Area Under The Curves

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{aligned} & R=\left\{(x, y): \frac{y^2}{2} \leq x \leq y+4\right\} \\ & x \geq \frac{y^2}{2} \Rightarrow y^2 \leq 2 x \\ & \text { and } x \leq y+4 \Rightarrow y-x+4 \geq 0 \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Solving } y^2=2 x \text { and } y-x+4=0\\ &\begin{aligned} & \quad y-\frac{y^2}{2}+4=0 \\ & \Rightarrow \quad y^2-2 y-8=0 \\ & \Rightarrow(y-4)(y+2)=0 \\ & \Rightarrow \quad y=-2,4 \\ & \text { Area required }=\int_{-2}^4\left(x_L-x_p\right) d y \\ & =\int_{-2}^4\left(y+4-\frac{y^2}{2}\right) d y \\ & =\left[\frac{y^2}{2}+4 y-\frac{y^3}{6}\right]_{-2}^4 \\ & =\left(8+16-\frac{32}{3}\right)-\left(2-8+\frac{4}{3}\right) \\ & =24-\frac{32}{3}+6-\frac{4}{3} \\ & =30-\frac{36}{3}=30-12=18 \end{aligned} \end{aligned} $

Given:

The circle is $x^2 + y^2 = 16$.

The parabola is $y^2 = 6x$.

Finding Intersection Points:

To find where these curves meet, put $y^2$ from the parabola into the circle's equation:

$ \begin{array}{ll} \rightarrow & x^2 + y^2 = 16 \\ & x^2 + 6x = 16 \\ & x^2 + 6x - 16 = 0 \\ & (x + 8)(x - 2) = 0 \\ & x = -8,\ 2 \end{array} $

We only use $x = 2$ (since $x = -8$ is outside the region for $y^2 = 6x$ and gives no real $y$ for $x^2 + y^2 = 16$).

For $x = 2$, $y^2 = 6 \cdot 2 = 12$. So $y = \sqrt{12}$ and $y = -\sqrt{12}$.

Finding the Area:

We want the area inside the circle and outside the parabola, between $x = 0$ and $x = 4$.

To get this area, we split it into two parts:

• From $x = 0$ to $x = 2$ (where the parabola is below the circle), the top curve is the parabola: $y = \sqrt{6x}$
• From $x = 2$ to $x = 4$ (where only the circle remains), the top curve is the circle: $y = \sqrt{16 - x^2}$

So, the area is:

Area $= 2 \left[\int_0^2 y_\text{parabola} \,dx + \int_2^4 y_\text{circle} \,dx \right]$

So, $ = 2 \int_0^2 \sqrt{6x}\,dx + 2 \int_2^4 \sqrt{16 - x^2}\,dx $

Solving the Integrals:

First, $\int_0^2 \sqrt{6x}\,dx$:

$\sqrt{6x} = \sqrt{6} \sqrt{x}$, so

$ \int_0^2 \sqrt{6x} dx = \sqrt{6} \int_0^2 x^{1/2} dx = \sqrt{6} \cdot \left[\frac{2}{3}x^{3/2}\right]_0^2 = \sqrt{6} \cdot \frac{2}{3} \left[(2)^{3/2} - 0\right] $

Second, $\int_2^4 \sqrt{16 - x^2}\,dx$:

The formula for this is:

$ \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) $

Here $a = 4$. Plug in $x = 4$ and $x = 2$ to get the answer.

Now, combine both solutions:

$ \begin{aligned} &= 2 \sqrt{6} \cdot \frac{2}{3} \left(2^{3/2}\right) + 2\left[ \frac{x}{2}\sqrt{16 - x^2} + 8 \sin^{-1}\left(\frac{x}{4}\right)\Big|_{x=2}^{x=4} \right] \\ &= \frac{4}{3}\sqrt{6}\cdot 2\sqrt{2} + 2 \left[ \left(0 + 8 \frac{\pi}{2}\right) - \left(2\sqrt{3} + 8 \frac{\pi}{6}\right) \right] \\ &= \frac{16\sqrt{3}}{3} + 2\left(\frac{8\pi}{3} - 2\sqrt{3}\right) \\ &= \frac{16\sqrt{3}}{3} + \frac{16\pi}{3} - 4\sqrt{3} \\ &= \frac{4\sqrt{3}}{3} + \frac{16\pi}{3} \\ &= \frac{4}{3}(4\pi + \sqrt{3}) \end{aligned} $

Given, curves are $y=x^2$ and

$ y=8-x^2 $

Equating these two equations, we get

$ x^2=8-x^2 \Rightarrow 2 x^2=8 $

$ \begin{aligned} & \Rightarrow \quad x^2=4 \Rightarrow x= \pm 2 \\ & \text { Thus, area }=\int_{-2}^2\left[\left(8-x^2\right)-x^2\right] d x \\ & \qquad \begin{array}{l} \quad=\int_{-2}^2\left(8-x^2-x^2\right) d x=\int_{-2}^2\left(8-2 x^2\right) d x \\ \quad=\left[8 x-\frac{2}{3} x^3\right]_{-2}^2 \\ \quad=\left\{8(2)-\frac{2}{3}(2)^3\right\}-\left\{8(-2)-\frac{2}{3}(-2)^3\right\} \\ \quad=\left(16-\frac{16}{3}\right)-\left(-16+\frac{16}{3}\right) \\ \quad=16-\frac{16}{3}+16-\frac{16}{3} \\ \quad=32-\frac{32}{3}=\frac{96-32}{3}=\frac{64}{3} \end{array} \end{aligned} $

$\therefore$ The area of the region bounded by the curves is $\frac{64}{3}$ sq. units.

$ y=x^2-5 x+4=(x-1)(x-4) $

$ \begin{aligned} A=A_1+A_2=\int_0^1\left(x^2\right. & -5 x+4) d x +\int_1^2-\left(x^2-5 x+4\right) \\ =\left[\frac{x^3}{3}-\frac{5 x^2}{2}+4 x\right]_0^1 & +\left[\frac{-x^3}{3}+\frac{5 x^2}{2}-4 x\right]^2_1 \end{aligned} $

$ \begin{aligned} & =\left(\frac{1}{3}-\frac{5}{2}+4\right)+\left(-\frac{8}{3}+10-8+\frac{1}{3}-\frac{5}{2}+4\right) \\ & =\left(\frac{3}{2}+\frac{1}{3}\right)+\left(6-\frac{7}{3}-\frac{5}{2}\right) \\ & =\frac{9+2}{6}+\left(\frac{36-14-15}{6}\right) \\ & =\frac{11}{6}+\frac{7}{6}=\frac{18}{6}=3 \end{aligned} $

$ \begin{aligned} &\text { Required area (A) }\\ &=\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x \end{aligned} $

$ \begin{aligned} = & (\sin x+\cos x)_0^{\frac{\pi}{4}}+(-\cos x-\sin x)_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ = & {\left[\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)-(\sin 0+\cos 0)\right] }-\left[\left(\cos \frac{\pi}{2}+\sin \frac{\pi}{2}\right)-\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)\right] \\ = & (\sqrt{2}-1)-(1-\sqrt{2}) \\ = & \sqrt{2}-1-1+\sqrt{2} \\ = & 2 \sqrt{2}-2=2(\sqrt{2}-1) \text { sq. unit } \end{aligned} $

$ \text { } y=\sqrt{4-x^2} \text { and } y^2=3 x $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & 3 x=4-x^2 \\ \Rightarrow & x^2+3 x-4=0 \\ \Rightarrow & (x+4)(x-1)=0 \\ & x=1,-4 \Rightarrow y=\sqrt{3} \end{array}\\ &\therefore \text { Required area, }\\ &\begin{aligned} & =\frac{1}{3} \int_0^{\sqrt{3}} y^2 d y+\int_{\sqrt{3}}^2 \sqrt{4-y^2} d y \\ & =\frac{1}{3}\left[\frac{y^3}{3}\right]_0^{\sqrt{3}}+\left[\frac{y}{2} \sqrt{4-y^2}+\frac{4}{2} \sin ^{-1} \frac{y}{2}\right]_{\sqrt{3}}^2 \\ & =\frac{1}{9}(3 \sqrt{3})+2 \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2\left(\frac{\pi}{3}\right) \\ & =\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{2}+\frac{\pi}{3}=\frac{\pi}{3}-\frac{1}{2 \sqrt{3}} \end{aligned} \end{aligned} $

$ \text { Area of enclosed region, } $

$ \begin{aligned} & =2\left(\frac{1}{2} \times 1 \times 1\right)+(x) \\ & =1+1=2 \end{aligned} $

$ \begin{aligned} &\text { As }(\alpha, \beta) \text { is the stationary point of }\\ &\begin{aligned} & & y & =2 x-x^2, \frac{d y}{d x}=0 \\ & & & 2-2 x \\ \Rightarrow & & x & =0 \\ \Rightarrow & & \alpha & =1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore \text { Required area }=\int_0^1\left(y_2-y_1\right) d x \\ & \left.=\int_0^1\left(2^x-2 x+x^2\right)\right) d x=\left[\frac{2^x}{\log 2}-x^2+\frac{x^3}{3}\right]_0^1 \\ & =\left[\frac{2^1}{\log 2}-1+\frac{1}{3}-\frac{1}{\log 2}\right] \end{aligned} $

$ \begin{aligned} & =\frac{2}{\log 2}-1+\frac{1}{3}-\frac{1}{\log 2} \\ & =\frac{1}{\log 2}-\frac{2}{3} \\ & \Rightarrow \frac{3-\log 4}{3 \log 2} \end{aligned} $

$ \begin{aligned} & x=y^2, x=3-2 y^2 \\ & y=\sqrt{x} \\ & y=\sqrt{\frac{3-x}{2}} \end{aligned} $

Vertices $(0,0)$ and $(3,0)$

intersect at ( 1,1 ) and (1, -1)

Area of shaded region

$=2 \times($ Area of $P Q M)$

$ \begin{aligned} &\begin{aligned} & =2\left\{\left[\int_0^1 \sqrt{x} d x\right]+\left[\int_1^3 \sqrt{\frac{3-x}{2}} d x\right]\right\} \\ & \left.=2\left[\left(\frac{2}{3} x^{3 / 2}\right)_0^1+\left(\frac{-1}{\sqrt{2}} \cdot \frac{2}{3}(3-x)^{3 / 2}\right)\right)_1^3\right] \\ & =2\left[\frac{2}{3}-\left(0-\frac{1}{\sqrt{2}} \cdot \frac{2}{3} 2^{3 / 2}\right)\right] \\ & =2\left(\frac{2}{3}+\frac{4}{3}\right)=2 \times \frac{6}{3}=4 \end{aligned}\\ &\text { Area of required region }=4 \mathrm{sq} \text { units } \end{aligned} $

Given curves are

$ \begin{array}{r} 3 x^2-y^2-2 x y+4 x+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } 3 x^2-y^2-2 x y+6 x+2 y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Eqs. (i) and (ii) can be written as

$(3 x+y+1)(x-y+1)=0$, and $(3 x+y)(x-y+2)=0$, respectively.

Thus, the curve (i) represents the pair of lines

$ \begin{array}{r} L_1: 3 x+y+1=0 \\ L_2: x-y+1=0 \end{array} $

and the curve (ii) represents the pair of lines

$ \begin{array}{r} L_3: 3 x+y=0 \\ L_4: x-y+2=0 \end{array} $

Here, $L_1 \| L_3$ and $L_2 \| L_4$

So, the region closed by the given curves will be a parallelogram.

Now, required area $=$ Area of parallelogram $A B C D$ $=2 \times($ area of $\triangle A B C)$

$ \begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc} -\frac{1}{2} & \frac{1}{2} & 1 \\ -\frac{1}{4} & \frac{3}{4} & 1 \\ -\frac{1}{2} & \frac{3}{2} & 1 \end{array}\right| \\ & =\left[-\frac{1}{2}\left(\frac{3}{4}-\frac{3}{2}\right)-\frac{1}{2}\left(-\frac{1}{4}+\frac{1}{2}\right)+1\left(-\frac{3}{8}+\frac{3}{8}\right)\right] \\ & =-\frac{1}{2}\left(-\frac{3}{4}\right)-\frac{1}{2}\left(\frac{1}{4}\right)+0 \\ & =\frac{3}{8}-\frac{1}{8}=\frac{2}{8}=\frac{1}{4} \end{aligned} $

To find the area under the curve $ y = |\sin x - \cos x| $ from $ x = 0 $ to $ x = \frac{\pi}{2} $, and above the $ X $-axis, we proceed as follows:

The expression $|\sin x - \cos x|$ changes based on $x$. Solving $|\sin x - \cos x| = 0$:

$ \begin{align*} \sin x - \cos x &= 0 \\ \sin x &= \cos x \\ \tan x &= 1 \\ x &= \frac{\pi}{4} \end{align*} $

This implies that $ \sin x - \cos x $ is negative from $ 0 $ to $ \frac{\pi}{4} $ and positive from $ \frac{\pi}{4} $ to $ \frac{\pi}{2} $.

Thus, the function can be defined piecewise as:

$ |\sin x - \cos x| = \begin{cases} \cos x - \sin x & \text{for } 0 \leq x < \frac{\pi}{4} \\ \sin x - \cos x & \text{for } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \end{cases} $

Integrating from $ x = 0 $ to $ x = \frac{\pi}{4} $:

$ \int_0^{\pi/4} (\cos x - \sin x) \, dx = \int_0^{\pi/4} \cos x \, dx - \int_0^{\pi/4} \sin x \, dx $

Calculating each integral:

$ \int_0^{\pi/4} \cos x \, dx = [\sin x]_0^{\pi/4} = \sin\frac{\pi}{4} - \sin 0 = \frac{\sqrt{2}}{2} $

$ \int_0^{\pi/4} \sin x \, dx = [-\cos x]_0^{\pi/4} = -\cos\frac{\pi}{4} + \cos 0 = -\frac{\sqrt{2}}{2} + 1 $

Thus, the area from $ x = 0 $ to $ x = \frac{\pi}{4} $ is:

$ \int_0^{\pi/4} (\cos x - \sin x) \, dx = \frac{\sqrt{2}}{2} - \left( -\frac{\sqrt{2}}{2} + 1 \right) = \sqrt{2} - 1 $

Integrating from $ x = \frac{\pi}{4} $ to $ x = \frac{\pi}{2} $:

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

Calculating each integral:

$ \int_{\pi/4}^{\pi/2} \sin x \, dx = [-\cos x]_{\pi/4}^{\pi/2} = -\cos\frac{\pi}{2} + \cos\frac{\pi}{4} = 0 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} $

$ \int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin\frac{\pi}{2} - \sin\frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} $

Thus, the area from $ x = \frac{\pi}{4} $ to $ x = \frac{\pi}{2} $ is:

$ \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = \frac{\sqrt{2}}{2} - \left( 1 - \frac{\sqrt{2}}{2} \right) = \sqrt{2} - 1 $

Total Area:

Adding the two parts:

$ (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1) $

Thus, the total area under the curve is $ 2(\sqrt{2} - 1) $ square units.

$ \text { We have, } $

$ y^2=8(x+2) \text { and } y^2=4(1-x) $

The intersecting point of the above 2 parabola are $(-1, \pm 2 \sqrt{2})$

$\therefore$ Required area

$ \begin{aligned} & =\left\lvert\, 2\left[\int_2^{2 \sqrt{2}}\left(1-\frac{y^2}{4}\right) d y+\int_{2 \sqrt{2}}^4\left(\frac{y^2}{8}-2\right) d y\right]\right. \\ & =\left\lvert\, 2\left[\left[y-\frac{y^3}{12}\right]_2^{2 \sqrt{2}}+\left[\frac{y^3}{24}-2 y\right]_{2 \sqrt{2}}^4\right]\right. \\ & =\left\lvert\, 2\left[\left(2 \sqrt{2}-\frac{16 \sqrt{2}}{12}\right)-\left(2-\frac{8}{12}\right)+\left(\frac{64}{24}-8\right)\right.\right. \\ & \left.-\left(\frac{16 \sqrt{2}}{24}-4 \sqrt{2}\right)\right] \end{aligned} $

$ \begin{aligned} & =\left\lvert\, 2\left[\frac{2 \sqrt{2}}{3}-\frac{4}{3}-\frac{16}{3}+\frac{10 \sqrt{2}}{3}\right]\right. \\ & =\left|2\left[\frac{12 \sqrt{2}}{3}-\frac{20}{3}\right]\right| \\ & =\left|\frac{8}{3}[3 \sqrt{2}-5]\right|=\frac{8}{3}(5-3 \sqrt{2}) \end{aligned} $

Given the problem of finding the area of a region bounded by a circle and a parabola, let's explore how to calculate it.

The equation of the circle is:

$ x^2 + y^2 = 2ax $

This can be rewritten by completing the square for $x$:

$ (x-a)^2 + y^2 = a^2 $

The equation of the parabola is:

$ y^2 = ax $

Finding Points of Intersection:

To find the intersection points between the circle and the parabola, we set:

$ x^2 + ax = 2ax $

$ x^2 - ax = 0 $

$ x(x-a) = 0 $

Thus, $x = 0$ or $x = a$.

For $x = 0$, substituting in the parabola $y^2 = ax$, we get $y = 0$.

For $x = a$, substituting in the parabola $y^2 = ax$, we get $y = \pm a$.

Thus, the points of intersection are $(0, 0)$, $(a, a)$, and $(a, -a)$.

Calculating the Area:

We need to integrate the area between the circle and the parabola from $x = 0$ to $x = a$:

$ \text{Required Area} = \int_0^a \left( \sqrt{a^2 - (x-a)^2} - \sqrt{a} \sqrt{x} \right) \, dx $

This evaluates to:

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

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

$ = -\frac{2}{3} a^2 + \frac{a^2}{2} \times \frac{\pi}{2} $

Therefore, the area of the smaller region lying above the $X$-axis is:

$ a^2 \left(\frac{\pi}{4} - \frac{2}{3}\right) $

$\begin{aligned} & (y)=x^3-19 x+30 \\ & =(x-2)(x-3)(x+5)\end{aligned}$

$\begin{aligned} & \therefore \text { Required area } \\ & \begin{aligned} &=\int_{-5}^2\left(x^3-19 x+30\right) d x+\left|\int_2^3 x^3-19 x+30 d x\right| \\ &= {\left[\frac{x^4}{4}-\frac{19 x^2}{2}+30 x\right]_{-5}^2 } \\ &+\left|\left[\frac{x^4}{4}-\frac{19 x^2}{2}+30 x\right]_2^3\right| \\ &=(4-38+60)-\left(\frac{625}{4}-\frac{475}{2}-150\right) \\ &+\left|\left(\frac{81}{4}-\frac{171}{2}+90\right)-(4-38+60)\right| \\ &=26+\frac{925}{4}+\left|\frac{99}{4}-26\right|=\frac{1029}{4}+\frac{5}{4} \\ &= \frac{1034}{4}=\frac{517}{2}\end{aligned}\end{aligned}$