86984
Area lying in the first quadrant and bounded by the circle \(x^{2}+y^{2}=4\), the line \(x=\sqrt{3} y\) and \(x-\) axis is
1 \(\pi\) sq units
2 \(\frac{\pi}{2}\) squnits
3 \(\frac{\pi}{3}\) squnits
4 None of these
Explanation:
(C) : Given, \(x^{2}+y^{2}=4\) \(x=\sqrt{3} y\) Area \(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}\) Area \(=\frac{1}{2 \sqrt{3}}[3-0]+\left\{\left[2 \sin ^{-1} 1\right]-\left[\frac{\sqrt{3}}{2} \times 1+2 \times \sin ^{-1} \frac{\sqrt{3}}{2}\right]\right\}\) Area \(=\frac{\sqrt{3}}{2}+2 \times \sin ^{-1}\left(\sin \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) Area \(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\) Area \(=\pi-\frac{2 \pi}{3} \Rightarrow\) Area \(=\frac{\pi}{3}\) square units.
VITEEE-2012
Application of the Integrals
86977
If the area between \(y=\mathbf{m x}^{2}\) and \(x=m y^{2}(m>0)\) is \(\frac{1}{4}\) sq. units, then value of \(m\) is
1 \(\pm 3 \sqrt{2}\)
2 \(\pm \frac{2}{\sqrt{3}}\)
3 \(\sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
(B) : Given, \(y=m x^{2}\) And \(x=m y^{2}(m>0)\) The point of intersection of both curves, \(x=m\left(m x^{2}\right)^{2} \Rightarrow x=m^{3} x^{4}\) \(m^{3} x^{4}-x=0 \Rightarrow x\left(m^{3} x^{3}-1\right)=0\) \(x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0\) Now using the zero property of product we get \(\mathrm{x}=0, \mathrm{x}=\frac{1}{\mathrm{~m}} \text { and } \mathrm{y}=0, \mathrm{y}=1 / \mathrm{m}\) So the point of intersection are : \((0,0),\left(\frac{1}{\mathrm{~m}}, \frac{1}{\mathrm{~m}}\right)\) Now required area, \(A=\int_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) \cdot d x\) \(\frac{1}{4}=\left[\frac{1}{\sqrt{m}} \cdot \frac{x^{3 / 2}}{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m} \frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{\mathrm{~m}^{3 / 2}}-\frac{\mathrm{m}}{3} \cdot \frac{1}{\mathrm{~m}^{3}}\right]\) \(\frac{1}{4}=\left[\frac{2}{3 \mathrm{~m}^{2}}-\frac{1}{3 \mathrm{~m}^{3}}\right] \frac{1}{4}=\frac{1}{3 \mathrm{~m}^{2}}\) \(\therefore \mathrm{m}= \pm \frac{2}{\sqrt{3}}\)
Karnataka CET-2011
Application of the Integrals
86978
The area of the region bounded by \(y=|x-1|\) and \(y=1\) is
1 2
2 1
3 \(1 / 2\)
4 \(1 / 4\)
Explanation:
(B) : The given region is represented by the equations \(\mathrm{y}=1-\mathrm{x}, \mathrm{x} \leq 1=\mathrm{x}-1, \mathrm{x} \geq 1\) and \(\mathrm{y}=1 ; \mathrm{C}=(2,1)\) and \(\mathrm{B}=(0,1)\) \(\therefore\) the shaded area in the figure \(\mathrm{A}=\frac{1}{2} \mathrm{BC} \cdot \mathrm{AC}=\frac{1}{2} 2 \cdot 1=1\)
BITSAT-2005
Application of the Integrals
86979
The area of the region \(R=\{(x, y):|x| \leq|y|\) and \(\left.\mathbf{x}^{2}+\mathbf{y}^{2} \leq 1\right\}\) is
1 \(\frac{3 \pi}{8}\) sq. units
2 \(\frac{5 \pi}{8}\) sq. units
3 \(\frac{\pi}{2}\) sq. units
4 \(\frac{\pi}{8}\) sq. units.
Explanation:
(C) : We have, Required area \(=4\) (Area of the shaded region in first quadrant) \(=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x\) \(=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}\) \(=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]\) \(=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}\) sq units
BITSAT-2015
Application of the Integrals
86980
The area of the region bounded by the curve \(\mathbf{y}=\mathbf{x}|\mathbf{x}|, \mathbf{x}\)-axis and the ordinates \(\mathrm{x}=1, \mathbf{x}=-1\) is given by :
1 zero
2 \(\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 1
Explanation:
(C) : The area of the region bounded by the curve \(y=f(x)\) and the ordinates \(x=a, x=b\) is given Area \(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx}\right|\) According to the question, \(y=x|x|= \begin{cases}x^{2}, x \geq 0 \\ -x^{2}, x\lt 0\end{cases}\) Required area \(=\) area of region \(\mathrm{OAB}+\) area of region \(\mathrm{OCD}\) \(=2 \times\) Area of region \(\mathrm{OAB}\) \(=2 \int_{0}^{1} x^{2} \mathrm{dx}=\frac{2}{3}\) sq. units
86984
Area lying in the first quadrant and bounded by the circle \(x^{2}+y^{2}=4\), the line \(x=\sqrt{3} y\) and \(x-\) axis is
1 \(\pi\) sq units
2 \(\frac{\pi}{2}\) squnits
3 \(\frac{\pi}{3}\) squnits
4 None of these
Explanation:
(C) : Given, \(x^{2}+y^{2}=4\) \(x=\sqrt{3} y\) Area \(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}\) Area \(=\frac{1}{2 \sqrt{3}}[3-0]+\left\{\left[2 \sin ^{-1} 1\right]-\left[\frac{\sqrt{3}}{2} \times 1+2 \times \sin ^{-1} \frac{\sqrt{3}}{2}\right]\right\}\) Area \(=\frac{\sqrt{3}}{2}+2 \times \sin ^{-1}\left(\sin \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) Area \(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\) Area \(=\pi-\frac{2 \pi}{3} \Rightarrow\) Area \(=\frac{\pi}{3}\) square units.
VITEEE-2012
Application of the Integrals
86977
If the area between \(y=\mathbf{m x}^{2}\) and \(x=m y^{2}(m>0)\) is \(\frac{1}{4}\) sq. units, then value of \(m\) is
1 \(\pm 3 \sqrt{2}\)
2 \(\pm \frac{2}{\sqrt{3}}\)
3 \(\sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
(B) : Given, \(y=m x^{2}\) And \(x=m y^{2}(m>0)\) The point of intersection of both curves, \(x=m\left(m x^{2}\right)^{2} \Rightarrow x=m^{3} x^{4}\) \(m^{3} x^{4}-x=0 \Rightarrow x\left(m^{3} x^{3}-1\right)=0\) \(x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0\) Now using the zero property of product we get \(\mathrm{x}=0, \mathrm{x}=\frac{1}{\mathrm{~m}} \text { and } \mathrm{y}=0, \mathrm{y}=1 / \mathrm{m}\) So the point of intersection are : \((0,0),\left(\frac{1}{\mathrm{~m}}, \frac{1}{\mathrm{~m}}\right)\) Now required area, \(A=\int_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) \cdot d x\) \(\frac{1}{4}=\left[\frac{1}{\sqrt{m}} \cdot \frac{x^{3 / 2}}{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m} \frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{\mathrm{~m}^{3 / 2}}-\frac{\mathrm{m}}{3} \cdot \frac{1}{\mathrm{~m}^{3}}\right]\) \(\frac{1}{4}=\left[\frac{2}{3 \mathrm{~m}^{2}}-\frac{1}{3 \mathrm{~m}^{3}}\right] \frac{1}{4}=\frac{1}{3 \mathrm{~m}^{2}}\) \(\therefore \mathrm{m}= \pm \frac{2}{\sqrt{3}}\)
Karnataka CET-2011
Application of the Integrals
86978
The area of the region bounded by \(y=|x-1|\) and \(y=1\) is
1 2
2 1
3 \(1 / 2\)
4 \(1 / 4\)
Explanation:
(B) : The given region is represented by the equations \(\mathrm{y}=1-\mathrm{x}, \mathrm{x} \leq 1=\mathrm{x}-1, \mathrm{x} \geq 1\) and \(\mathrm{y}=1 ; \mathrm{C}=(2,1)\) and \(\mathrm{B}=(0,1)\) \(\therefore\) the shaded area in the figure \(\mathrm{A}=\frac{1}{2} \mathrm{BC} \cdot \mathrm{AC}=\frac{1}{2} 2 \cdot 1=1\)
BITSAT-2005
Application of the Integrals
86979
The area of the region \(R=\{(x, y):|x| \leq|y|\) and \(\left.\mathbf{x}^{2}+\mathbf{y}^{2} \leq 1\right\}\) is
1 \(\frac{3 \pi}{8}\) sq. units
2 \(\frac{5 \pi}{8}\) sq. units
3 \(\frac{\pi}{2}\) sq. units
4 \(\frac{\pi}{8}\) sq. units.
Explanation:
(C) : We have, Required area \(=4\) (Area of the shaded region in first quadrant) \(=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x\) \(=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}\) \(=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]\) \(=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}\) sq units
BITSAT-2015
Application of the Integrals
86980
The area of the region bounded by the curve \(\mathbf{y}=\mathbf{x}|\mathbf{x}|, \mathbf{x}\)-axis and the ordinates \(\mathrm{x}=1, \mathbf{x}=-1\) is given by :
1 zero
2 \(\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 1
Explanation:
(C) : The area of the region bounded by the curve \(y=f(x)\) and the ordinates \(x=a, x=b\) is given Area \(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx}\right|\) According to the question, \(y=x|x|= \begin{cases}x^{2}, x \geq 0 \\ -x^{2}, x\lt 0\end{cases}\) Required area \(=\) area of region \(\mathrm{OAB}+\) area of region \(\mathrm{OCD}\) \(=2 \times\) Area of region \(\mathrm{OAB}\) \(=2 \int_{0}^{1} x^{2} \mathrm{dx}=\frac{2}{3}\) sq. units
86984
Area lying in the first quadrant and bounded by the circle \(x^{2}+y^{2}=4\), the line \(x=\sqrt{3} y\) and \(x-\) axis is
1 \(\pi\) sq units
2 \(\frac{\pi}{2}\) squnits
3 \(\frac{\pi}{3}\) squnits
4 None of these
Explanation:
(C) : Given, \(x^{2}+y^{2}=4\) \(x=\sqrt{3} y\) Area \(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}\) Area \(=\frac{1}{2 \sqrt{3}}[3-0]+\left\{\left[2 \sin ^{-1} 1\right]-\left[\frac{\sqrt{3}}{2} \times 1+2 \times \sin ^{-1} \frac{\sqrt{3}}{2}\right]\right\}\) Area \(=\frac{\sqrt{3}}{2}+2 \times \sin ^{-1}\left(\sin \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) Area \(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\) Area \(=\pi-\frac{2 \pi}{3} \Rightarrow\) Area \(=\frac{\pi}{3}\) square units.
VITEEE-2012
Application of the Integrals
86977
If the area between \(y=\mathbf{m x}^{2}\) and \(x=m y^{2}(m>0)\) is \(\frac{1}{4}\) sq. units, then value of \(m\) is
1 \(\pm 3 \sqrt{2}\)
2 \(\pm \frac{2}{\sqrt{3}}\)
3 \(\sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
(B) : Given, \(y=m x^{2}\) And \(x=m y^{2}(m>0)\) The point of intersection of both curves, \(x=m\left(m x^{2}\right)^{2} \Rightarrow x=m^{3} x^{4}\) \(m^{3} x^{4}-x=0 \Rightarrow x\left(m^{3} x^{3}-1\right)=0\) \(x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0\) Now using the zero property of product we get \(\mathrm{x}=0, \mathrm{x}=\frac{1}{\mathrm{~m}} \text { and } \mathrm{y}=0, \mathrm{y}=1 / \mathrm{m}\) So the point of intersection are : \((0,0),\left(\frac{1}{\mathrm{~m}}, \frac{1}{\mathrm{~m}}\right)\) Now required area, \(A=\int_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) \cdot d x\) \(\frac{1}{4}=\left[\frac{1}{\sqrt{m}} \cdot \frac{x^{3 / 2}}{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m} \frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{\mathrm{~m}^{3 / 2}}-\frac{\mathrm{m}}{3} \cdot \frac{1}{\mathrm{~m}^{3}}\right]\) \(\frac{1}{4}=\left[\frac{2}{3 \mathrm{~m}^{2}}-\frac{1}{3 \mathrm{~m}^{3}}\right] \frac{1}{4}=\frac{1}{3 \mathrm{~m}^{2}}\) \(\therefore \mathrm{m}= \pm \frac{2}{\sqrt{3}}\)
Karnataka CET-2011
Application of the Integrals
86978
The area of the region bounded by \(y=|x-1|\) and \(y=1\) is
1 2
2 1
3 \(1 / 2\)
4 \(1 / 4\)
Explanation:
(B) : The given region is represented by the equations \(\mathrm{y}=1-\mathrm{x}, \mathrm{x} \leq 1=\mathrm{x}-1, \mathrm{x} \geq 1\) and \(\mathrm{y}=1 ; \mathrm{C}=(2,1)\) and \(\mathrm{B}=(0,1)\) \(\therefore\) the shaded area in the figure \(\mathrm{A}=\frac{1}{2} \mathrm{BC} \cdot \mathrm{AC}=\frac{1}{2} 2 \cdot 1=1\)
BITSAT-2005
Application of the Integrals
86979
The area of the region \(R=\{(x, y):|x| \leq|y|\) and \(\left.\mathbf{x}^{2}+\mathbf{y}^{2} \leq 1\right\}\) is
1 \(\frac{3 \pi}{8}\) sq. units
2 \(\frac{5 \pi}{8}\) sq. units
3 \(\frac{\pi}{2}\) sq. units
4 \(\frac{\pi}{8}\) sq. units.
Explanation:
(C) : We have, Required area \(=4\) (Area of the shaded region in first quadrant) \(=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x\) \(=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}\) \(=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]\) \(=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}\) sq units
BITSAT-2015
Application of the Integrals
86980
The area of the region bounded by the curve \(\mathbf{y}=\mathbf{x}|\mathbf{x}|, \mathbf{x}\)-axis and the ordinates \(\mathrm{x}=1, \mathbf{x}=-1\) is given by :
1 zero
2 \(\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 1
Explanation:
(C) : The area of the region bounded by the curve \(y=f(x)\) and the ordinates \(x=a, x=b\) is given Area \(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx}\right|\) According to the question, \(y=x|x|= \begin{cases}x^{2}, x \geq 0 \\ -x^{2}, x\lt 0\end{cases}\) Required area \(=\) area of region \(\mathrm{OAB}+\) area of region \(\mathrm{OCD}\) \(=2 \times\) Area of region \(\mathrm{OAB}\) \(=2 \int_{0}^{1} x^{2} \mathrm{dx}=\frac{2}{3}\) sq. units
86984
Area lying in the first quadrant and bounded by the circle \(x^{2}+y^{2}=4\), the line \(x=\sqrt{3} y\) and \(x-\) axis is
1 \(\pi\) sq units
2 \(\frac{\pi}{2}\) squnits
3 \(\frac{\pi}{3}\) squnits
4 None of these
Explanation:
(C) : Given, \(x^{2}+y^{2}=4\) \(x=\sqrt{3} y\) Area \(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}\) Area \(=\frac{1}{2 \sqrt{3}}[3-0]+\left\{\left[2 \sin ^{-1} 1\right]-\left[\frac{\sqrt{3}}{2} \times 1+2 \times \sin ^{-1} \frac{\sqrt{3}}{2}\right]\right\}\) Area \(=\frac{\sqrt{3}}{2}+2 \times \sin ^{-1}\left(\sin \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) Area \(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\) Area \(=\pi-\frac{2 \pi}{3} \Rightarrow\) Area \(=\frac{\pi}{3}\) square units.
VITEEE-2012
Application of the Integrals
86977
If the area between \(y=\mathbf{m x}^{2}\) and \(x=m y^{2}(m>0)\) is \(\frac{1}{4}\) sq. units, then value of \(m\) is
1 \(\pm 3 \sqrt{2}\)
2 \(\pm \frac{2}{\sqrt{3}}\)
3 \(\sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
(B) : Given, \(y=m x^{2}\) And \(x=m y^{2}(m>0)\) The point of intersection of both curves, \(x=m\left(m x^{2}\right)^{2} \Rightarrow x=m^{3} x^{4}\) \(m^{3} x^{4}-x=0 \Rightarrow x\left(m^{3} x^{3}-1\right)=0\) \(x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0\) Now using the zero property of product we get \(\mathrm{x}=0, \mathrm{x}=\frac{1}{\mathrm{~m}} \text { and } \mathrm{y}=0, \mathrm{y}=1 / \mathrm{m}\) So the point of intersection are : \((0,0),\left(\frac{1}{\mathrm{~m}}, \frac{1}{\mathrm{~m}}\right)\) Now required area, \(A=\int_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) \cdot d x\) \(\frac{1}{4}=\left[\frac{1}{\sqrt{m}} \cdot \frac{x^{3 / 2}}{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m} \frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{\mathrm{~m}^{3 / 2}}-\frac{\mathrm{m}}{3} \cdot \frac{1}{\mathrm{~m}^{3}}\right]\) \(\frac{1}{4}=\left[\frac{2}{3 \mathrm{~m}^{2}}-\frac{1}{3 \mathrm{~m}^{3}}\right] \frac{1}{4}=\frac{1}{3 \mathrm{~m}^{2}}\) \(\therefore \mathrm{m}= \pm \frac{2}{\sqrt{3}}\)
Karnataka CET-2011
Application of the Integrals
86978
The area of the region bounded by \(y=|x-1|\) and \(y=1\) is
1 2
2 1
3 \(1 / 2\)
4 \(1 / 4\)
Explanation:
(B) : The given region is represented by the equations \(\mathrm{y}=1-\mathrm{x}, \mathrm{x} \leq 1=\mathrm{x}-1, \mathrm{x} \geq 1\) and \(\mathrm{y}=1 ; \mathrm{C}=(2,1)\) and \(\mathrm{B}=(0,1)\) \(\therefore\) the shaded area in the figure \(\mathrm{A}=\frac{1}{2} \mathrm{BC} \cdot \mathrm{AC}=\frac{1}{2} 2 \cdot 1=1\)
BITSAT-2005
Application of the Integrals
86979
The area of the region \(R=\{(x, y):|x| \leq|y|\) and \(\left.\mathbf{x}^{2}+\mathbf{y}^{2} \leq 1\right\}\) is
1 \(\frac{3 \pi}{8}\) sq. units
2 \(\frac{5 \pi}{8}\) sq. units
3 \(\frac{\pi}{2}\) sq. units
4 \(\frac{\pi}{8}\) sq. units.
Explanation:
(C) : We have, Required area \(=4\) (Area of the shaded region in first quadrant) \(=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x\) \(=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}\) \(=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]\) \(=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}\) sq units
BITSAT-2015
Application of the Integrals
86980
The area of the region bounded by the curve \(\mathbf{y}=\mathbf{x}|\mathbf{x}|, \mathbf{x}\)-axis and the ordinates \(\mathrm{x}=1, \mathbf{x}=-1\) is given by :
1 zero
2 \(\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 1
Explanation:
(C) : The area of the region bounded by the curve \(y=f(x)\) and the ordinates \(x=a, x=b\) is given Area \(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx}\right|\) According to the question, \(y=x|x|= \begin{cases}x^{2}, x \geq 0 \\ -x^{2}, x\lt 0\end{cases}\) Required area \(=\) area of region \(\mathrm{OAB}+\) area of region \(\mathrm{OCD}\) \(=2 \times\) Area of region \(\mathrm{OAB}\) \(=2 \int_{0}^{1} x^{2} \mathrm{dx}=\frac{2}{3}\) sq. units
86984
Area lying in the first quadrant and bounded by the circle \(x^{2}+y^{2}=4\), the line \(x=\sqrt{3} y\) and \(x-\) axis is
1 \(\pi\) sq units
2 \(\frac{\pi}{2}\) squnits
3 \(\frac{\pi}{3}\) squnits
4 None of these
Explanation:
(C) : Given, \(x^{2}+y^{2}=4\) \(x=\sqrt{3} y\) Area \(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}\) Area \(=\frac{1}{2 \sqrt{3}}[3-0]+\left\{\left[2 \sin ^{-1} 1\right]-\left[\frac{\sqrt{3}}{2} \times 1+2 \times \sin ^{-1} \frac{\sqrt{3}}{2}\right]\right\}\) Area \(=\frac{\sqrt{3}}{2}+2 \times \sin ^{-1}\left(\sin \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\sin \frac{\pi}{3}\right)\) Area \(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\) Area \(=\pi-\frac{2 \pi}{3} \Rightarrow\) Area \(=\frac{\pi}{3}\) square units.
VITEEE-2012
Application of the Integrals
86977
If the area between \(y=\mathbf{m x}^{2}\) and \(x=m y^{2}(m>0)\) is \(\frac{1}{4}\) sq. units, then value of \(m\) is
1 \(\pm 3 \sqrt{2}\)
2 \(\pm \frac{2}{\sqrt{3}}\)
3 \(\sqrt{2}\)
4 \(\sqrt{3}\)
Explanation:
(B) : Given, \(y=m x^{2}\) And \(x=m y^{2}(m>0)\) The point of intersection of both curves, \(x=m\left(m x^{2}\right)^{2} \Rightarrow x=m^{3} x^{4}\) \(m^{3} x^{4}-x=0 \Rightarrow x\left(m^{3} x^{3}-1\right)=0\) \(x(m x-1)\left(m^{2} x^{2}+1+m x\right)=0\) Now using the zero property of product we get \(\mathrm{x}=0, \mathrm{x}=\frac{1}{\mathrm{~m}} \text { and } \mathrm{y}=0, \mathrm{y}=1 / \mathrm{m}\) So the point of intersection are : \((0,0),\left(\frac{1}{\mathrm{~m}}, \frac{1}{\mathrm{~m}}\right)\) Now required area, \(A=\int_{0}^{1 / m}\left(\sqrt{\frac{x}{m}}-m x^{2}\right) \cdot d x\) \(\frac{1}{4}=\left[\frac{1}{\sqrt{m}} \cdot \frac{x^{3 / 2}}{3 / 2}-m \cdot \frac{x^{3}}{3}\right]_{0}^{1 / m} \frac{1}{4}=\left[\frac{2}{3 \sqrt{m}} \cdot \frac{1}{\mathrm{~m}^{3 / 2}}-\frac{\mathrm{m}}{3} \cdot \frac{1}{\mathrm{~m}^{3}}\right]\) \(\frac{1}{4}=\left[\frac{2}{3 \mathrm{~m}^{2}}-\frac{1}{3 \mathrm{~m}^{3}}\right] \frac{1}{4}=\frac{1}{3 \mathrm{~m}^{2}}\) \(\therefore \mathrm{m}= \pm \frac{2}{\sqrt{3}}\)
Karnataka CET-2011
Application of the Integrals
86978
The area of the region bounded by \(y=|x-1|\) and \(y=1\) is
1 2
2 1
3 \(1 / 2\)
4 \(1 / 4\)
Explanation:
(B) : The given region is represented by the equations \(\mathrm{y}=1-\mathrm{x}, \mathrm{x} \leq 1=\mathrm{x}-1, \mathrm{x} \geq 1\) and \(\mathrm{y}=1 ; \mathrm{C}=(2,1)\) and \(\mathrm{B}=(0,1)\) \(\therefore\) the shaded area in the figure \(\mathrm{A}=\frac{1}{2} \mathrm{BC} \cdot \mathrm{AC}=\frac{1}{2} 2 \cdot 1=1\)
BITSAT-2005
Application of the Integrals
86979
The area of the region \(R=\{(x, y):|x| \leq|y|\) and \(\left.\mathbf{x}^{2}+\mathbf{y}^{2} \leq 1\right\}\) is
1 \(\frac{3 \pi}{8}\) sq. units
2 \(\frac{5 \pi}{8}\) sq. units
3 \(\frac{\pi}{2}\) sq. units
4 \(\frac{\pi}{8}\) sq. units.
Explanation:
(C) : We have, Required area \(=4\) (Area of the shaded region in first quadrant) \(=4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x\) \(=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}\) \(=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]\) \(=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}\) sq units
BITSAT-2015
Application of the Integrals
86980
The area of the region bounded by the curve \(\mathbf{y}=\mathbf{x}|\mathbf{x}|, \mathbf{x}\)-axis and the ordinates \(\mathrm{x}=1, \mathbf{x}=-1\) is given by :
1 zero
2 \(\frac{1}{3}\)
3 \(\frac{2}{3}\)
4 1
Explanation:
(C) : The area of the region bounded by the curve \(y=f(x)\) and the ordinates \(x=a, x=b\) is given Area \(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx}\right|\) According to the question, \(y=x|x|= \begin{cases}x^{2}, x \geq 0 \\ -x^{2}, x\lt 0\end{cases}\) Required area \(=\) area of region \(\mathrm{OAB}+\) area of region \(\mathrm{OCD}\) \(=2 \times\) Area of region \(\mathrm{OAB}\) \(=2 \int_{0}^{1} x^{2} \mathrm{dx}=\frac{2}{3}\) sq. units
