Area Under The Curves

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

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} $

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 = $\pi$]

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

The given equation ${x^2} + {y^2} = 4$ is equation of circle of radius 2 centred at origin and equation ${y^2} = 3x$ is the equation of parabola.

${x^2} + {y^2} = 4$ ..... (1)

${y^2} = 3x$ ..... (2)

Substituting Eq. (2) in Eq. (1), we get

${x^2} + 3x - 4 = 0$

$ \Rightarrow {x^2} + 4x - x - 4 = 0$

$ \Rightarrow x(x + 4) - 1(x + 4) = 0$

$ \Rightarrow (x - 1)(x + 4) = 0$

$ \Rightarrow (x - 1) = 0$ and $(x + 4) = 0$

Therefore, x = 1, $-$4. Considering x = 1, then from Eq. (2), we get $y = \sqrt 3 , - \sqrt 3 $.

Therefore $(1,\sqrt 3 )$ and $(1, - \sqrt 3 )$ are the points of intersection of parabola and circle.

The required area (A) is the area of the shaded region shown in the figure. Therefore,

$A = 2\left[ {\int\limits_0^1 {{y_2}dx + \int\limits_1^2 {{y_1}dx} } } \right]$

From Eq. (1), we get ${y_1} = \sqrt {4 - {x^2}} $

From Eq. (2), we get ${y_2} = \sqrt {3x} $

Therefore, $A = 2\left[ {\int\limits_0^1 {\sqrt {3x} dx + \int\limits_1^2 {\sqrt {4 - {x^2}} dx} } } \right]$

$ = 2\left[ {\int\limits_0^1 {\sqrt 3 {x^{1/2}}dx + \int\limits_1^2 {\sqrt {{2^2} - {x^2}} dx} } } \right]$

Using standard integral : $\int {{x^n}dx = {{{x^{n - 1}}} \over {n + 1}}} $, we have

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

Therefore,

$A = 2\left[ {\left. {\left( {\sqrt 3 {{{x^{3/2}}} \over {3/2}}} \right)} \right|_0^1 + \left. {\left( {{x \over 2}\sqrt {4 - {x^2}} + {4 \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {4 - {x^2}} }}} \right)} \right|_1^2} \right]$

$ = 2\left[ {\sqrt 3 .{1 \over {3/2}} - 0 + {2 \over 2}\sqrt {4 - 4} + 2{{\tan }^{ - 1}}{2 \over {\sqrt {4 - 4} }} - {1 \over 2}\sqrt {4 - 1} - 2{{\tan }^{ - 1}}{1 \over {\sqrt {4 - 1} }}} \right]$

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

Now, $\tan {\pi \over 2} = \infty $ and $\tan {\pi \over 6} = {1 \over {\sqrt 3 }}$. Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:

$A = 2\left[ {{{2\sqrt 3 } \over 3} - {{\sqrt 3 } \over 2} + 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 2}} \right) - 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 6}} \right)} \right]$

$ = 2\left[ {\sqrt 3 {1 \over 6} + 2{\pi \over 2} - 2{\pi \over 6}} \right]$

$ = 2\left[ {\sqrt 3 + {1 \over 6} + 2{{2\pi } \over 6}} \right] = 2\left[ {{{\sqrt 3 } \over 6} + {{4\pi } \over 6}} \right]$

$ = {{\sqrt 3 } \over 3} + {{4\pi } \over 3} = \left( {{1 \over {\sqrt 3 }} + {{4\pi } \over 3}} \right)$ sq. units