Area Under The Curves

Here, $x^4=x^{\frac{1}{3}} \Rightarrow x^{12}=x$

$ \Rightarrow x\left(x^{11}-1\right)=0 $

So, $x=0$ or $x^{11}=1$ which gives $x=1$

$ \begin{aligned} \therefore \quad A & =\int_0^1\left(x^{\frac{1}{3}}-x^4\right) d x=\left[\frac{3}{4} x^{\frac{4}{3}}-\frac{1}{5} x^5\right]_0^1 \\ & =\left(\frac{3}{4}-\frac{1}{5}\right)=\frac{15-4}{20}=\frac{11}{20} \end{aligned} $

Given, $ y=1+4 x-x^{2} $

$ \Rightarrow \quad \frac{d y}{d x}=4-2 x=0 \Rightarrow x=2 $ [point of maxima]

$ \Rightarrow \quad y_{\max }=1+4 \times 2-4=5 $

when, $ x=0, y=1 $

$ \therefore $ Graph according to question and above information

$ \Rightarrow $ If $ y=m x $ bisects the area bounded

$ \Rightarrow A_{1}=A_{2} $

$ \Rightarrow \quad \int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) d x=2 \int_{0}^{3 / 2} m x d x $

$ \Rightarrow\left[x+2 x^{2}-\frac{x^{3}}{3}\right]_{0}^{3 / 2}=\left[m x^{2}\right]_{0}^{3 / 2} $

$ \Rightarrow \frac{3}{2}+2 \times \frac{9}{4}-\frac{27}{8} \times \frac{1}{3}=m \times \frac{9}{4} $

$ \therefore \quad m=\frac{13}{6} $

Given curves are $ y=x^{3} $ and $ y=\sqrt{x} $

Both curves intersects at $ x=0 $ and $ x=1 $

Shaded region is required area

$ \begin{aligned} A & =\left|\int_{0}^{1}\left(x^{3}-\sqrt{x}\right) d x\right|=\left|\left[\frac{x^{4}}{4}-\frac{2 x^{3 / 2}}{3}\right]_{0}^{1}\right| \\ & =\left|\left[\frac{1}{4}-\frac{2}{3}\right]\right| \\ & =\frac{5}{12} \text { sq. unit } \end{aligned} $

The described region is :

It is clear that

$\begin{aligned} & A=\left(\frac{1}{2}, 0\right), B=(2,0), C=(2,3), D=(1,2) \\ & E=\left(\frac{1}{2}, \frac{3}{2}\right) \text { and } F=\left(\frac{1}{2}, \frac{5}{4}\right) \end{aligned}$

$\therefore$ Required area $=$ Area of region A B C D E A $-$ Area of region EDFE

$\begin{aligned} & =\frac{1}{2}\left(\frac{3}{2}+3\right)\left(2-\frac{1}{2}\right)-\int_\limits{1 / 2}^1\left[(x+1)-\left(x^2+1\right)\right] d x \\ & =\frac{1}{2} \times \frac{9}{2} \times \frac{3}{2}-\int_\limits{1 / 2}^1\left(x-x^2\right) d x \\ & =\frac{27}{8}-\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{1 / 2}^1 \\ & =\frac{27}{8}-\left[\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{8}-\frac{1}{24}\right)\right] \\ & =\frac{27}{8}-\left(\frac{1}{6}-\frac{1}{12}\right) \\ & =\frac{27}{8}-\frac{1}{12}=\frac{79}{24} \end{aligned}$

Given, $f(x)=e^x-e^{-|x-1|}$

$\therefore \quad f(x)=\left\{\begin{array}{cl}0 & \text { if } x < 1 \\ e^{x-1}-e^{1-x} & , \text { if } x \geq 1\end{array}\right\}$

The graph of $f$ and $g$ can be plotted in the first quadrant as given below.

The point of intersection, $A$ is given by $f(x)=g(x)$

$\begin{aligned} & \Rightarrow \quad e^{x-1}-e^{1-x}=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right) \\ & \Rightarrow \quad e^{x-1}=3 e^{1-x} \\ & \Rightarrow \quad\left(e^{x-1}\right)^2=3 \Rightarrow e^{2 x-2}=3 \\ & \Rightarrow \quad 2 x-2=\ln 3 \\ & \Rightarrow \quad x=1+\frac{1}{2} \ln 3 \\ & \therefore \text { Required area }=\int_\limits0^1\left(\frac{1}{2} e^{x-1}+e^{1-x}\right) d x +\int_\limits1^{1+\frac{1}{2} \ln 3}\left[\left(\frac{1}{2} e^{x-1}+e^{1-x}\right)-\left(e^{x-1}-e^{1-x}\right)\right] d x \end{aligned}$

$\begin{aligned} & =\frac{1}{2}\left[e^{x-1}-e^{1-x}\right]_0^1-\left[\frac{1}{2} e^{x-1}+\frac{3}{2} e^{1-x}\right]_1^{1+\frac{1}{2} \ln 3} \\ & =\frac{1}{2}\left(0-\frac{1}{e}+e\right)-\frac{1}{2}\left[e^{\ln \sqrt{3}}+3 e^{-\ln \sqrt{3}}-4\right] \\ & =\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(\sqrt{3}+\frac{3}{\sqrt{3}}-4\right) \\ & =-\frac{1}{2}(2 \sqrt{3}-4)+\frac{1}{2}\left(e-\frac{1}{e}\right) \\ & =(2-\sqrt{3})+\frac{1}{2}\left(e-\frac{1}{e}\right) \end{aligned}$

The equation of the parabola ${y^2} - 4y - x + 5 = 0$

the equation of tangent at (2, 3) is

$3y - 2(y + 3) - {{x + 2} \over 2} + 5 = 0$

$2y - x - 4 = 0$

$\therefore$ Required area A is given by

$A = \int_0^3 {({x_2} - {x_1})dy} $

$ \Rightarrow A = \int_0^3 {[\{ {{(y - 2)}^3} + 1\} - 2(2y - 4\} ]dy} $

$ \Rightarrow A = \int_0^3 {({y^2} - 6y + 9)dy = \int_0^3 {{{(3 - y)}^2}dy} } $

$ = - \left[ {{{{{(3 - y)}^3}} \over 3}} \right]_0^3 = 9$

Given, $y = \sin x + \cos x,\,\,x \in [0,\pi /2]$

${{dy} \over {dx}} = \cos x - \sin x$

$y = |\cos x - \sin x|$

$ = \left[ {\matrix{ {\cos x - \sin x} & {x \in [0,\pi /4]} \cr {\sin x - \cos x} & {x \in [\pi /4,\pi /2]} \cr } } \right]$

$\therefore$ Required area

$ = \int_0^{\pi /4} {|(\sin x + \cos x) - (\cos x - \sin x)|dx + \int_{\pi /4}^{\pi /2} {|2\cos x|dx} } $

$ = \int_0^{\pi /4} {|2\sin x|dx + \int_{\pi /4}^{\pi /2} {|2\cos x|dx} } $

$ = 2( - \cos x)_0^{\pi /4} + 2(\sin x)_{\pi /4}^{\pi /2}$

$ = 2\left[ { - {1 \over {\sqrt 2 }} + 1 + 1 - {1 \over {\sqrt 2 }}} \right]$

$ = 2\left( {2 - {2 \over {\sqrt 2 }}} \right)$

$ = 2(2 - \sqrt 2 )$

$ = 4 - 2\sqrt 2 $

$ = 2\sqrt 2 (\sqrt 2 - 1)$

Required area

$\int_0^{\pi /4} {\sin xdx + \int_{\pi /4}^{\pi /2} {\cos xdx} } $

$ = - [\cos ]_0^{\pi /4} + [\sin x]_{\pi /4}^{\pi /2} = 2\left( {1 - {1 \over {\sqrt 2 }}} \right) = (2 - \sqrt 2 )$ sq units

Graph of circle whose centre is origin and radius is $\pi$ units and y = sin x is

Area of shaded region = ${1 \over 2}$ Area of circle

= ${1 \over 2}$($\pi$r²) = ${1 \over 2}$$\pi$($\pi$²) [$\because$ r = $\pi$]

= ${{{\pi ^3}} \over 2}$