NCERT Solutions for Class 12 Maths Chapter 8 (Ex 8.1)
Exercise 8.1
1. Find the Area of the region bounded by the curve \[{{y}^{2}}=x\] and the lines \[x=1,\text{ }x=4\] and the x-axis in the first quadrant.
Ans:
The area of the region bounded by the curve \[{{y}^{2}}=x\] and the lines \[x = 1,{\text{ }}x = 4\]
and the x-axis
Area of the region \[\text{ABCDA}=\int\limits_{1}^{4}{\sqrt{x}dx}\]
Area of the region \[\text{ABCDA}=\left[\frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}}\right]_{1}^{4}\]
Area of the region \[\text{ABCDA}=\left[ \frac{{{4}^{\frac{3}{2}}}}{\frac{3}{2}}-\frac{{{1}^{\frac{3}{2}}}}{\frac{3}{2}} \right]\]
Area of the region \[\text{ABCDA}=\frac{2}{3}\left[ {{4}^{\frac{3}{2}}}-1 \right]\]
Area of the region \[\text{ABCDA}=\frac{2}{3}\left[ 8-1 \right]\]
Area of the region \[\text{ABCDA}=\frac{2}{3}\left[ 7 \right]\]
Area of the region \[\text{ABCDA}=\frac{14}{3}\]square units
2. Find the Area of the region bounded by the curve \[{{y}^{2}}=9x\], \[x=2,\text{ }x=4\] and the x-axis in the first quadrant.
Ans:
The area of the region bounded by the curve and the lines \[x = 2,{\text{ }}x = 4\] and the x-axis is the area of ABCDA
Area of \[\text{ABCDA=}\int\limits_{\text{2}}^{\text{4}}{\text{3}\sqrt{\text{x}}\text{dx}}\]
Area of \[\text{ABCDA=3}\left[ \frac{{{\text{x}}^{\frac{\text{3}}{\text{2}}}}}{\frac{\text{3}}{\text{2}}} \right]_{\text{2}}^{\text{4}}\]
Area of \[\text{ABCDA=2}\left[ {{\left( \text{x} \right)}^{\frac{\text{3}}{\text{2}}}} \right]_{\text{2}}^{\text{4}}\]
Area of \[\text{ABCDA=2}\left[ {{\left( 4 \right)}^{\frac{\text{3}}{\text{2}}}}-{{\left( 2 \right)}^{\frac{\text{3}}{\text{2}}}} \right]\]
Area of \[\text{ABCDA=2}\left[ {{\left( {{2}^{2}} \right)}^{\frac{\text{3}}{\text{2}}}}-{{\left( 2 \right)}^{\frac{\text{3}}{\text{2}}}} \right]\]
Area of \[\text{ABCDA=2}\left[ {{2}^{3}}-{{\left( 2 \right)}^{\frac{\text{3}}{\text{2}}}} \right]\]
Area of \[\text{ABCDA=2}\left[ {{2}^{3}}-{{8}^{\frac{1}{\text{2}}}} \right]\]
Area of \[\text{ABCDA=2}\left[ 8-2\sqrt{2} \right]\]
Area of \[ABCDA=16-4\sqrt{2}\] sq.units
3. Find the area of the region which is bounded \[{{x}^{2}}=4y,\text{ }y=2,\text{ }y=4\] and the y-axis in the first quadrant.
Ans:
The area of the region bounded by the curve \[{x^2} = 4y,{\text{ }}y = 2,{\text{ }}y = 4\] the y-axis is the area ABCD.
Area of ABCDA \[=\int\limits_{2}^{4}{xdy}\]
\[{{x}^{2}}=4y\]
\[x\text{ }=\text{2}\sqrt{y}\]
Area of ABCDA \[=\int\limits_{2}^{4}{\text{2}\sqrt{y}dy}\]
Area of ABCDA \[=2\int\limits_{2}^{4}{\sqrt{y}dy}\]
Area of ABCDA \[=2\left[ \frac{{{y}^{\frac{3}{2}}}}{\frac{3}{2}} \right]_{2}^{4}\]
Area of ABCDA \[=\frac{4}{3}\left[ {{\left( 4 \right)}^{\frac{3}{2}}}-{{\left( 2 \right)}^{\frac{3}{2}}} \right]_{2}^{4}\]
Area of ABCDA \[=\frac{4}{3}\left[ 8-2\sqrt{2} \right]\]
Area of ABCDA \[=\left( \frac{32-8\sqrt{2}}{3} \right)\text{ sq}\text{. units}\]
4. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\].
Ans: Rewrite the given equation of ellipse
\[\frac{{{x}^{2}}}{{{4}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]
The intersection point of the ellipse and the x-axis are at \[x=\pm 4\].
The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].
The given equation of the ellipse can be represented as
If we change \[x\] to \[-x\] or \[y\] to \[y\] the equation remains the same.
So, the ellipse is symmetrical about the x-axis and the y-axis.
Area bounded by ellipse \[=4\times\] Area of OABO
Area of OABO \[=\int\limits_{0}^{4}{ydx}\]
\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\]
\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{16}\]
\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{16} \right)\]
\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{16}}\]
Area of OABO \[=\int\limits_{0}^{4}{3\sqrt{1-\frac{{{x}^{2}}}{16}}dx}\]
Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}dx}\]
Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{{{4}^{2}}-{{x}^{2}}}dx}\]
Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation
Area of OABO \[=\frac{3}{4}\left[ \frac{x}{2}\sqrt{16-{{x}^{2}}}+\frac{16}{2}{{\sin }^{-1}}\frac{x}{4} \right]_{0}^{4}\]
Area of OABO \[=\frac{3}{4}\left[ 0+8{{\sin }^{-1}}(1)-0-8{{\sin }^{-1}}(0) \right]_{0}^{4}\]
Area of OABO \[=\frac{3}{4}\left( \frac{8\pi }{2} \right)\]
Area of OABO \[=3\pi \]
Therefore, the area bounded by ellipse \[=\text{4}\times \text{3}\pi\]
The area bounded by ellipse \[=\text{12}\pi \text{ sq}\text{. units}\]
5. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\].
Ans: Rewrite the given equation of the ellipse
\[\frac{{{x}^{2}}}{{{2}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]
The intersection point of the ellipse and the x-axis are at \[x=\pm 2\].
The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].
The given equation of the ellipse can be represented as a vertical ellipse.
It can be observed that the ellipse is symmetrical about the x-axis and y-axis.
Since, if we change \[x\] to \[-x\] or \[y\] to \[-y\] the equation remains the same.
Area bounded by ellipse \[=4\times\]Area of OABO
Area of OABO \[=\int\limits_{0}^{2}{ydx}\]…………………….(A)
\[\Rightarrow \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\]
\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{4}\]
\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{4} \right)\]
\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{4}}\text{ }\]
\[\Rightarrow y=\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }\]…………………..(1)
Area of OABO \[=\int\limits_{0}^{2}{\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }dx}\] using equation (1) in (A)
Area of OABO \[=\frac{3}{2}\int\limits_{0}^{2}{\sqrt{{{2}^{2}}-{{x}^{2}}}dx}\]
Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation
Area of OABO \[=\frac{3}{2}\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]
Area of OABO \[=\frac{3}{2}\left[ \frac{2\pi }{2} \right]\]
Area of OABO \[=\frac{3\pi }{2}\]
Area bound by the ellipse \[=\text{4}\times \frac{\text{3}\pi }{2}\]
Area bound by the ellipse \[=\text{6}\pi \text{ sq}\text{. units}\]
6. Find the area of the region in the first quadrant enclosed by the x-axis, line \[x=\sqrt{3y}\] and the circle \[{{x}^{2}}+{{y}^{2}}=4\].
Ans: The area of the region bounded by the circle \[{{\text{x}}^{2}}+{{y}^{2}}=4\],\[\text{x}=\sqrt{3}y\] and the x-axis is the area OAB
Substituting \[\text{x}=\sqrt{3}y\] in \[{{\text{x}}^{2}}+{{y}^{2}}=4\] for finding the point of intersection.
\[\therefore \left( \sqrt{3y} \right)+{{y}^{2}}=4\]
\[\Rightarrow {{y}^{2}}=1\]
Put \[{{y}^{2}}=1\] in equation \[\text{x}=\sqrt{3y}\]
\[{{\text{x}}^{2}}+1=4\]
\[{{\text{x}}^{2}}=4-1\]
\[{{\text{x}}^{2}}=3\]
\[\text{x}=\pm \sqrt{3}\]
So,
\[\Rightarrow y=\pm 1,\text{ x}=\pm \sqrt{3}\]
The line \[\text{x}=\sqrt{3}y\] can be rewritten as
\[\text{y}=\frac{x}{\sqrt{3}}\]
The point of intersection of the line and the circle in the first quadrant is \[\left( \sqrt{3},1 \right)\].
Area of OABO = Area of \[\Delta \text{OCA }+\]Area of ABCA
Now, the shaded area OAC is between the segment OM of line \[\text{x}=\sqrt{3}y\] and x-axis.
Area of \[\Delta \text{OCA}=\left| \int\limits_{0}^{\sqrt{3}}{ydx} \right|\]
Put \[\text{y}=\frac{x}{\sqrt{3}}\] in above equation
Area of \[\Delta \text{OCA}=\int\limits_{0}^{\sqrt{3}}{\frac{x}{\sqrt{3}}dx}\]
Area of \[\Delta \text{OCA}=\frac{1}{\sqrt{3}}\int\limits_{0}^{\sqrt{3}}{xdx}\]
Area of \[\Delta \text{OCA}=\frac{1}{\sqrt{3}}\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{\sqrt{3}}\]
Area of \[\Delta \text{OCA}=\frac{1}{\sqrt{3}}\left[ \frac{{{(\sqrt{3})}^{2}}}{2}-0 \right]\]
Area of \[\Delta \text{OCA}=\frac{1}{\sqrt{3}}\left[ \frac{3}{2} \right]\]
Area of \[\Delta \text{OCA}=\frac{3}{2\sqrt{3}}\]
Area of \[\Delta \text{OCA}=\frac{3}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\]
Area of \[\Delta \text{OCA}=\frac{3\sqrt{3}}{6}\]
Area of \[\Delta \text{OCA}=\frac{\sqrt{3}}{2}\]……………………..(1)
Now,
Area of \[\text{ABCA}=\int\limits_{\sqrt{3}}^{2}{ydx}\]
Area of \[\text{ABCA}=\int\limits_{\sqrt{3}}^{2}{\sqrt{4-{{x}^{2}}}dx}\]
When, \[x=2\sin \theta\]
Then,
\[\theta ={{\sin }^{-1}}\left( \frac{x}{2} \right)\]
When, \[\text{x}=2\text{ , }\theta =\frac{\pi }{2}\]
\[\text{ x}=\sqrt{3},\text{ }\theta =\frac{\pi }{3}\]
Area of \[ABCA=\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\sqrt{4-4{{\sin }^{2}}\theta }\left( 2\cos \theta \right)d\theta }\]
Area of \[ABCA=4\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}}{{{\cos }^{2}}\theta d\theta }\]
Area of \[ABCA=\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}}{1+\frac{\left( \cos 2\theta \right)}{2}d\theta }\]
Area of \[ABCA=2\int\limits_{\frac{\pi }{3}}^{\frac{\pi }{2}}{(1+\cos 2\theta )d\theta }\]
Area of \[ABCA=2\left[ \theta +\frac{\sin 2\theta }{2} \right]_{\frac{\pi }{3}}^{\frac{\pi }{2}}\]
Area of \[ABCA=2\left[ \frac{\pi }{2}-\frac{\pi }{3}-\frac{1}{2}\times \frac{\sqrt{3}}{2} \right]\]
Area of \[ABCA=2\left[ \frac{\pi }{6}-\frac{\sqrt{3}}{4} \right]\text{ }\]………………………(2)
From (1) and (2)
Area of OAB\[=\frac{\sqrt{3}}{2}+2\left[ \frac{\pi }{6}-\frac{\sqrt{3}}{4} \right]\]
Area of OAB\[=\frac{\pi }{3}\]
Therefore, the area enclosed by x-axis, the line \[\text{x}=\sqrt{3}y\]and the circle \[{{\text{x}}^{2}}+{{y}^{2}}=4\]in the first quadrant is\[\text{ }\frac{\pi }{3}sq.units\].
7. Find the area of the smaller part of the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] cut off by the line \[x=\frac{a}{\sqrt{2}}\].
Ans:
The area of the smaller part of the circle, \[{{\text{x}}^{2}}+{{y}^{2}}={{a}^{2}}\], cut off by the line \[\text{x}=\frac{a}{\sqrt{2}}\] is area of the region ABCDA.
It can be observed that the area ABCDA is symmetrical about the x-axis.
Area of \[\text{ABCDA}=\text{ 2}\times\] Area of ABCA
Area of \[\text{ABCA}=\int\limits_{\frac{a}{\sqrt{2}}}^{a}{ydx}\]
Area of \[ABCA=\int\limits_{\frac{a}{\sqrt{2}}}^{a}{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}\]
\[x=a\sin \theta \text{ , dx}=a\cos \theta d\theta\]
\[\text{x}=\frac{a}{\sqrt{2}}\] and \[\theta ={{\sin }^{-1}}\left( \frac{x}{a} \right)\]
\[\theta ={{\sin }^{-1}}\left( \frac{1}{\sqrt{2}} \right)\]
\[\theta =\frac{\pi }{4}\]
\[\text{x}=a\] and \[\theta ={{\sin }^{-1}}\left( \frac{a}{a} \right)=\frac{\pi }{2}\]
Area of \[ABCA=\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta }\left( a\cos \theta \right)d\theta }\]
Area of \[ABCA={{a}^{2}}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{{{\cos }^{2}}\theta d\theta }\]
Area of \[ABCA=\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\frac{1+\left( \cos 2\theta \right)}{2}d\theta }\]
Area of \[ABCA=\frac{{{a}^{2}}}{2}\left[ \theta +\frac{\sin 2\theta }{2} \right]_{\frac{\pi }{4}}^{\frac{\pi }{2}}\]
Area of \[ABCA=\frac{{{a}^{2}}}{2}\left[ \frac{\pi }{2}+\frac{\sin \pi }{2}-\frac{\pi }{4}-\frac{\sin \frac{\pi }{2}}{2} \right]\]
Area of \[ABCA=\frac{{{a}^{2}}}{2}\left[ \frac{\pi }{4}-\frac{1}{2} \right]\]
Area of \[ABCA=\frac{{{a}^{2}}}{4}\left[ \frac{\pi }{2}-1 \right]\]
Area of \[\text{ABCD}=2\left[ \frac{{{a}^{2}}}{4}\left( \frac{\pi }{2}-1 \right) \right]\]
Area of \[ABCD=\left[ \frac{{{a}^{2}}}{2}\left( \frac{\pi }{2}-1 \right) \right]\]
Therefore, the area of the smaller part of the circle, \[{{\text{x}}^{2}}+{{y}^{2}}={{a}^{2}}\], cut off by the line \[\text{x}=\frac{a}{\sqrt{2}}\] is\[\frac{{{a}^{2}}}{2}\left( \frac{\pi }{2}-1 \right)\text{ sq}\text{. units}\].
8. The area between \[x={{y}^{2}}\] and \[x=4\] is divided into two equal parts by the line \[x=a\], find the value of a.
Ans:
The line \[x=a\] divides the area bounded by the parabola \[x={{y}^{2}}\] and \[x=4\] into two equal parts.
Area OADO =Area ABCDA
It can be observed that the given area is symmetrical about the x-axis.
Area of \[\text{OEDO}={\scriptstyle{}^{1}/{}_{2}}\left( Area\text{ of OADO} \right)\]
Area of \[\text{EFCDE}={}^{1}/{}_{2}\left( Area\text{ of ABCDA} \right)\]
\[\therefore Area\text{ OEDO}=Area\text{ EFCDE}\]
Area of \[\text{OEDO}=\int\limits_{0}^{a}{ydx}\]
Area of \[\text{OEDO}=\int\limits_{0}^{a}{\sqrt{x}dx}\]
Area of \[\text{OEDO}=\left[ \frac{\frac{{{x}^{\frac{3}{2}}}}{3}}{2} \right]_{0}^{a}\]
Area of \[\text{OEDO}=\frac{2}{3}{{\left( a \right)}^{\frac{3}{2}}}\]…………………………..(1)
Area of \[\text{EFCDE}=\int\limits_{0}^{4}{ydx}\]
Area of \[\text{EFCDE}=\int\limits_{0}^{4}{\sqrt{x}dx}\]
Area of \[\text{EFCDE}=\left[ \frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}} \right]_{0}^{4}\]
Area of \[\text{EFCDE}=\frac{2}{3}\left[ {{4}^{\frac{3}{2}}}-{{a}^{\frac{3}{2}}} \right]\]
Area of \[\text{EFCDE}=\frac{2}{3}\left[ 8-{{a}^{\frac{3}{2}}} \right]\]
From (1) and (2), we obtain
\[\frac{2}{3}{{\left( a \right)}^{\frac{3}{2}}}=\frac{2}{3}\left[ 8-{{a}^{\frac{3}{2}}} \right]\text{ }\]
\[\Rightarrow 2.\left( a \right)\frac{3}{2}=4\]
\[\Rightarrow {{\left( a \right)}^{\frac{3}{2}}}=4\]
\[\Rightarrow a={{\left( 4 \right)}^{\frac{2}{3}}}\]
Therefore, the value of \[\text{a}\]is \[{{\left( 4 \right)}^{\frac{2}{3}}}\].
9. Find the Area of the region bounded by the parabola curve \[{{x}^{2}}=4y\], and \[y=\left| x \right|\].
Ans: Here, the required area is the area between the parabola \[{{x}^{2}}=4y\] and the modulus function \[y=\left| x \right|\].
The modulus function can be written as
$ y= x \text{ if } x > 0 $
\[y=\left\{ \begin{align} & x,\text{ if }x\ge 0 \\ & -x,\text{ if }x\le 0 \\ \end{align} \right.\]
The parabola curve \[{{x}^{2}}=4y\] represents an upward parabola with vertex at origin.
Since, the parabola is symmetric about its axis.
\[{{x}^{2}}=y\] is symmetric about the y-axis.
Area of the shaded region \[=2\times (\]Area of \[OAD)\]
Since, point B is the point of intersection of \[y=x\] and the parabola
Put the value of \[y=x\] in the equation of the parabola that is \[y={{x}^{2}}\]
\[x={{x}^{2}}\]
\[{{x}^{2}}-x=0\]
\[x(x-1)=0\]
So,
\[x=0\] or \[x=1\]
For \[x=0\]
\[y=x\]
\[y=0\]
For \[x=1\]
\[y=x\]
\[y=1\]
So, the point is \[(0,0)\] and \[(1,1)\].
So, \[A=(1,1)\]
Now, the area of OAD
Area of \[OAD=\] Area of \[OAP-\] Area of ODAP
Area of \[OAD=\left[ \int_{0}^{1}{{{y}_{1}}dx-}\int_{0}^{1}{{{y}_{2}}dx} \right]\]
Area of \[OAD=\left[ \int_{0}^{1}{x.dx-}\int_{0}^{1}{{{x}^{2}}dx} \right]\]
Area of shaded region \[=2\times\](Area of OAD)
Area of shaded region \[=2\times \left[ \int_{0}^{1}{x.dx-}\int_{0}^{1}{{{x}^{2}}dx} \right]\]
Area of shaded region \[=2\left[ \left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}-\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{1} \right]\]
Area of shaded region \[=2\left[ \frac{1-0}{2}-\frac{1-0}{3} \right]\]
Area of shaded region \[=2\left[ \frac{1}{2}-\frac{1}{3} \right]\]
Area of shaded region \[=2\left[ \frac{3-2}{6} \right]\]
Area of shaded region \[=2\left[ \frac{1}{6} \right]\]
Area of shaded region \[=\frac{1}{3}\] square units
10. The Area of the region bounded by the curve \[{{x}^{2}}=4y\], and the line \[x=4y-2\] is
Ans:
The area bounded by the curve \[{{x}^{2}}=4y\] and the line \[x=4y-2\] is represented by the shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.
\[{{\left( 4y-2 \right)}^{2}}=4y\]
\[16{{y}^{2}}-16y+4=4y\]
\[16{{y}^{2}}-20y+4=0\]
\[4{{y}^{2}}-5y+1=0\]
\[\left( 4y-1 \right)\left( y-1 \right)=0\]
\[y=\frac{1}{4}\]or \[y=1\]
For \[y=\frac{1}{4}\]
\[x=4y-2\]
\[x=4\times \frac{1}{4}-2\]
\[x=1-2\]
\[x=-1\]
For \[y=1\]
\[x=4y-2\]
\[x=4(1)-2\]
\[x=4-2\]
\[x=2\]
So, the points are \[\left( -1,\frac{1}{4} \right)\] and \[(2,1)\].
So, the required area \[=\]Area of \[ALBM-\] Area of \[ALOMBA\]
Area of \[ALMB=\int_{-1}^{2}{ydx}\]
Since \[x=4y-2\] Equation of the line
\[x+2=4y\]
\[y=\frac{x+2}{4}\]
Area of \[ALMB=\int_{-1}^{2}{\frac{x+2}{4}dx}\]
Area of \[ALMB=\frac{1}{4}\int_{-1}^{2}{(x+2)dx}\]
Area of \[ALMB=\frac{1}{4}\left[ \frac{{{x}^{2}}}{2}+2x \right]_{-1}^{2}\]
Area of \[ALMB=\frac{1}{4}\left[ \left( \frac{{{2}^{2}}}{2}+2(2) \right)-\left( \frac{{{(-1)}^{2}}}{2}+2(-1) \right) \right]\]
Area of \[ALMB=\frac{1}{4}\left[ \left( \frac{4}{2}+4 \right)-\left( \frac{1}{2}-2 \right) \right]\]
Area of \[ALMB=\frac{1}{4}\left[ \left( 2+4 \right)-\left( \frac{1}{2}-2 \right) \right]\]
Area of \[ALMB=\frac{1}{4}\left[ 6-\frac{1}{2}+2 \right]\]
Area of \[ALMB=\frac{1}{4}\times \frac{15}{2}\]
Area of \[ALMB=\frac{15}{8}\]
Area of \[ALOMBA=\int_{-1}^{2}{ydx}\]
\[{{x}^{2}}=4y\] Equation of the parabola
\[y=\frac{1}{4}{{x}^{2}}\]
Area of \[ALOMBA=\int_{-1}^{2}{\frac{1}{4}{{x}^{2}}dx}\]
Area of \[ALOMBA=\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{-1}^{2}\]
Area of \[ALOMBA=\frac{1}{4}\left[ \frac{{{2}^{3}}-{{(-1)}^{3}}}{3} \right]\]
Area of \[ALOMBA=\frac{1}{4}\left[ \frac{8-(-1)}{3} \right]\]
Area of \[ALOMBA=\frac{1}{4}\left[ \frac{8+1}{3} \right]\]
Area of \[ALOMBA=\frac{1}{4}\left[ \frac{9}{3} \right]\]
Area of \[ALOMBA=\frac{3}{4}\]
Now,
The required area \[=\] Area of \[ALMB\text{ }\] Area of ALOMBA
The required area \[=\frac{15}{8}-\frac{3}{4}\]
The required area \[=\frac{15-6}{8}\]
The required area \[=\frac{9}{8}\] square units.
11. Find the area of the region bounded by the curve \[{{y}^{2}}=4x\], and the line \[x=3\] is
Ans:
The area of OACO is the region bounded by the parabola,\[{{y}^{2}}=4x\] and the line, \[x=3\].
Since, the area of the region OACO is symmetrical about the x-axis.
Area of OACO \[=2\](Area of OAB)
Area of \[OACO=2\left[ \int_{0}^{3}{ydx} \right]\]
Area of \[OACO=2\int_{0}^{3}{2\sqrt{x}dx}\]
Area of \[OACO=4\int_{0}^{3}{\sqrt{x}dx}\]
Area of \[OACO=4\left[ \frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}} \right]_{0}^{3}\]
Area of \[OACO=4\left[ \frac{{{3}^{\frac{3}{2}}}}{\frac{3}{2}}-0 \right]\]
Area of \[OACO=\frac{8}{3}\left[ {{3}^{\frac{3}{2}}} \right]\]
Area of \[OACO=8\sqrt{3}\] square units
Hence, the required area is \[8\sqrt{3}\] square units.
12. Area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is
\[\pi \]
\[\frac{\pi }{2}\]
\[\frac{\pi }{3}\]
\[\frac{\pi }{4}\]
Ans:
Area of \[OAB=\int_{0}^{2}{ydx}\]
Since, \[{{\text{x}}^2} + {y^2} = 4\] $ [\text {Equation of circle}]$
\[{{y}^{2}}=4-{{\text{x}}^{2}}\]
\[y=\sqrt{4-{{\text{x}}^{2}}}\]
So,
Area of \[OAB=\int_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\]
Area of \[OAB=\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]
Area of \[OAB=\left[ \frac{2}{2}\sqrt{4-{{2}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{2}{2}-\left( \frac{0}{2}\sqrt{4-{{0}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{0}{2} \right) \right]\]
Area of \[OAB=\left[ \sqrt{4-4}+2{{\sin }^{-1}}1-0 \right]\]
Area of \[OAB=\left[ 2\times \frac{\pi }{2} \right]\]
Area of \[OAB=\pi \] square units
Hence, the area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is \[\pi \] square units.
Hence, option (A) is the correct answer.
13. Area of the region bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is
2
\[\frac{9}{4}\]
\[\frac{9}{3}\]
\[\frac{9}{2}\]
Ans:
The area bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is the area of the region \[OAB\].
Area of \[OAB=\int_{0}^{3}{xdy}\]
Area of \[OAB=\int_{0}^{3}{\frac{{{y}^{2}}}{4}dy}\]
Area of \[OAB=\frac{1}{4}\int_{0}^{3}{{{y}^{2}}dy}\]
Area of \[OAB=\frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{0}^{3}\]
Area of \[OAB=\frac{1}{4}\left[ \frac{{{3}^{3}}}{3}-0 \right]\]
Area of \[OAB=\frac{1}{4}\left[ \frac{27}{3} \right]\]
Area of \[OAB=\frac{1}{4}\left[ 9 \right]\]
Area of \[OAB=\frac{9}{4}\]square units
Hence, option (B) is the correct answer.
Access NCERT Solutions for Class 12 Maths Chapter 08-Application of Integrals
Opting for the NCERT solutions for Ex 8.1 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.1 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Topics Covered in Class 12 Maths Chapter 8 Exercise 8.1
The questions asked in Chapter 8 Exercise 8.1 are based on the following topics:
Area under Simple Curves
Area of a region bounded by a curve and a line.
How to Find the Area Under a Simple Curve?
The area under a curve between two points can be found by performing a definite integral between the two points.
To find the area under the curve y = f(x) from x = a and x = b, integrate y = f(x) w.r.t. X between the limits a and b.
NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 12 Maths Chapter 8 Exercises
Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 8 Exercise 8.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 12 Maths Chapter 8 Exercise 8.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
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