86824
The area under the curve \(y=|\cos x-\sin x|\), \(0 \leq \mathrm{x} \leq \frac{\boldsymbol{\pi}}{\mathbf{2}}\), and above \(\mathrm{x}\)-axis is :
1 \(2 \sqrt{2}\)
2 \(2 \sqrt{2}-2\)
3 \(2 \sqrt{2}+2\)
4 0
Explanation:
(B) : Given, \(\mathrm{y}=|\cos \mathrm{x}-\sin \mathrm{x}| \quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) Required area \(=2 \int_{0}^{\pi / 4}(\cos x-\sin x) d x=2[\sin x+\cos x]_{0}^{\pi / 4}\) \(=2\left[\frac{2}{\sqrt{2}}-1\right]=2\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]=\sqrt{2}(2-\sqrt{2})\) Area \(=(2 \sqrt{2}-2)\) sq. unit
BITSAT-2017
Application of the Integrals
86825
The area of the region bounded by the curve \(x\) \(=2 y+3\) and lines \(y=1\) and \(y=-1\) is
1 4 sq. units
2 \(\frac{3}{2}\) sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(C) : Given, \(x=2 y+3\), Required area = Area of shaded region \(=\int_{-1}^{1}(2 y+3) d y \quad=\left[y^{2}+3 y\right]_{-1}^{1}\) \(=[(1+3)-(1-3)]\) \(=[4-1+3]=6\) sq. units.
VITEEE-2018
Application of the Integrals
86906
The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2}\) and \(x=\) \(\pi\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{8}\)
4 \(\frac{\pi}{16}\)
5 \(2 \pi\)
Explanation:
(B) : The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2} \text { and } x=\pi\) According to question - Required area \(=\int_{\frac{\pi}{2}}^{\pi} \sin ^{2} x d x\) \(=\int_{\frac{\pi}{2}}^{\pi}\left[\frac{1-\cos 2 x}{2}\right] d x=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi}(1-\cos 2 x) d x\) \(=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{\frac{\pi}{2}}^{\pi}=\frac{1}{2}\left[\frac{\pi}{2}\right]=\frac{\pi}{4}\)
Kerala CEE-2018
Application of the Integrals
86826
The area bounded by \(f(x)=x^{2}, 0 \leq x \leq 1, g(x)=\) \(-x+2,1 \leq x \leq 2\) and \(x-\) axis is
1 \(\frac{3}{2}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 None of these
Explanation:
(D) : Required area \(=\) Area of \(\mathrm{OAB}+\) Area of \(\mathrm{ABC}\) Now, Area of OAB \(=\int_{0}^{1} f(x) d x+\int_{1}^{2} g(x) d x\) \(=\int_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}+\left[\frac{-\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{1}^{2}=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}+2\right)\right]\) \(=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\) sq unit
VITEEE-2017
Application of the Integrals
86828
The area bounded by the curve \(y=|\sin x|, x\)-axis and the lines \(|\mathbf{x}|=\pi\). is
1 2 sq unit
2 1 sq unit
3 4 sq unit
4 None of these
Explanation:
(C) : Required area \(=\) Shaded area \(=\int_{-\pi}^{\pi}|\sin \mathrm{x}| \mathrm{dx}=2 \int_{0}^{\pi}|\sin \mathrm{x}| \mathrm{dx}\) \(=-2[\cos x]_{0}^{\pi}=-2(\cos \pi-\cos 0)=4\) sq units
86824
The area under the curve \(y=|\cos x-\sin x|\), \(0 \leq \mathrm{x} \leq \frac{\boldsymbol{\pi}}{\mathbf{2}}\), and above \(\mathrm{x}\)-axis is :
1 \(2 \sqrt{2}\)
2 \(2 \sqrt{2}-2\)
3 \(2 \sqrt{2}+2\)
4 0
Explanation:
(B) : Given, \(\mathrm{y}=|\cos \mathrm{x}-\sin \mathrm{x}| \quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) Required area \(=2 \int_{0}^{\pi / 4}(\cos x-\sin x) d x=2[\sin x+\cos x]_{0}^{\pi / 4}\) \(=2\left[\frac{2}{\sqrt{2}}-1\right]=2\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]=\sqrt{2}(2-\sqrt{2})\) Area \(=(2 \sqrt{2}-2)\) sq. unit
BITSAT-2017
Application of the Integrals
86825
The area of the region bounded by the curve \(x\) \(=2 y+3\) and lines \(y=1\) and \(y=-1\) is
1 4 sq. units
2 \(\frac{3}{2}\) sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(C) : Given, \(x=2 y+3\), Required area = Area of shaded region \(=\int_{-1}^{1}(2 y+3) d y \quad=\left[y^{2}+3 y\right]_{-1}^{1}\) \(=[(1+3)-(1-3)]\) \(=[4-1+3]=6\) sq. units.
VITEEE-2018
Application of the Integrals
86906
The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2}\) and \(x=\) \(\pi\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{8}\)
4 \(\frac{\pi}{16}\)
5 \(2 \pi\)
Explanation:
(B) : The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2} \text { and } x=\pi\) According to question - Required area \(=\int_{\frac{\pi}{2}}^{\pi} \sin ^{2} x d x\) \(=\int_{\frac{\pi}{2}}^{\pi}\left[\frac{1-\cos 2 x}{2}\right] d x=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi}(1-\cos 2 x) d x\) \(=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{\frac{\pi}{2}}^{\pi}=\frac{1}{2}\left[\frac{\pi}{2}\right]=\frac{\pi}{4}\)
Kerala CEE-2018
Application of the Integrals
86826
The area bounded by \(f(x)=x^{2}, 0 \leq x \leq 1, g(x)=\) \(-x+2,1 \leq x \leq 2\) and \(x-\) axis is
1 \(\frac{3}{2}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 None of these
Explanation:
(D) : Required area \(=\) Area of \(\mathrm{OAB}+\) Area of \(\mathrm{ABC}\) Now, Area of OAB \(=\int_{0}^{1} f(x) d x+\int_{1}^{2} g(x) d x\) \(=\int_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}+\left[\frac{-\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{1}^{2}=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}+2\right)\right]\) \(=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\) sq unit
VITEEE-2017
Application of the Integrals
86828
The area bounded by the curve \(y=|\sin x|, x\)-axis and the lines \(|\mathbf{x}|=\pi\). is
1 2 sq unit
2 1 sq unit
3 4 sq unit
4 None of these
Explanation:
(C) : Required area \(=\) Shaded area \(=\int_{-\pi}^{\pi}|\sin \mathrm{x}| \mathrm{dx}=2 \int_{0}^{\pi}|\sin \mathrm{x}| \mathrm{dx}\) \(=-2[\cos x]_{0}^{\pi}=-2(\cos \pi-\cos 0)=4\) sq units
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Application of the Integrals
86824
The area under the curve \(y=|\cos x-\sin x|\), \(0 \leq \mathrm{x} \leq \frac{\boldsymbol{\pi}}{\mathbf{2}}\), and above \(\mathrm{x}\)-axis is :
1 \(2 \sqrt{2}\)
2 \(2 \sqrt{2}-2\)
3 \(2 \sqrt{2}+2\)
4 0
Explanation:
(B) : Given, \(\mathrm{y}=|\cos \mathrm{x}-\sin \mathrm{x}| \quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) Required area \(=2 \int_{0}^{\pi / 4}(\cos x-\sin x) d x=2[\sin x+\cos x]_{0}^{\pi / 4}\) \(=2\left[\frac{2}{\sqrt{2}}-1\right]=2\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]=\sqrt{2}(2-\sqrt{2})\) Area \(=(2 \sqrt{2}-2)\) sq. unit
BITSAT-2017
Application of the Integrals
86825
The area of the region bounded by the curve \(x\) \(=2 y+3\) and lines \(y=1\) and \(y=-1\) is
1 4 sq. units
2 \(\frac{3}{2}\) sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(C) : Given, \(x=2 y+3\), Required area = Area of shaded region \(=\int_{-1}^{1}(2 y+3) d y \quad=\left[y^{2}+3 y\right]_{-1}^{1}\) \(=[(1+3)-(1-3)]\) \(=[4-1+3]=6\) sq. units.
VITEEE-2018
Application of the Integrals
86906
The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2}\) and \(x=\) \(\pi\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{8}\)
4 \(\frac{\pi}{16}\)
5 \(2 \pi\)
Explanation:
(B) : The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2} \text { and } x=\pi\) According to question - Required area \(=\int_{\frac{\pi}{2}}^{\pi} \sin ^{2} x d x\) \(=\int_{\frac{\pi}{2}}^{\pi}\left[\frac{1-\cos 2 x}{2}\right] d x=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi}(1-\cos 2 x) d x\) \(=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{\frac{\pi}{2}}^{\pi}=\frac{1}{2}\left[\frac{\pi}{2}\right]=\frac{\pi}{4}\)
Kerala CEE-2018
Application of the Integrals
86826
The area bounded by \(f(x)=x^{2}, 0 \leq x \leq 1, g(x)=\) \(-x+2,1 \leq x \leq 2\) and \(x-\) axis is
1 \(\frac{3}{2}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 None of these
Explanation:
(D) : Required area \(=\) Area of \(\mathrm{OAB}+\) Area of \(\mathrm{ABC}\) Now, Area of OAB \(=\int_{0}^{1} f(x) d x+\int_{1}^{2} g(x) d x\) \(=\int_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}+\left[\frac{-\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{1}^{2}=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}+2\right)\right]\) \(=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\) sq unit
VITEEE-2017
Application of the Integrals
86828
The area bounded by the curve \(y=|\sin x|, x\)-axis and the lines \(|\mathbf{x}|=\pi\). is
1 2 sq unit
2 1 sq unit
3 4 sq unit
4 None of these
Explanation:
(C) : Required area \(=\) Shaded area \(=\int_{-\pi}^{\pi}|\sin \mathrm{x}| \mathrm{dx}=2 \int_{0}^{\pi}|\sin \mathrm{x}| \mathrm{dx}\) \(=-2[\cos x]_{0}^{\pi}=-2(\cos \pi-\cos 0)=4\) sq units
86824
The area under the curve \(y=|\cos x-\sin x|\), \(0 \leq \mathrm{x} \leq \frac{\boldsymbol{\pi}}{\mathbf{2}}\), and above \(\mathrm{x}\)-axis is :
1 \(2 \sqrt{2}\)
2 \(2 \sqrt{2}-2\)
3 \(2 \sqrt{2}+2\)
4 0
Explanation:
(B) : Given, \(\mathrm{y}=|\cos \mathrm{x}-\sin \mathrm{x}| \quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) Required area \(=2 \int_{0}^{\pi / 4}(\cos x-\sin x) d x=2[\sin x+\cos x]_{0}^{\pi / 4}\) \(=2\left[\frac{2}{\sqrt{2}}-1\right]=2\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]=\sqrt{2}(2-\sqrt{2})\) Area \(=(2 \sqrt{2}-2)\) sq. unit
BITSAT-2017
Application of the Integrals
86825
The area of the region bounded by the curve \(x\) \(=2 y+3\) and lines \(y=1\) and \(y=-1\) is
1 4 sq. units
2 \(\frac{3}{2}\) sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(C) : Given, \(x=2 y+3\), Required area = Area of shaded region \(=\int_{-1}^{1}(2 y+3) d y \quad=\left[y^{2}+3 y\right]_{-1}^{1}\) \(=[(1+3)-(1-3)]\) \(=[4-1+3]=6\) sq. units.
VITEEE-2018
Application of the Integrals
86906
The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2}\) and \(x=\) \(\pi\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{8}\)
4 \(\frac{\pi}{16}\)
5 \(2 \pi\)
Explanation:
(B) : The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2} \text { and } x=\pi\) According to question - Required area \(=\int_{\frac{\pi}{2}}^{\pi} \sin ^{2} x d x\) \(=\int_{\frac{\pi}{2}}^{\pi}\left[\frac{1-\cos 2 x}{2}\right] d x=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi}(1-\cos 2 x) d x\) \(=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{\frac{\pi}{2}}^{\pi}=\frac{1}{2}\left[\frac{\pi}{2}\right]=\frac{\pi}{4}\)
Kerala CEE-2018
Application of the Integrals
86826
The area bounded by \(f(x)=x^{2}, 0 \leq x \leq 1, g(x)=\) \(-x+2,1 \leq x \leq 2\) and \(x-\) axis is
1 \(\frac{3}{2}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 None of these
Explanation:
(D) : Required area \(=\) Area of \(\mathrm{OAB}+\) Area of \(\mathrm{ABC}\) Now, Area of OAB \(=\int_{0}^{1} f(x) d x+\int_{1}^{2} g(x) d x\) \(=\int_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}+\left[\frac{-\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{1}^{2}=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}+2\right)\right]\) \(=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\) sq unit
VITEEE-2017
Application of the Integrals
86828
The area bounded by the curve \(y=|\sin x|, x\)-axis and the lines \(|\mathbf{x}|=\pi\). is
1 2 sq unit
2 1 sq unit
3 4 sq unit
4 None of these
Explanation:
(C) : Required area \(=\) Shaded area \(=\int_{-\pi}^{\pi}|\sin \mathrm{x}| \mathrm{dx}=2 \int_{0}^{\pi}|\sin \mathrm{x}| \mathrm{dx}\) \(=-2[\cos x]_{0}^{\pi}=-2(\cos \pi-\cos 0)=4\) sq units
86824
The area under the curve \(y=|\cos x-\sin x|\), \(0 \leq \mathrm{x} \leq \frac{\boldsymbol{\pi}}{\mathbf{2}}\), and above \(\mathrm{x}\)-axis is :
1 \(2 \sqrt{2}\)
2 \(2 \sqrt{2}-2\)
3 \(2 \sqrt{2}+2\)
4 0
Explanation:
(B) : Given, \(\mathrm{y}=|\cos \mathrm{x}-\sin \mathrm{x}| \quad 0 \leq \mathrm{x} \leq \frac{\pi}{2}\) Required area \(=2 \int_{0}^{\pi / 4}(\cos x-\sin x) d x=2[\sin x+\cos x]_{0}^{\pi / 4}\) \(=2\left[\frac{2}{\sqrt{2}}-1\right]=2\left[\frac{2-\sqrt{2}}{\sqrt{2}}\right]=\sqrt{2}(2-\sqrt{2})\) Area \(=(2 \sqrt{2}-2)\) sq. unit
BITSAT-2017
Application of the Integrals
86825
The area of the region bounded by the curve \(x\) \(=2 y+3\) and lines \(y=1\) and \(y=-1\) is
1 4 sq. units
2 \(\frac{3}{2}\) sq. units
3 6 sq. units
4 8 sq. units
Explanation:
(C) : Given, \(x=2 y+3\), Required area = Area of shaded region \(=\int_{-1}^{1}(2 y+3) d y \quad=\left[y^{2}+3 y\right]_{-1}^{1}\) \(=[(1+3)-(1-3)]\) \(=[4-1+3]=6\) sq. units.
VITEEE-2018
Application of the Integrals
86906
The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2}\) and \(x=\) \(\pi\) is
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{8}\)
4 \(\frac{\pi}{16}\)
5 \(2 \pi\)
Explanation:
(B) : The area bounded by \(y=\sin ^{2} x, x=\frac{\pi}{2} \text { and } x=\pi\) According to question - Required area \(=\int_{\frac{\pi}{2}}^{\pi} \sin ^{2} x d x\) \(=\int_{\frac{\pi}{2}}^{\pi}\left[\frac{1-\cos 2 x}{2}\right] d x=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi}(1-\cos 2 x) d x\) \(=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{\frac{\pi}{2}}^{\pi}=\frac{1}{2}\left[\frac{\pi}{2}\right]=\frac{\pi}{4}\)
Kerala CEE-2018
Application of the Integrals
86826
The area bounded by \(f(x)=x^{2}, 0 \leq x \leq 1, g(x)=\) \(-x+2,1 \leq x \leq 2\) and \(x-\) axis is
1 \(\frac{3}{2}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 None of these
Explanation:
(D) : Required area \(=\) Area of \(\mathrm{OAB}+\) Area of \(\mathrm{ABC}\) Now, Area of OAB \(=\int_{0}^{1} f(x) d x+\int_{1}^{2} g(x) d x\) \(=\int_{0}^{1} x^{2} d x+\int_{1}^{2}(-x+2) d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}+\left[\frac{-\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{1}^{2}=\frac{1}{3}+\left[\left(\frac{-4}{2}+4\right)-\left(\frac{-1}{2}+2\right)\right]\) \(=\frac{1}{3}+\left[(-2+4)-\left(\frac{3}{2}\right)\right]=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\) sq unit
VITEEE-2017
Application of the Integrals
86828
The area bounded by the curve \(y=|\sin x|, x\)-axis and the lines \(|\mathbf{x}|=\pi\). is
1 2 sq unit
2 1 sq unit
3 4 sq unit
4 None of these
Explanation:
(C) : Required area \(=\) Shaded area \(=\int_{-\pi}^{\pi}|\sin \mathrm{x}| \mathrm{dx}=2 \int_{0}^{\pi}|\sin \mathrm{x}| \mathrm{dx}\) \(=-2[\cos x]_{0}^{\pi}=-2(\cos \pi-\cos 0)=4\) sq units
