Area Under The Curves

$ \begin{aligned} \text { Area }= & \int_{-2}^0-x^3 d x+\int_0^4 x^3 d x=\frac{1}{4} \times 2^4+\frac{4^4}{4} \\ & =4+64=68 \end{aligned} $

Given, curves, $y=x^3, y=x^2$

for point of intersection

$ \begin{array}{ll} & x^3=x^2 \\ \Rightarrow \quad & x^2(x-1)=0 \Rightarrow x=0,1 \\ \therefore \quad & y=0,1 \end{array} $

point of intersection are $(0,0)$ and $(1,1)$

$\therefore$ Required area

$ \begin{aligned} & =\int_0^1\left(x^2-x^3\right) d x+\int_1^2\left(x^3-x^2\right) d x \\ & =\left(\frac{x^3}{3}-\frac{x^4}{4}\right)_0^1+\left(\frac{x^4}{4}-\frac{x^3}{3}\right)_1^2 \\ & =\left(\frac{1}{3}-\frac{1}{4}\right)+\left[\left(\frac{16}{4}-\frac{8}{3}\right)-\left(\frac{1}{4}-\frac{1}{3}\right)\right] \\ & =\frac{1}{12}+\frac{15}{4}-\frac{7}{3} \\ & =\frac{1+45-28}{12}=\frac{18}{12}=\frac{3}{2} \text { sq. units } \end{aligned} $

$ A=\int_{-1}^{\frac{1}{2}}\left(8 x^{3}-1\right) d x+\int_{\frac{1}{2}}^{1} 8 x^{3}-1 $

$ \begin{aligned} A= & \left(\frac{8 x^{4}}{4}-x\right)_{-1}^{1 / 2}+\left(8 \frac{x^{4}}{4}-x\right)_{1 / 2}^{1} \\\\ & =\left|\left(2 x^{4}-x\right)_{-1}^{1 / 2}\right|+\left(2 x^{4}-x\right)_{1 / 2}^{1} \\\\ & =\left|\left(2 \cdot \frac{1}{16}-\frac{1}{2}\right)-(2+1)\right| +\left[(2-1)-\left(\frac{2}{16}-\frac{1}{2}\right)\right] \\\\ & =\left|\frac{1}{8}-\frac{1}{2}-3\right|+1-\frac{1}{8}+\frac{1}{2} \\\\ & =\left|\frac{1-4-24}{8}\right|+\frac{8-1+4}{8} \\\\ & =\frac{27}{8}+\frac{11}{8}=\frac{38}{8}=\frac{19}{4} \end{aligned} $

Let the value of $ a=1 $

then area between curves $ x^{2}=y $ and $ x+y=2 $ is $ k $

$ \begin{aligned} & y=2-x \\\\ \Rightarrow \quad & x^{2}=2-x \\\\ & x^{2}+x-2=0 \\\\ & x^{2}+2 x-x-2=0 \\\\ & x=-2, x=1 \end{aligned} $

$ \text { Area }=\int_{-2}^{1}\left(x^{2}-2+x\right) d x $

$ =\left[\frac{x^{3}}{3}-2 x+\frac{x^{2}}{2}\right]_{-2}^{1} $

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

$ =\frac{5}{6}-2+\frac{8}{3}-6 $

$ =\frac{16+5}{6}-8 $

$ =\frac{21}{6}-8 $

$ =\frac{7}{2}-8=\frac{-9}{2} $

Area $ =\frac{9}{2}=k $

We have,

$ \begin{array}{ll} & y^{2}=4 x+4 \text { and } y^{2}=5 x-20 \\\\ & \Rightarrow 4 x+4=5 x-20 \\\\ & x=24 \Rightarrow y^{2}=4(x+1) \\\\ & y^{2}=4 \times 25 \Rightarrow y= \pm 10 \end{array} $

Shaded area

$ \begin{array}{l} =2\left[\int_{-1}^{24} 2 \sqrt{x+1}-\int_{4}^{24} \sqrt{5} \sqrt{x-4}\right] \\\\ =2\left[2\left[\frac{(x+1)^{3 / 2}}{3 / 2}\right)_{-1}^{24}-\sqrt{5}\left(\frac{(x-4)^{3 / 2}}{3 / 2}\right)_{4}^{24}\right] \\\\ =2\left[\frac{4}{3}(25)^{3 / 2}-\frac{2 \sqrt{5}}{3}(\sqrt{20})^{3}\right] \end{array} $

$ =\frac{2}{3}[4 \times 125-2 \times \sqrt{5} \times 20 \times 2 \sqrt{5}] $

$ =\frac{2}{3}[500-400]=\frac{200}{3} $ sq. units

We have,

$ \begin{aligned} & C_1: y^2=4(x+7) \\\\ & C_2: y^2=5(2-x)=-5(x-2) \end{aligned} $

$\begin{aligned} & \therefore \text { Required area }=\left|\int\left(x_1-x_2\right) d y\right| \\\\ & =\left|\int_{-2 \sqrt{5}}^{2 \sqrt{5}}\left[\left(2-\frac{y^2}{5}\right)-\left(\frac{y^2}{4}-7\right)\right] d y\right| \\\\ & =\left|\int_{-2 \sqrt{5}}^{2 \sqrt{5}}\left(-\frac{9 y^2}{20}+9\right) d y\right| \\\\ & =\left|\left[-\frac{3 y^3}{20}+9 y\right]_{-2 \sqrt{5}}^{2 \sqrt{5}}\right|=24 \sqrt{5}\end{aligned}$

$ x^2+y^2=64 \Rightarrow y=\sqrt{3} x $

$ \begin{aligned} &\begin{aligned} & \Rightarrow 3 x^2+x^2=64 \Rightarrow x=4 \\ & \Rightarrow \quad y=4 \sqrt{3} \end{aligned}\\ &\text { Area of shaded region } \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \times 4 \times 4 \sqrt{3}+\int_4^8 \sqrt{64-x^2} d x \\ & =8 \sqrt{3}+\left[\frac{x}{2} \sqrt{64-x^2}+\frac{64}{2} \sin ^{-1}\left(\frac{x}{8}\right)\right]_4^8 \\ & =8 \sqrt{3}+\left[0-2 \sqrt{48}+32 \sin ^{-1}(1)-32 \sin ^{-1}\left(\frac{1}{2}\right)\right] \\ & =32\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\frac{32 \pi}{3} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=|\sin 2 x| \\ & \therefore A=4 \int_0^{\pi / 2} \sin 2 x d x \Rightarrow A=4\left[\frac{-\cos 2 x}{2}\right]_0^{\pi / 2} \\ & \Rightarrow A=-2\left[\cos 2 \times \frac{\pi}{2}-\cos 2 \times 0\right] \\ & \Rightarrow A=-2[-1-1] \Rightarrow A=4 \end{aligned} \end{aligned} $

$y=2-x-3 x^2$

$\Rightarrow \quad\left(x+\frac{1}{6}\right)^2=-\frac{1}{3}\left(y-\frac{25}{12}\right)$ is a parabola of form $X^2=-4 A Y$

Intersection point with $X$-axis is $y=0$

$ \left(x+\frac{1}{6}\right)^2=\frac{25}{36} \Rightarrow x= \pm \frac{5}{6}-\frac{1}{6} \Rightarrow x=-1, \frac{2}{3} $

Intersection point with $Y$-axis i.e. $x=0$

$ \frac{1}{36}=-\frac{1}{3}\left(y-\frac{25}{12}\right) \Rightarrow y=2 $

Vertex of parabola $=\left(-\frac{1}{6}, \frac{25}{12}\right)$

$ \text { ∴ Required region is not possible. } $

$ \begin{aligned} &\text { The required area is }\\ &\begin{aligned} & =\int_0^3\left\{\left(2 x-x^2\right)-(-x)\right\} d x \\ & =\int_0^3\left(2 x-x^2+x\right) d x=\int_0^3\left(3 x-x^2\right) d x \\ & =\left(\frac{3 x^2}{2}-\frac{x^3}{3}\right)_0^3=\frac{27}{2}-\frac{27}{3}-0 \\ & =\frac{81-54}{6}=\frac{27}{6}=\frac{9}{2} \end{aligned} \end{aligned} $

Curves,

$ y=2 x-x^2 \Rightarrow y=x^2-2 x-6 $

Intersecting points,

$ 2 x-x^2=x^2-2 x-6 $

$ \begin{aligned} & \quad 2 x^2-4 x-6=0 \text { or } x^2-2 x-3=0 \\ & \text { or }(x-3)(x+1)=0 \text { or } x=-1, \quad x=3 \\ & \text { At } x=-1, y=2 \times(-1)-(-1)^2=-3 \\ & \text { At } x=3, y=-3 \end{aligned} $

So, both the curves are intersecting each other at points $(-1,-3)$ and $(3,-3)$

$ \begin{aligned} &\text { ∴ Required area, }\\ &\begin{aligned} A & =\int y_1 d x-\int y_2 d x=\int_{-1}^3\left(y_1-y_2\right) d x \\ & =\int_{-1}^3\left(2 x-x^2-x^2+2 x+6\right) d x \\ & =\int_{-1}^3\left(4 x-2 x^2+6\right) d x \\ & =\left[2 x^{-2}-\frac{2}{3} x^3+6 x\right]_{-1}^3 \\ & =\left\{2(3)^2-\frac{2}{3}(3)^3+6(3)\right\} \\ & \quad-\left\{2(-1)^2-\frac{2}{3}(-1)^3+6(-1)\right\} \\ & =18+\frac{10}{3}=\frac{64}{3} \text { sq units } \end{aligned} \end{aligned} $

(a) Given parabola,

$ y=x^2+3 $

Tangent of parabola at $(3,12)$ is

$ \begin{aligned} & & \frac{Y+12}{2} & =3 x+3 \\ \Rightarrow & & Y & =6 x-6 \end{aligned} $

$ \begin{aligned} &\text { Required area }\\ &\begin{aligned} & =\int_0^3\left(x^2+3\right) d x-\int_1^3(6 x-6) d x \\ & =\left[\frac{x^3}{3}+3 x\right]_0^3-\left[3 x^2-6 x\right]_1^3 \\ & =(9+9)-((27-18)+3) \\ & =18-12=6 \end{aligned} \end{aligned} $

Given equation of parabola $y^2=2 x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and equation of line $y=4 x-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Intersecting points of Eqs. (i) and (ii)

$ \text { are }\left(\frac{1}{8}, \frac{-1}{2}\right) \text { and }\left(\frac{1}{2}, 1\right) $

$ \begin{aligned} &\text { Required area } A O B\\ &\begin{aligned} & =\int_{-1 / 2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y \\ & =\frac{1}{4} \int_{-1 / 2}^1(y+1) d y-\frac{1}{2} \int_{-1 / 2}^1 y^2 d y \\ & =\frac{1}{4}\left[\frac{y^2}{2}+y\right]_{-1 / 2}^1-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-1 / 2}^1 \\ & =\frac{1}{4}\left[\left(\frac{1}{2}+1\right)-\left(\frac{1}{8}-\frac{1}{2}\right)\right]-\frac{1}{6}\left[1-\left(-\frac{1}{8}\right)\right] \\ & =\frac{1}{4}\left[\frac{3}{2}+\frac{3}{8}\right]-\frac{1}{6}\left(\frac{9}{8}\right)=\frac{1}{4} \times \frac{15}{8}-\frac{3}{16} \\ & =\frac{15-6}{32}=\frac{9}{32} \text { sq units } \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { According to the question, }\\ &\begin{aligned} & \int_{\frac{\pi}{4}}^a(\sin x-\cos x) d x=\int_a^\pi(\sin x-\cos x) d x \\ & {[-\cos x-\sin x]_{\frac{\pi}{4}}^a=[-\cos x-\sin x]_a^\pi} \\ & -\cos a-\sin a+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=1+\cos a+\sin a \end{aligned} \end{aligned} $

$ \begin{aligned} & \sqrt{2}-1=2(\cos a+\sin a) \\ & \frac{\sqrt{2}-1}{2}=\sin a+\cos a \\ & \frac{\sqrt{2}-1}{2 \sqrt{2}}=\frac{1}{\sqrt{2}} \sin a+\frac{1}{\sqrt{2}} \cos a \end{aligned} $

$ \begin{gathered} \frac{\sqrt{2}-1}{2 \sqrt{2}}=\sin a \cdot \cos \frac{\pi}{4}+\cos a \sin \frac{\pi}{4} \\ \frac{\sqrt{2}-1}{2 \sqrt{2}}=\sin \left(a+\frac{\pi}{4}\right) \end{gathered} $