}$

$ \begin{aligned} & \text { Or } g(x)=-\frac{1}{5}\left(2 x+\frac{3}{x}\right) \\ & \int_1^2 f(x) d x=\int_1^2 \frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) d x \end{aligned}$

$ \begin{aligned} &\begin{aligned} & =\frac{1}{3}\left[-\frac{2}{x}-\frac{x^3}{3}+5 x\right]_1^2 \\ & =\frac{1}{3}\left[\left(-\frac{2}{2}-\frac{8}{3}+10\right)-\left(-2-\frac{1}{3}+5\right)\right] \\ & =\frac{1}{3}\left[-1-\frac{8}{3}+10+2+\frac{1}{3}-5\right] \\ & \alpha=\frac{11}{9} \end{aligned}\\ &\text { Now, } 2 g(x)=x+3 g\left(\frac{1}{2}\right)\\ &\begin{aligned} & 2 g\left(\frac{1}{2}\right)=\frac{1}{2}+3 g\left(\frac{1}{2}\right) \\ & g\left(\frac{1}{2}\right)=-\frac{1}{2} \end{aligned} \end{aligned}$

$\begin{aligned} & \therefore \beta=\int_1^2 g(x) d x \\ & =\frac{1}{2} \int_1^2\left(x+3 g\left(\frac{1}{2}\right)\right) d x \\ & =\frac{1}{2}\left[\frac{x^2}{2}+3 g\left(\frac{1}{2}\right) x\right]_1^2 \\ & =0 \\ & \therefore 9 \alpha+\beta=11 \end{aligned}$

$\begin{aligned} & I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x \\ & =\int_0^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} \end{aligned}$

$\begin{aligned} & +\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^x}\right) d x \\ = & \int_0^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^x+e^{-x}\right)}{e^x+e^{-x}} d x \\ = & \int_0^1(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x \end{aligned}$

$\begin{aligned} & =\int_0^1(1+\sqrt{2 x}) d x=\left.x\right|_0 ^1+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_0 ^1 \\ & =1+\frac{2 \sqrt{2}}{3} \end{aligned}$

$\begin{aligned} & I=\int_0^\pi \frac{8 x}{4 \cos ^2 x+\sin ^2 x} d x \ldots(1) \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2(\pi-x)+\sin ^2(\pi-x)} d x \\ & I=\int_0^\pi \frac{8(\pi-x)}{4 \cos ^2 x+\sin ^2 x} d x \ldots(2) \end{aligned}$

Adding (1) and (2)

$\begin{aligned} & 2 I=8 \pi \int_0^\pi \frac{1}{4 \cos ^2 x+\sin ^2 x} d x \\ & I=4 \pi \times 2 \int_0^\pi \frac{\sec ^2 x}{4 \tan ^2 x} d x \end{aligned}$

Put $\tan x=t$

$\sec ^2 x d x=d t$

$I=8 \pi \int_0^{\infty} \frac{d t}{4+t^2}$

$\begin{aligned} & I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_0^{\infty} \\ & I=4 \pi\left(\frac{\pi}{2}\right) \\ & I=2 \pi^2 \end{aligned}$

Step 1: Ensure the innermost function is greater than zero:

$ \log_6(3 + 4x - x^2) > 1 $

Step 2: Simplify the inequality from Step 1:

$ 3 + 4x - x^2 > 6 \implies x^2 - 4x + 3 < 0 $

Step 3: Solve the quadratic inequality:

$ (x - 1)(x - 3) < 0 $

This inequality indicates that $ x $ must lie between the roots, giving the interval $ x \in (1, 3) $.

With the domain of $ f(x) $ identified as $ (a, b) = (1, 3) $, we calculate the definite integral over $[0, b-a]$:

Calculate the Integral:

Given:

$ \int_0^{b-a} [x^2] \, dx $

For $ b - a = 2 $, we compute:

$ \int_1^2 [x^2] \, dx = \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx $

Computation of Each Integral Segment:

$\int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1$

$\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2})$

$\int_{\sqrt{3}}^2 3 \, dx = 3(2 - \sqrt{3})$

Summing these, we have:

$ 5 - \sqrt{2} - \sqrt{3} $

Conclusion:

The values for $ p, q, $ and $ r $ are $ p = 5 $, $ q = 2 $, and $ r = 3 $, with the greatest common divisor of these numbers being 1. Therefore, adding them together gives:

$ p + q + r = 5 + 2 + 3 = 10 $

To solve the problem, we start with the given equation:

$ 10 \int_1^x f(t) \, dt = 5x f(x) - x^5 - 9 $

By differentiating both sides with respect to $ x $, we have:

$ \frac{d}{dx}\left(10 \int_1^x f(t) \, dt\right) = \frac{d}{dx}(5x f(x) - x^5 - 9) $

The left side simplifies to $ 10 f(x) $. For the right side, using the product rule and the power rule, we get:

$ 10 f(x) = 5f(x) + 5x \frac{d}{dx} f(x) - 5x^4 $

Rearranging terms, we obtain:

$ 5 f(x) = 5x \frac{d}{dx} f(x) - 5x^4 $

Let $ y = f(x) $. Thus:

$ 5 y = 5x \frac{dy}{dx} - 5x^4 $

Dividing by 5, we have:

$ y = x \frac{dy}{dx} - x^4 $

Rewriting, we get:

$ \frac{dy}{dx} - \frac{y}{x} = x^3 $

This is a linear differential equation. The integrating factor (I.F.) is calculated as:

$ \text{I.F.} = e^{\int \frac{-1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} $

Multiplying through by the integrating factor, we have:

$ y \cdot \frac{1}{x} = \int x^3 \cdot \frac{1}{x} \, dx = \int x^2 \, dx = \frac{x^3}{3} + C $

Thus, we solve for $ y $:

$ y = \frac{x^4}{3} + Cx $

Substituting back, we need to find $ C $ using the condition at $ x = 1 $:

Since $ \int_1^1 f(t) \, dt = 0 $, we substitute:

$ 10 \cdot 0 = 5 \cdot 1 \cdot f(1) - 1^5 - 9 \quad \Rightarrow \quad 0 = 5f(1) - 1 - 9 $

$ 5f(1) = 10 \quad \Rightarrow \quad f(1) = 2 $

Now use $ f(1) = 2 $ in the function:

$ 2 = \frac{1}{3} + C \cdot 1 \quad \Rightarrow \quad C = \frac{5}{3} $

The function $ f(x) $ is given by:

$ f(x) = \frac{x^4}{3} + \frac{5}{3}x $

To find $ f(3) $:

$ f(3) = \frac{3^4}{3} + \frac{5}{3} \cdot 3 = 27 + 5 = 32 $

Therefore, the value of $ f(3) $ is 32.

$\begin{aligned} & x^2=2 y \text { and } y-2 x-6=0 \\ & \frac{x^2}{2}-2 x-6=0 \\ & x^2-4 x-12=0 \\ & x^2-6 x+2 x-12=0 \\ & x(x-6)+2(x-6)=0 \\ & (x-6)(x+2)=0 \end{aligned}$

Point of intersection are $(6,18)$ and $(-2,2)$

$(-2,2)$ is in second quadrant

$\begin{aligned} & a=-2, b=2 \\ & I=\int_{-2}^2 \frac{9 x^2}{1+5^x} d x\quad\text{..... (i)} \end{aligned}$

$I=\int_{-2}^2 \frac{9 x^2}{1+5^{-x}} d x\quad\text{..... (ii)}$

$\begin{aligned} &\text { Adding (i) and (ii) }\\ &\begin{aligned} & 2 I=\int_{-2}^2 9 x^2 d x \\ & I=9 \int_0^2 x^2 d x \\ & I=9\left(\frac{x^3}{3}\right)_0^2 \Rightarrow I=24 \end{aligned} \end{aligned}$

$\begin{aligned} & I=4 \int_0^1 \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} d x \\ & =2 \int_0^1 \sqrt{3+x^2}-\sqrt{1+x^2} d x \end{aligned}$

$\begin{aligned} & =2\left[\int_0^1 \sqrt{3+x^2} d x-\int_0^1 \sqrt{1+x^2} d x\right] \\ & =2\left[\left(\frac{1}{2} x \sqrt{x^2+3}+\frac{3}{2} \ln \left|\sqrt{3+x^2}+x\right|\right)-\right. \\ & \left.\quad\left(\frac{1}{2} x \sqrt{1+x^2}+\frac{1}{2} \ln \left|\sqrt{1+x^2}+x\right|\right)\right]_0^1 \end{aligned}$

$\begin{aligned} & =2\left[\left(1+\frac{3}{2} \ln 3-\frac{3}{2} \ln \sqrt{3}\right)-\left(\frac{\sqrt{2}}{2}+\frac{1}{2} \ln (\sqrt{2}+1)\right)\right] \\ & =2\left(1+\frac{3}{4} \ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} \ln (\sqrt{2}+1)\right) \\ & =3 \ln \sqrt{3}+2-\sqrt{2}-\ln (\sqrt{2}+1) \\ & I-3 \ln \sqrt{3}=2-\sqrt{2}-\ln (\sqrt{2}+1) \end{aligned}$

$f^{\prime}(x)=x^3-9 x^2+20 x=x(x-4)(x-5)$

$\begin{aligned} & \therefore f(x)=\frac{x^4}{4}-\frac{9 x^3}{3}+\frac{20 x^2}{2} \\ & f(1)=\frac{1}{4}-3+10=\frac{29}{4}=\alpha \\ & \left.f(4)=\frac{256}{4}-3(64) \right\rvert\,+10(16)=32=\beta \\ & 4(\alpha+\beta)=4\left(\frac{29}{4}+32\right)=157 \end{aligned}$

$\begin{aligned} & I=80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16(2 \sin \theta \cdot \cos \theta)}\right) d \theta \\ & =80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9-16(1-2 \sin \theta \cdot \cos \theta-1)} d \theta \end{aligned}$

$\begin{aligned} &=80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9+16-16(\sin \theta-\cos \theta)^2} \mathrm{~d} \theta\\ &\text { Let } \sin \theta-\cos \theta=\mathrm{t}\\ &(\cos \theta+\sin \theta) \mathrm{d} \theta=\mathrm{dt}\\ &=80 \int_{-1}^0 \frac{\mathrm{dt}}{25-16 \mathrm{t}^2}\\ &=\frac{80}{16} \int_{-1}^0 \frac{\mathrm{dt}}{\left(\frac{5}{4}\right)^2-\mathrm{t}^2}\\ &\left.=\frac{5}{2\left(\frac{5}{4}\right)} \ln \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]_{-1}^0\\ &=2 \ln (1)+4 \ln 3\\ &=4 \ln 3 \end{aligned}$

$\begin{aligned} &\begin{aligned} & \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \\ & =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \\ & =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \\ & \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 \quad \ldots(1) \\ & \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2\quad \ldots(2) \end{aligned}\\ &\text { Also }\\ &\begin{aligned} & \int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x F^{\prime}(x) d x=40 \\ & =4 F^{\prime}(2)-2 \times 6=40 \\ & F^{\prime}(2)=13 \\ & \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11 \end{aligned} \end{aligned}$

$\begin{aligned} & g(x)=x 2+x^{\frac{7}{3}} \\ & g^{\prime}(x)=2 x+\frac{7}{3} x^{\frac{4}{3}} \\ & f(x)=\frac{g^{\prime}(x)}{x} \\ & f(x)=2+\frac{7}{3} x^{\frac{1}{3}} \\ & f\left(r^3\right)=2+\frac{7 r}{3} \\ & \sum_{r=1}^{15}\left(1+\frac{7}{3} r\right)=310 \end{aligned}$

$\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} d x \quad$ (Apply King Property)

$\begin{aligned} & \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x=48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\ & 48\left[\left(\frac{x^3}{3}\right)_0^{\pi / 2}+\int_0^{\frac{\pi}{2}} x_1^2 \cos _{I I} 2 x d x\right] \\ & \Rightarrow \text { On solving } \pi\left(2 \pi^2-12\right) \\ & \Rightarrow \alpha=2, \beta=-12 \\ & \Rightarrow(\alpha+\beta)^2=100 \end{aligned}$

$\begin{aligned} & \mathrm{I}(\mathrm{~m}, \mathrm{~m})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx} \\ & \text { Let } \mathrm{x}=\sin ^2 \theta \quad \mathrm{dx}=2 \sin \theta \cos \theta \mathrm{~d} \theta \\ & \mathrm{I}(\mathrm{~m}, \mathrm{n})=2 \int_0^{\pi / 2}(\sin \theta)^{2 \mathrm{~m}-1}(\cos \theta)^{2 \mathrm{n}-1} \mathrm{~d} \theta \\ & \mathrm{I}(9,14)+\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} \mathrm{~d} \theta \\ & +2 \int_0^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =2 \int_0^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} \mathrm{~d} \theta \\ & =\mathrm{I}(9,13) \end{aligned}$

For I

Apply king ($\mathrm{P}-5$) and add

$\begin{aligned} & 2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \\ & I_2=\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \end{aligned}$

Apply king and add

$\begin{aligned} & \mathrm{I}_2=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan \mathrm{xsec}{ }^2 \mathrm{xdx}}{\tan ^4 \mathrm{x}+1} \\ & \text { put } \tan ^2 \mathrm{x}=\mathrm{t} \\ & \frac{\pi}{8} \int_0^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\ & =\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16} \end{aligned}$

$\begin{aligned} &\text { Let } \ln \mathrm{x}=\mathrm{t} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\\ &\begin{aligned} & I=\int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}}+e^{\frac{1}{1+(6-t)^2}}} d t \\ & I=\int_2^4 \frac{\frac{1}{e^{1+(6-t)^2}}}{\frac{1}{e^{1+(6-t)^2}}+e^{\frac{1}{1+t^2}}} d t \\ & 2 I=\int_2^4 d t=(t)_2^4=4-2=2 \\ & I=1 \end{aligned} \end{aligned}$

$\begin{aligned} f(x) & =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \\ & =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x\end{aligned}$

$\begin{aligned} I_1= & \int_0^{\frac{\pi}{4}} f(x) d x=\int_0^{\frac{\pi}{4}}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.\left(\frac{7 \tan ^7 x}{7}-\frac{3 \tan ^3 x}{3}\right)\right|_0 ^{\frac{\pi}{4}}=1-1=0\end{aligned}$

$\begin{aligned} I_2= & \int_0^{\frac{\pi}{4}} x f(x) d x=\int_0^{\frac{\pi}{4}} x\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.x\left(\tan ^7 x-\tan ^3 x\right)\right|_0 ^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot\left(\tan ^7 x-\tan ^3 x\right) d x \\ & =0-\int_0^{\frac{\pi}{4}} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x\end{aligned}$

$ \begin{aligned} & =\int_0^{\frac{\pi}{4}}\left(\tan ^3 x-\tan ^5 x\right) \sec ^2 x d x=\frac{\tan ^4 x}{4}-\left.\frac{\tan ^6 x}{6}\right|_0 ^{\frac{\pi}{4}} \\ & =\frac{1}{12} \end{aligned} $

Hence $7 I_1+12 I_2=1$

$\begin{aligned} & \int_\limits{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\ & x=\cos 2 \theta \\ & \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta) \end{aligned}$

Take limit as $\alpha$ and $\beta$

$\begin{aligned} & -2 \int_\limits\alpha^\beta \cos 2 \theta \cdot \sin 2 \theta d \theta \\ & =\int_\limits\alpha^\beta \sin 4 \theta d \theta \\ & =\left.\frac{-\cos 4 \theta}{4}\right|_\alpha ^\beta \\ & =-\left.\frac{1}{4}\left(2 \cdot\left(x^2\right)-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\left.\frac{1}{4}\left(2 x^2-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1\right) \\ & =-\frac{1}{4} \end{aligned}$

$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$

Using Newton Leibniz theorem

$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{\sin \left(2 \times \frac{\pi}{2}\right) \cdot 0-\sin (2 x) \cdot 3 x^2+\left(\cos \frac{\pi}{2}\right) \cdot 0-\cos x \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\right] \\ & =\lim _\limits w{x \rightarrow \frac{\pi}{2}} \frac{-3 x^2 \sin 2 x-3 x^2 \cos x}{2\left(x-\frac{\pi}{2}\right)}\left(\frac{0}{0}\right) \text { form } \\ & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{-6 x \sin 2 x-6 x^2 \cos 2 x-6 x \cos x+3 x^2 \sin x}{2}\right] \\ & =\frac{6 \times \frac{\pi^2}{4}+3 \times \frac{\pi^2}{4}}{2} \\ & =\frac{9 \pi^2}{8} \end{aligned}$

$\begin{aligned} & \int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\ & =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int_\limits{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\ & =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int_\limits{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\ & =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\ & =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\ & =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right] \end{aligned}$

$\begin{aligned} & \text { Let } \sqrt{e^x-1}=t \\ & e^x-1=t^2 \\ & e^x=1+t^2 \\ & e^x=0+2 t-\frac{d t}{d x} \\ & \frac{d t}{d x}=\frac{e^x}{2 t}=\frac{t^2+1}{2 t} \\ & I=\int \frac{2 t}{t\left(1+t^2\right)} d t=2 \tan ^{-1} t \\ & \Rightarrow \quad I=\int_\limits\alpha^{\log ^4} \frac{d x}{\sqrt{e^x-1}} \\ & I=\left.2 \tan ^{-1} \sqrt{e^x-1}\right|_\alpha ^{\log 4} \\ & =2\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{e^\alpha-1}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{\pi}{3}-\tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{12} \\ & \tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{4} \\ & \Rightarrow e^\alpha-1=1 \\ & e^\alpha=2 \Rightarrow e^{-\alpha}=\frac{1}{2} \end{aligned}$

$\therefore \quad$ Quadratic equation whose roots are $e^a$ & $e^{-\alpha}$ is

$\begin{aligned} & x^2-\left(e^\alpha+e^{-\alpha}\right) x+e^\alpha \times e^{-\alpha}=0 \\ & x^2-\left(2+\frac{1}{2}\right) x+1=0 \\ & 2 x^2-5 x+2=0 \end{aligned}$

$\begin{aligned} & I(21)=\int_\limits0^1\left(1-x^k\right)^{21} d x \\ & =\int_\limits0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\ & =\int_\limits0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\ & I(21)=I(20)-\int_\limits0^1 x^k\left(1-x^k\right)^{20} d x \end{aligned}$

$I(21)=I(20)-\left\lfloor\frac{\left(1-x^k\right)^{21}}{-21 k} x-\int_\limits0^1 \frac{(1-x^k)^{21}}{-21 k} d x\right\rfloor$

$\begin{aligned} & I(21)=I(20)-\frac{1}{21 k} I(20) \\ & \Rightarrow[I(21)](21 k+1)=21 K I(20) \\ & \Rightarrow 21 K=147 \Rightarrow K=7 \end{aligned}$

$\begin{aligned} & \int_\limits0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\ & =\int_\limits0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x \end{aligned}$

Let $\tan x=t$

$\int_\limits0^1 \frac{t^2 d t}{\left(1+t^3\right)^2}$

Let $1+t^3=\mathrm{z}$

$\begin{gathered} 3 t^2 d t=d z \\ \frac{1}{3} \int_\limits1^2 \frac{d z}{z^2}=\left.\frac{1}{3}\left(-\frac{1}{z}\right)\right|_1 ^2 \\ =-\frac{1}{3}\left(\frac{1}{2}-1\right)=\frac{1}{6}\end{gathered}$

First, let's rewrite the given integral using the given form of the Beta function. The given integral is:

$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$

To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \frac{1}{10}t^{-\frac{9}{10}} dt $ or $ dx = \frac{1}{10} t^{-\frac{9}{10}} dt $. The limits of integration change as follows: when $ x = 0 $, $ t = 0 $, and when $ x = 1 $, $ t = 1 $.

Substituting these into the integral, we have:

$\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt$

which simplifies to:

$\frac{1}{10} \int_\limits0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt$

We recognize this integral as a Beta function $ \beta(m, n) $ where $ m = 1 - \frac{9}{10} = \frac{1}{10} $ and $ n = 20 + 1 = 21 $.

Therefore, we can write this as:

$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$

Comparing this to $ a \times \beta(b, c) $, we have $ a = \frac{1}{10} $, $ b = \frac{1}{10} $, and $ c = 21 $.

Now we calculate $ 100(a + b + c) $:

$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$

So, the answer is Option D, 2120.

$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$

$\begin{aligned} & \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & \begin{array}{l} 3 A-5 B=1 \\ 5 A+3 B=0 \end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34} \end{aligned}$

$\begin{aligned} & \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\ & \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\ & 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\ & 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\ & \Rightarrow 3 \pi-50 \ln 2+20 \ln 5 \end{aligned}$

$\begin{aligned} & I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\ & =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\ & =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y \end{aligned}$

$I=4 \int_0^\pi\left(\frac{y \sin }{1+\cos ^2 y}\right) d y \quad \text{... (1)}$

$\begin{aligned} & I=4 \int_\limits0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\ & I=4\left\lfloor\int_\limits0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int_\limits0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)} \end{aligned}$

Adding equation (1) and (2)

$\begin{aligned} & 2 I=4 \int_\limits0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\ & I=2 \pi \int_\limits0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi \times \frac{\pi}{2} \\ & =\pi^2 \end{aligned}$

Given $f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$

Now, $\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \\ & =\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{1+\cos \left(1-e^x\right)\left(-e^x\right)}{6 x}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{-\sin \left(1-e^x\right)\left(e^x\right)^2+\cos \left(1-e^x\right)\left(-e^x\right)}{6} \\ & =-\frac{1}{6} \end{aligned}$

$\begin{aligned} & \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\ & \begin{aligned} I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\ \Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\ & =\int_\limits0^1 \cos \alpha x d x \\ & =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\ & =\frac{\sin \alpha}{\alpha} \\ \Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\ \Rightarrow & \alpha=\frac{\pi}{2} \end{aligned} \end{aligned}$

$f(x)=\left\{\begin{array}{cc} -2 & -2 \leq x \leq 0 \\ x-2 & 0< x \leq 2 \end{array} h(x)=f|x|+|f(x)|\right.$

$\begin{aligned} & h(x)=\left\{\begin{array}{cc} -x-2+2=-x & -2 \leq x \leq 0 \\ 0 & 0< x \leq 2 \end{array}\right. \\ & \therefore \int_\limits{-2}^2 h(x) d x=\int_\limits{-2}^0-x d x+\int_\limits0^2 0 d x \\ & \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2 \end{aligned}$

To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$, we can write it in terms of double angles as:

$ \cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2 $

We can then expand and simplify the integral using this formula. Let's proceed with this:

$ \int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x $

Now, let's expand the integrand and then integrate term by term:

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x $

For the $\cos^2(2x)$ term, we again use the power reduction formula:

$ \cos^2(2x) = \frac{1 + \cos(4x)}{2} $

Let's substitute this into the integral and continue:

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x $

Simplify and integrate:

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x $

$ = \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x $

$ = \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}} $

Evaluating this from $0$ to $\frac{\pi}{3}$:

$ = \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right) $

$ = \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right) $

$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$:

$ = \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2} $

$ = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} $

Now, combining terms we get the final result:

$ = \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64} $

$ = \frac{\pi}{8} + \frac{7\sqrt{3}}{64} $

Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$:

$ a = \frac{1}{8}, \quad b = \frac{7}{64} $

Now we find $9a + 8b$:

$ 9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64} $

$ = \frac{9}{8} + \frac{7}{8} $

$ = \frac{16}{8} $

$ = 2 $

Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\int_\limits{-\mathrm{x}}^{\mathrm{x}}\left(|\mathrm{t}|-\mathrm{t}^2\right) \mathrm{e}^{-\mathrm{t}^2} \mathrm{dt} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \cdot\left(|\mathrm{x}|-\mathrm{x}^2\right) \mathrm{e}^{-\mathrm{x}^2} \ldots \ldots \ldots . .(1) \\ & \mathrm{g}(\mathrm{x})=\int_\limits0^{\mathrm{x}^2} \mathrm{t}^{\frac{1}{2}} \mathrm{e}^{-t} \mathrm{dt} \\ & \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xe}^{-\mathrm{x}^2}(2 \mathrm{x})-0 \\ & \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{xe}^{-\mathrm{x}^2}-2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2}+2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2} \end{aligned}$

Integrating both sides w.r.t. $\mathrm{x}$

$\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\int_\limits0^\alpha 2 \mathrm{xe}^{-\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{x}^2=\mathrm{t} \\ & \Rightarrow \int_0^{\sqrt{\alpha}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\left[-\mathrm{e}^{-\mathrm{t}}\right]_0^{\sqrt{\alpha}} \\ & =-\mathrm{e}^{\left(\log _c(9)^{-1}\right)+1} \\ & \Rightarrow 9(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=\left(1-\frac{1}{9}\right) 9=8 \end{aligned}$

$\begin{aligned} &f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\\ &f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\\ &f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}} \end{aligned}$

$18 \int_\limits0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x$

$\begin{aligned} & \text { Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \\ & 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \\ & \frac{18}{4} \int_\limits1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \\ & =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \\ & =\frac{3}{2}[26]=39 \end{aligned}$

$\begin{aligned} & y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \\ & \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \\ & \left.\frac{d y}{d x}\right)_{(3, f(3))}=f^{\prime}(3)=\tan \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3)=1 \\ & 27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \\ & I=\int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t \\ & f^{\prime}(t)=z \Rightarrow f^{\prime \prime}(t) d t=d z \\ & z=f^{\prime}(3)=1 \\ & z=f^{\prime}(1)=\frac{1}{\sqrt{3}} \end{aligned}$

$\begin{aligned} & I=\int_\limits{1 / \sqrt{3}}^1\left(z^2+1\right) d z=\left(\frac{z^3}{3}+z\right)_{1 / \sqrt{3}}^1 \\ & =\left(\frac{1}{3}+1\right)-\left(\frac{1}{3} \cdot \frac{1}{3 \sqrt{3}}+\frac{1}{\sqrt{3}}\right) \\ & =\frac{4}{3}-\frac{10}{9 \sqrt{3}}=\frac{4}{3}-\frac{10}{27} \sqrt{3} \\ & \alpha+\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27} \sqrt{3}\right)=36-10 \sqrt{3} \\ & \alpha=36, \beta=-10 \\ & \alpha+\beta=36-10=26 \end{aligned}$

To determine the integral of the piecewise function $f$ over the interval $[-2, 2]$, we first ensure that $f$ is differentiable on $\mathbb{R}$, as given in the problem statement. Differentiability implies continuity, so $f$ must also be continuous at $x=1$.

The condition for continuity at $x=1$ is:

$x^2 + 3x + a = bx + 2$ at $x=1$.

This simplifies to:

$1 + 3 + a = b(1) + 2$

$\Rightarrow a = b - 2$

The first derivative of $f$ gives us two different expressions depending on the value of $x$:

For $x \leq 1$, $f'(x) = 2x + 3$; and for $x > 1$, $f'(x) = b$.

For $f$ to be differentiable at $x=1$, these derivatives must be equal at that point. Setting $f'(1)$ from both expressions equal to each other gives $2(1) + 3 = b$, thus $b = 5$. And from $a = b - 2$, we have $a = 3$.

Now, we can calculate the integral of $f$ over the specified interval:

$\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx$

$\begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\ & =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\ & =6+\frac{3}{2}+12-\frac{5}{2}=17 \end{aligned}$

$\begin{array}{ll} \mathrm{f}(\mathrm{x})=a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{cx} & \mathrm{f}(0)=-1 \\\\ & \mathrm{a}+\mathrm{b}=-1 \\\\ \mathrm{f}^{\prime}(\mathrm{x})=2 a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{c} & \mathrm{f}^{\prime}(\ln 2)=21 \\\\ & 8 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=21 \\\\ \int_\limits0^{\ln 4}\left(a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}=\frac{39}{2} & \end{array}$

$\begin{aligned} & {\left[\frac{\mathrm{ae}^{2 \mathrm{x}}}{2}+\mathrm{be}^{\mathrm{x}}\right]_0^{\ln 4}=\frac{39}{2} \Rightarrow 8 \mathrm{a}+4 \mathrm{~b}-\frac{\mathrm{a}}{2}-\mathrm{b}=\frac{39}{2}} \\\\ & 15 a+6 b=39 \\\\ & 15 a-6 a-6=39 \\\\ & 9 \mathrm{a}=45 \Rightarrow \mathrm{a}=5 \\\\ & b=-6 \\\\ & c=21-40+12=-7 \\\\ & \mathrm{a}+\mathrm{b}+\mathrm{c}-8 \\\\ & |\mathrm{a}+\mathrm{b}+\mathrm{c}|=8 \\\\ \end{aligned}$

$\begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^3}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \end{aligned}$

$=\int_\limits0^1 \frac{d x}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}$

$\begin{aligned} & =\int_\limits0^1 \frac{1}{3} \times \frac{3}{2} \frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)} d x \\ & =\frac{1}{2} \int_\limits0^1\left[\frac{1}{x^2+\left(\frac{1}{\sqrt{3}}\right)^2}-\frac{1}{1+x^2}\right] d x \\ & =\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_0^1-\frac{1}{2}\left(\tan ^{-1} x\right)_0^1 \\ & =\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8} \\ & =\frac{13 \pi}{8 \cdot(4 \sqrt{3}+3)} \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}\right) \times x^2} \\\\ & \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\\\ & =\lim _{x \rightarrow 0} \frac{f(x)}{1} \text { (using L Hospital) } \\\\ & f(0)=\frac{1}{2} \\\\ & \alpha=\frac{1}{2} \\\\ & 8 \alpha^2=2 \end{aligned}$

Using L'hospital rule

$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\ & =\frac{3 \pi^2}{8} \end{aligned}$

$\begin{aligned} & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x \\ & I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^{-x}}+\frac{1+\sin ^2 x}{1+e^{\sin (-x)^{2023}}}\right) d x \end{aligned}$

On Adding, we get

$2 I=\int_\limits{-\pi / 2}^{\pi / 2}\left(x^2 \cos x+1+\sin ^2 x\right) d x$

On solving

$\begin{aligned} & \mathrm{I}=\frac{\pi^2}{4}+\frac{3 \pi}{4}-2 \\ & \mathrm{a}=3 \end{aligned}$

$\begin{aligned} & I=\int_\limits0^\pi \frac{d x}{1-2 a \cos x+a^2} ; 0< a<1 \\ & I=\int_\limits0^\pi \frac{d x}{1+2 a \cos x+a^2} \\ & 2 I=2 \int_\limits0^{\pi / 2} \frac{2\left(1+a^2\right)}{\left(1+a^2\right)^2-4 a^2 \cos ^2 x} d x \\ & \Rightarrow I=\int_\limits0^{\pi / 2} \frac{2\left(1+a^2\right) \cdot \sec ^2 x}{\left(1+a^2\right)^2 \cdot \sec ^2 x-4 a^2} d x \\ & \Rightarrow I=\int_\limits0^{\pi / 2} \frac{2 \cdot\left(1+a^2\right) \cdot \sec ^2 x}{\left(1+a^2\right)^2 \cdot \tan ^2 x+\left(1-a^2\right)^2} d x \end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{I}=\int_\limits0^{\pi / 2} \frac{\frac{2 \cdot \sec ^2 \mathrm{x}}{1+\mathrm{a}^2} \cdot \mathrm{dx}}{\tan ^2 \mathrm{x}+\left(\frac{1-\mathrm{a}^2}{1+\mathrm{a}^2}\right)^2} \\ & \Rightarrow \mathrm{I}=\frac{2}{\left(1-\mathrm{a}^2\right)}\left[\frac{\pi}{2}-0\right] \\ & \mathrm{I}=\frac{\pi}{1-\mathrm{a}^2} \end{aligned}$