86838
The area bounded by the curve \(y=2 x-x^{2}\) and the straight line \(y=-x\) is given by:
1 \(\frac{9}{2}\) squnit
2 \(\frac{43}{6}\) sq unit
3 \(\frac{35}{6}\) sq unit
4 none of these
Explanation:
(A) : Given that, \(y=2 x-x^{2}\) \(y=-x \tag{i}\) Solving equation (i) and (ii)- \(-x=2 x-x^{2} \Rightarrow \quad x^{2}-3 x=0 \tag{ii}\) \(x(x-3)=0\) \(x=0 \text { and } x=3\) putting the value of \(x\) in equation (ii) we get intersection point \((0,0)\) and \((3,-3)\) So, required area, \(A=\int_{0}^{3}\left(2 x-x^{2}\right) d x-\int_{0}^{3}(-x) d x\) \(A=\int_{0}^{3}\left(2 x-x^{2}+x\right) d x=\int_{0}^{3}\left(3 x-x^{2}\right) d x\) \(A=\left[\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3}=\left[\frac{27}{2}-\frac{27}{3}\right]=\frac{27}{6}=\frac{9}{2}\) sq. units
AMU-2017
Application of the Integrals
86839
The area bounded by \(y=\sin ^{-1} x, x=\frac{1}{\sqrt{2}}\) and \(\mathrm{X}\)-axis is
1 \(\frac{1}{\sqrt{2}}+1\)
2 \(1-\frac{1}{\sqrt{2}}\)
3 \(\frac{\pi}{4 \sqrt{2}}\)
4 \(\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
Explanation:
(D) : Given that, \(\mathrm{y}=\sin ^{-1} \mathrm{x} \quad \mathrm{x}=\frac{1}{\sqrt{2}}\) From \(y=\sin ^{-1}(x)\), at \(x=0, \quad y=0\) and at \(x=\frac{1}{\sqrt{2}}, y=\) \(\frac{\pi}{4}\) at \(x=1, \quad y=\pi / 2\) \(\because \mathrm{y}=\sin ^{-1} \mathrm{x}\) \(\therefore \mathrm{x}=\sin \mathrm{y}\) Now as per the figure area of shaded region- \(\mathrm{A}=\) area of rectangle \(\mathrm{OABC}-\) area of curve \(\mathrm{OBAO}\) \(\mathrm{A}=\left(\frac{1}{\sqrt{2}} \times \frac{\pi}{4}\right)-\int_{0}^{\pi / 4} \sin \mathrm{ydy}\) \(\mathrm{A}=\frac{\pi}{4 \sqrt{2}}+[\cos \mathrm{y}]_{0}^{\pi / 4}=\frac{\pi}{4 \sqrt{2}}+\left[\cos \frac{\pi}{4}-\cos 0^{\circ}\right]\) \(A=\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
BCECE-2014
Application of the Integrals
86840
The volume of the solid formed by rotating the area enclosed between the curve \(y=x^{2}\) and the line \(y=1\) about \(y=1\) is (in cubic unit):
1 \(9 \pi / 5\)
2 \(2 \pi / 5\)
3 \(8 \pi / 3\)
4 \(7 \pi / 5\)
Explanation:
(B) : The curve is \(y=x^{2}\) Given line, \(\quad y=1\) and \(y=1\) Putting the value of \(y\) in equation (i) \(y=x^{2} \Rightarrow \quad x^{2}=1\) \(x= \pm 1\) As per the figure required volume, \(V=\pi \int_{-1}^{1} y^{2} d x \Rightarrow V=2 \pi \int_{0}^{1} x^{4} d x\) \(\left(\because \int_{-9}^{9} f(x) d x=2 \int_{-9}^{9} f(x) d x\right)\) \(V=2 \pi \times\left(\frac{1^{5}-0^{5}}{5}\right)=\frac{2 \pi}{5}\)
[JCECE-2004]
Application of the Integrals
86841
The area bounded by the curve \(y=\sin x\) between \(x=0\) and \(x=2 \pi\) is
1 1 sq unit
2 2 sq unit
3 4 sq unit
4 8 sq unit
Explanation:
(C): The curve, \(\quad \mathrm{y}=\sin \mathrm{x} \quad \ldots .\). (i) From equation (i)- at. \(x=0, y=0\), at \(x=\frac{\pi}{2}, y=1\) at \(\mathrm{x}=\pi, \mathrm{y}=0, \quad\) at \(\mathrm{x}=\frac{3 \pi}{2}, \mathrm{y}=1\) So, as per the figure required area, \(\mathrm{A}=\) area of (OBAO \(+\mathrm{BDCB})\) \(\therefore \quad \mathrm{A}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{dx}+\left|\int_{\pi}^{2 \pi} \sin \mathrm{x} \mathrm{dx}\right|\) \(\mathrm{A}=[-\cos \mathrm{x}]_{0}^{\pi}+\left|[-\cos \mathrm{x}]_{\pi}^{2 \pi}\right|\) \(A=\left[-\cos \pi+\cos 0^{\circ}\right]+\mid[-\cos 2 \pi+\cos \pi]\) \(\mathrm{A}=1+1+|-1-1| \Rightarrow \mathrm{A}=4\) sq units
86838
The area bounded by the curve \(y=2 x-x^{2}\) and the straight line \(y=-x\) is given by:
1 \(\frac{9}{2}\) squnit
2 \(\frac{43}{6}\) sq unit
3 \(\frac{35}{6}\) sq unit
4 none of these
Explanation:
(A) : Given that, \(y=2 x-x^{2}\) \(y=-x \tag{i}\) Solving equation (i) and (ii)- \(-x=2 x-x^{2} \Rightarrow \quad x^{2}-3 x=0 \tag{ii}\) \(x(x-3)=0\) \(x=0 \text { and } x=3\) putting the value of \(x\) in equation (ii) we get intersection point \((0,0)\) and \((3,-3)\) So, required area, \(A=\int_{0}^{3}\left(2 x-x^{2}\right) d x-\int_{0}^{3}(-x) d x\) \(A=\int_{0}^{3}\left(2 x-x^{2}+x\right) d x=\int_{0}^{3}\left(3 x-x^{2}\right) d x\) \(A=\left[\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3}=\left[\frac{27}{2}-\frac{27}{3}\right]=\frac{27}{6}=\frac{9}{2}\) sq. units
AMU-2017
Application of the Integrals
86839
The area bounded by \(y=\sin ^{-1} x, x=\frac{1}{\sqrt{2}}\) and \(\mathrm{X}\)-axis is
1 \(\frac{1}{\sqrt{2}}+1\)
2 \(1-\frac{1}{\sqrt{2}}\)
3 \(\frac{\pi}{4 \sqrt{2}}\)
4 \(\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
Explanation:
(D) : Given that, \(\mathrm{y}=\sin ^{-1} \mathrm{x} \quad \mathrm{x}=\frac{1}{\sqrt{2}}\) From \(y=\sin ^{-1}(x)\), at \(x=0, \quad y=0\) and at \(x=\frac{1}{\sqrt{2}}, y=\) \(\frac{\pi}{4}\) at \(x=1, \quad y=\pi / 2\) \(\because \mathrm{y}=\sin ^{-1} \mathrm{x}\) \(\therefore \mathrm{x}=\sin \mathrm{y}\) Now as per the figure area of shaded region- \(\mathrm{A}=\) area of rectangle \(\mathrm{OABC}-\) area of curve \(\mathrm{OBAO}\) \(\mathrm{A}=\left(\frac{1}{\sqrt{2}} \times \frac{\pi}{4}\right)-\int_{0}^{\pi / 4} \sin \mathrm{ydy}\) \(\mathrm{A}=\frac{\pi}{4 \sqrt{2}}+[\cos \mathrm{y}]_{0}^{\pi / 4}=\frac{\pi}{4 \sqrt{2}}+\left[\cos \frac{\pi}{4}-\cos 0^{\circ}\right]\) \(A=\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
BCECE-2014
Application of the Integrals
86840
The volume of the solid formed by rotating the area enclosed between the curve \(y=x^{2}\) and the line \(y=1\) about \(y=1\) is (in cubic unit):
1 \(9 \pi / 5\)
2 \(2 \pi / 5\)
3 \(8 \pi / 3\)
4 \(7 \pi / 5\)
Explanation:
(B) : The curve is \(y=x^{2}\) Given line, \(\quad y=1\) and \(y=1\) Putting the value of \(y\) in equation (i) \(y=x^{2} \Rightarrow \quad x^{2}=1\) \(x= \pm 1\) As per the figure required volume, \(V=\pi \int_{-1}^{1} y^{2} d x \Rightarrow V=2 \pi \int_{0}^{1} x^{4} d x\) \(\left(\because \int_{-9}^{9} f(x) d x=2 \int_{-9}^{9} f(x) d x\right)\) \(V=2 \pi \times\left(\frac{1^{5}-0^{5}}{5}\right)=\frac{2 \pi}{5}\)
[JCECE-2004]
Application of the Integrals
86841
The area bounded by the curve \(y=\sin x\) between \(x=0\) and \(x=2 \pi\) is
1 1 sq unit
2 2 sq unit
3 4 sq unit
4 8 sq unit
Explanation:
(C): The curve, \(\quad \mathrm{y}=\sin \mathrm{x} \quad \ldots .\). (i) From equation (i)- at. \(x=0, y=0\), at \(x=\frac{\pi}{2}, y=1\) at \(\mathrm{x}=\pi, \mathrm{y}=0, \quad\) at \(\mathrm{x}=\frac{3 \pi}{2}, \mathrm{y}=1\) So, as per the figure required area, \(\mathrm{A}=\) area of (OBAO \(+\mathrm{BDCB})\) \(\therefore \quad \mathrm{A}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{dx}+\left|\int_{\pi}^{2 \pi} \sin \mathrm{x} \mathrm{dx}\right|\) \(\mathrm{A}=[-\cos \mathrm{x}]_{0}^{\pi}+\left|[-\cos \mathrm{x}]_{\pi}^{2 \pi}\right|\) \(A=\left[-\cos \pi+\cos 0^{\circ}\right]+\mid[-\cos 2 \pi+\cos \pi]\) \(\mathrm{A}=1+1+|-1-1| \Rightarrow \mathrm{A}=4\) sq units
86838
The area bounded by the curve \(y=2 x-x^{2}\) and the straight line \(y=-x\) is given by:
1 \(\frac{9}{2}\) squnit
2 \(\frac{43}{6}\) sq unit
3 \(\frac{35}{6}\) sq unit
4 none of these
Explanation:
(A) : Given that, \(y=2 x-x^{2}\) \(y=-x \tag{i}\) Solving equation (i) and (ii)- \(-x=2 x-x^{2} \Rightarrow \quad x^{2}-3 x=0 \tag{ii}\) \(x(x-3)=0\) \(x=0 \text { and } x=3\) putting the value of \(x\) in equation (ii) we get intersection point \((0,0)\) and \((3,-3)\) So, required area, \(A=\int_{0}^{3}\left(2 x-x^{2}\right) d x-\int_{0}^{3}(-x) d x\) \(A=\int_{0}^{3}\left(2 x-x^{2}+x\right) d x=\int_{0}^{3}\left(3 x-x^{2}\right) d x\) \(A=\left[\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3}=\left[\frac{27}{2}-\frac{27}{3}\right]=\frac{27}{6}=\frac{9}{2}\) sq. units
AMU-2017
Application of the Integrals
86839
The area bounded by \(y=\sin ^{-1} x, x=\frac{1}{\sqrt{2}}\) and \(\mathrm{X}\)-axis is
1 \(\frac{1}{\sqrt{2}}+1\)
2 \(1-\frac{1}{\sqrt{2}}\)
3 \(\frac{\pi}{4 \sqrt{2}}\)
4 \(\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
Explanation:
(D) : Given that, \(\mathrm{y}=\sin ^{-1} \mathrm{x} \quad \mathrm{x}=\frac{1}{\sqrt{2}}\) From \(y=\sin ^{-1}(x)\), at \(x=0, \quad y=0\) and at \(x=\frac{1}{\sqrt{2}}, y=\) \(\frac{\pi}{4}\) at \(x=1, \quad y=\pi / 2\) \(\because \mathrm{y}=\sin ^{-1} \mathrm{x}\) \(\therefore \mathrm{x}=\sin \mathrm{y}\) Now as per the figure area of shaded region- \(\mathrm{A}=\) area of rectangle \(\mathrm{OABC}-\) area of curve \(\mathrm{OBAO}\) \(\mathrm{A}=\left(\frac{1}{\sqrt{2}} \times \frac{\pi}{4}\right)-\int_{0}^{\pi / 4} \sin \mathrm{ydy}\) \(\mathrm{A}=\frac{\pi}{4 \sqrt{2}}+[\cos \mathrm{y}]_{0}^{\pi / 4}=\frac{\pi}{4 \sqrt{2}}+\left[\cos \frac{\pi}{4}-\cos 0^{\circ}\right]\) \(A=\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
BCECE-2014
Application of the Integrals
86840
The volume of the solid formed by rotating the area enclosed between the curve \(y=x^{2}\) and the line \(y=1\) about \(y=1\) is (in cubic unit):
1 \(9 \pi / 5\)
2 \(2 \pi / 5\)
3 \(8 \pi / 3\)
4 \(7 \pi / 5\)
Explanation:
(B) : The curve is \(y=x^{2}\) Given line, \(\quad y=1\) and \(y=1\) Putting the value of \(y\) in equation (i) \(y=x^{2} \Rightarrow \quad x^{2}=1\) \(x= \pm 1\) As per the figure required volume, \(V=\pi \int_{-1}^{1} y^{2} d x \Rightarrow V=2 \pi \int_{0}^{1} x^{4} d x\) \(\left(\because \int_{-9}^{9} f(x) d x=2 \int_{-9}^{9} f(x) d x\right)\) \(V=2 \pi \times\left(\frac{1^{5}-0^{5}}{5}\right)=\frac{2 \pi}{5}\)
[JCECE-2004]
Application of the Integrals
86841
The area bounded by the curve \(y=\sin x\) between \(x=0\) and \(x=2 \pi\) is
1 1 sq unit
2 2 sq unit
3 4 sq unit
4 8 sq unit
Explanation:
(C): The curve, \(\quad \mathrm{y}=\sin \mathrm{x} \quad \ldots .\). (i) From equation (i)- at. \(x=0, y=0\), at \(x=\frac{\pi}{2}, y=1\) at \(\mathrm{x}=\pi, \mathrm{y}=0, \quad\) at \(\mathrm{x}=\frac{3 \pi}{2}, \mathrm{y}=1\) So, as per the figure required area, \(\mathrm{A}=\) area of (OBAO \(+\mathrm{BDCB})\) \(\therefore \quad \mathrm{A}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{dx}+\left|\int_{\pi}^{2 \pi} \sin \mathrm{x} \mathrm{dx}\right|\) \(\mathrm{A}=[-\cos \mathrm{x}]_{0}^{\pi}+\left|[-\cos \mathrm{x}]_{\pi}^{2 \pi}\right|\) \(A=\left[-\cos \pi+\cos 0^{\circ}\right]+\mid[-\cos 2 \pi+\cos \pi]\) \(\mathrm{A}=1+1+|-1-1| \Rightarrow \mathrm{A}=4\) sq units
86838
The area bounded by the curve \(y=2 x-x^{2}\) and the straight line \(y=-x\) is given by:
1 \(\frac{9}{2}\) squnit
2 \(\frac{43}{6}\) sq unit
3 \(\frac{35}{6}\) sq unit
4 none of these
Explanation:
(A) : Given that, \(y=2 x-x^{2}\) \(y=-x \tag{i}\) Solving equation (i) and (ii)- \(-x=2 x-x^{2} \Rightarrow \quad x^{2}-3 x=0 \tag{ii}\) \(x(x-3)=0\) \(x=0 \text { and } x=3\) putting the value of \(x\) in equation (ii) we get intersection point \((0,0)\) and \((3,-3)\) So, required area, \(A=\int_{0}^{3}\left(2 x-x^{2}\right) d x-\int_{0}^{3}(-x) d x\) \(A=\int_{0}^{3}\left(2 x-x^{2}+x\right) d x=\int_{0}^{3}\left(3 x-x^{2}\right) d x\) \(A=\left[\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3}=\left[\frac{27}{2}-\frac{27}{3}\right]=\frac{27}{6}=\frac{9}{2}\) sq. units
AMU-2017
Application of the Integrals
86839
The area bounded by \(y=\sin ^{-1} x, x=\frac{1}{\sqrt{2}}\) and \(\mathrm{X}\)-axis is
1 \(\frac{1}{\sqrt{2}}+1\)
2 \(1-\frac{1}{\sqrt{2}}\)
3 \(\frac{\pi}{4 \sqrt{2}}\)
4 \(\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
Explanation:
(D) : Given that, \(\mathrm{y}=\sin ^{-1} \mathrm{x} \quad \mathrm{x}=\frac{1}{\sqrt{2}}\) From \(y=\sin ^{-1}(x)\), at \(x=0, \quad y=0\) and at \(x=\frac{1}{\sqrt{2}}, y=\) \(\frac{\pi}{4}\) at \(x=1, \quad y=\pi / 2\) \(\because \mathrm{y}=\sin ^{-1} \mathrm{x}\) \(\therefore \mathrm{x}=\sin \mathrm{y}\) Now as per the figure area of shaded region- \(\mathrm{A}=\) area of rectangle \(\mathrm{OABC}-\) area of curve \(\mathrm{OBAO}\) \(\mathrm{A}=\left(\frac{1}{\sqrt{2}} \times \frac{\pi}{4}\right)-\int_{0}^{\pi / 4} \sin \mathrm{ydy}\) \(\mathrm{A}=\frac{\pi}{4 \sqrt{2}}+[\cos \mathrm{y}]_{0}^{\pi / 4}=\frac{\pi}{4 \sqrt{2}}+\left[\cos \frac{\pi}{4}-\cos 0^{\circ}\right]\) \(A=\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1\)
BCECE-2014
Application of the Integrals
86840
The volume of the solid formed by rotating the area enclosed between the curve \(y=x^{2}\) and the line \(y=1\) about \(y=1\) is (in cubic unit):
1 \(9 \pi / 5\)
2 \(2 \pi / 5\)
3 \(8 \pi / 3\)
4 \(7 \pi / 5\)
Explanation:
(B) : The curve is \(y=x^{2}\) Given line, \(\quad y=1\) and \(y=1\) Putting the value of \(y\) in equation (i) \(y=x^{2} \Rightarrow \quad x^{2}=1\) \(x= \pm 1\) As per the figure required volume, \(V=\pi \int_{-1}^{1} y^{2} d x \Rightarrow V=2 \pi \int_{0}^{1} x^{4} d x\) \(\left(\because \int_{-9}^{9} f(x) d x=2 \int_{-9}^{9} f(x) d x\right)\) \(V=2 \pi \times\left(\frac{1^{5}-0^{5}}{5}\right)=\frac{2 \pi}{5}\)
[JCECE-2004]
Application of the Integrals
86841
The area bounded by the curve \(y=\sin x\) between \(x=0\) and \(x=2 \pi\) is
1 1 sq unit
2 2 sq unit
3 4 sq unit
4 8 sq unit
Explanation:
(C): The curve, \(\quad \mathrm{y}=\sin \mathrm{x} \quad \ldots .\). (i) From equation (i)- at. \(x=0, y=0\), at \(x=\frac{\pi}{2}, y=1\) at \(\mathrm{x}=\pi, \mathrm{y}=0, \quad\) at \(\mathrm{x}=\frac{3 \pi}{2}, \mathrm{y}=1\) So, as per the figure required area, \(\mathrm{A}=\) area of (OBAO \(+\mathrm{BDCB})\) \(\therefore \quad \mathrm{A}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{dx}+\left|\int_{\pi}^{2 \pi} \sin \mathrm{x} \mathrm{dx}\right|\) \(\mathrm{A}=[-\cos \mathrm{x}]_{0}^{\pi}+\left|[-\cos \mathrm{x}]_{\pi}^{2 \pi}\right|\) \(A=\left[-\cos \pi+\cos 0^{\circ}\right]+\mid[-\cos 2 \pi+\cos \pi]\) \(\mathrm{A}=1+1+|-1-1| \Rightarrow \mathrm{A}=4\) sq units
