86877
If the area (in sq. units) of the region \(\left\{(x, y): y^{2}\right.\) \(\leq 4 x, x+y \leq 1, x \leq 0, y \geq 0\}\) is \(a \sqrt{2}+b\) then a \(b\) is equal to
1 \(\frac{10}{3}\)
2 6
3 \(\frac{8}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
(B) : Given, Curve, \(C_{1}: y^{2} \leq 4 x\) \(\mathrm{C}_{2}: \mathrm{x}+\mathrm{y} \leq 1\) \(\mathrm{x}, \mathrm{y} \geq 0\) Point of intersection \(P\) these of two curve, \(y^{2}=4 x, x+y=1\) \(y^{2}-4(1-y)=0\) \(y^{2}+4 y-4=0\) \(y=\frac{-4 \pm \sqrt{16-4(-4)}}{2}=-2 \pm 2 \sqrt{2}\) \(P:(3-2 \sqrt{2},-2+2 \sqrt{2}), O:(0,0) R:(1,0)\) Required area \(=\int_{0}^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\int_{3-2 \sqrt{2}}^{1}(1-x) d x\) \(=\frac{4}{3}\left|x^{\frac{3}{2}}\right|_{0}^{3-2 \sqrt{2}}+\left|x-\frac{x^{2}}{2}\right|_{3-2 \sqrt{2}}^{1}=\frac{8 \sqrt{2}}{3}-\frac{10}{3}=a \sqrt{2}+b\) \(\mathrm{a}=\frac{8}{3}, \mathrm{~b}=-\frac{10}{3}\) \(\therefore \quad \mathrm{a}-\mathrm{b}=\frac{8+10}{3}=6\)
JEE Main-2019-12.04.2019
Application of the Integrals
86878
The area (in sq units) of the region bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{7}{8}\)
2 \(\frac{9}{8}\)
3 \(\frac{5}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(B) : Given, Curve, \(x^{2}=4 y\) and, Straight line \(x=4 y-2\) Point of intersection of the given parabola and line are, \(\left(-1, \frac{1}{4}\right) \text { and }(2,1)\) Required area \(=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x\) \(=\int_{-1}^{2}\left(\frac{x}{4}+\frac{1}{2}-\frac{x^{2}}{4}\right) d x\) \(=\frac{\mathrm{x}^{2}}{8}+\frac{\mathrm{x}}{2}-\left.\frac{\mathrm{x}^{3}}{12}\right|_{-1} ^{2}=\left(\frac{4}{8}+\frac{2}{2}-\frac{8}{12}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)\) \(=\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{3-12+2}{24}\right)\) \(=\left(\frac{3}{2}-\frac{2}{3}\right)-\left(\frac{-7}{24}\right)=\left(\frac{9-4}{6}\right)-\left(\frac{-7}{24}\right)\) \(=\frac{5}{6}+\frac{7}{24}=\frac{20+7}{24}=\frac{27}{24}=\frac{9}{8}\) square unit
JEE Main-2020-11.01.2020
Application of the Integrals
86879
The area of the region bounded by \(y-x=2\) and \(x^{2}=y\) is equal to
1 \(\frac{16}{3}\)
2 \(\frac{2}{3}\)
3 \(\frac{9}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(C) : : Given \(\mathrm{y}-\mathrm{x}=2\) and \(\mathrm{x}^{2}=\mathrm{y}\) Now, \(\mathrm{x}^{2}=2+\mathrm{x}\) \(\mathrm{x}^{2}-\mathrm{x}-2=0\) \((\mathrm{x}+1)(\mathrm{x}-2)=0\) \(\mathrm{x}=-1,2\) Area \(=\int_{-1}^{2}\left(2+x-x^{2}\right)\) \(=\left[2 x+\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-1}^{2}=\left(4+2-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right)\) \(=6-3+2-\frac{1}{2}=\frac{9}{2}\)
JEE Main-2021-27.07.2021
Application of the Integrals
86880
The area bounded by the curve \(y=\sin ^{-1} x\) and the line \(x=0,|y|=\frac{\pi}{2}\) is
1 1
2 2
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(B) : Given curve is, \(\mathrm{y}=\sin ^{-1} \mathrm{x}\) and \(\mathrm{y}-\) axis between \(\mathrm{y}=0\) and \(\mathrm{y}=\frac{\pi}{2}\) Hence, the required area \(\mathrm{A}=\int_{0}^{\pi / 2} 2 \sin \mathrm{ydy}=2 \int_{0}^{\pi / 2} \sin \mathrm{y} \mathrm{dy}=2[-\cos \mathrm{y}]_{0}^{\pi / 2}\) \(=-2\left[\cos \frac{\pi}{2}-\cos 0\right]=-2[0-1]=2\) square unit
Jamia Millia Islamia-2011
Application of the Integrals
86881
The area between \(y^{2}+4 x-8=0\), the \(x\)-axis and the line \(x=1\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{2}{3}\) squnit
3 \(\frac{1}{3}\) sq unit
4 1 sq unit
Explanation:
(A) : Given, \(\mathrm{y}^{2}+4 \mathrm{x}-8=0\) \(y^{2}=-4(x-2)\) \(\therefore\) Required area \(=2 \int_{1}^{2} y d x=\int_{1}^{2} \sqrt{8-4 x} d x\) \(=\left[\frac{(8-4 x)^{3 / 2}}{\frac{3}{2}(-4)}\right]_{1}^{2}=-\frac{1}{6}(0-8)=\frac{4}{3}\) sq unit
86877
If the area (in sq. units) of the region \(\left\{(x, y): y^{2}\right.\) \(\leq 4 x, x+y \leq 1, x \leq 0, y \geq 0\}\) is \(a \sqrt{2}+b\) then a \(b\) is equal to
1 \(\frac{10}{3}\)
2 6
3 \(\frac{8}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
(B) : Given, Curve, \(C_{1}: y^{2} \leq 4 x\) \(\mathrm{C}_{2}: \mathrm{x}+\mathrm{y} \leq 1\) \(\mathrm{x}, \mathrm{y} \geq 0\) Point of intersection \(P\) these of two curve, \(y^{2}=4 x, x+y=1\) \(y^{2}-4(1-y)=0\) \(y^{2}+4 y-4=0\) \(y=\frac{-4 \pm \sqrt{16-4(-4)}}{2}=-2 \pm 2 \sqrt{2}\) \(P:(3-2 \sqrt{2},-2+2 \sqrt{2}), O:(0,0) R:(1,0)\) Required area \(=\int_{0}^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\int_{3-2 \sqrt{2}}^{1}(1-x) d x\) \(=\frac{4}{3}\left|x^{\frac{3}{2}}\right|_{0}^{3-2 \sqrt{2}}+\left|x-\frac{x^{2}}{2}\right|_{3-2 \sqrt{2}}^{1}=\frac{8 \sqrt{2}}{3}-\frac{10}{3}=a \sqrt{2}+b\) \(\mathrm{a}=\frac{8}{3}, \mathrm{~b}=-\frac{10}{3}\) \(\therefore \quad \mathrm{a}-\mathrm{b}=\frac{8+10}{3}=6\)
JEE Main-2019-12.04.2019
Application of the Integrals
86878
The area (in sq units) of the region bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{7}{8}\)
2 \(\frac{9}{8}\)
3 \(\frac{5}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(B) : Given, Curve, \(x^{2}=4 y\) and, Straight line \(x=4 y-2\) Point of intersection of the given parabola and line are, \(\left(-1, \frac{1}{4}\right) \text { and }(2,1)\) Required area \(=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x\) \(=\int_{-1}^{2}\left(\frac{x}{4}+\frac{1}{2}-\frac{x^{2}}{4}\right) d x\) \(=\frac{\mathrm{x}^{2}}{8}+\frac{\mathrm{x}}{2}-\left.\frac{\mathrm{x}^{3}}{12}\right|_{-1} ^{2}=\left(\frac{4}{8}+\frac{2}{2}-\frac{8}{12}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)\) \(=\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{3-12+2}{24}\right)\) \(=\left(\frac{3}{2}-\frac{2}{3}\right)-\left(\frac{-7}{24}\right)=\left(\frac{9-4}{6}\right)-\left(\frac{-7}{24}\right)\) \(=\frac{5}{6}+\frac{7}{24}=\frac{20+7}{24}=\frac{27}{24}=\frac{9}{8}\) square unit
JEE Main-2020-11.01.2020
Application of the Integrals
86879
The area of the region bounded by \(y-x=2\) and \(x^{2}=y\) is equal to
1 \(\frac{16}{3}\)
2 \(\frac{2}{3}\)
3 \(\frac{9}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(C) : : Given \(\mathrm{y}-\mathrm{x}=2\) and \(\mathrm{x}^{2}=\mathrm{y}\) Now, \(\mathrm{x}^{2}=2+\mathrm{x}\) \(\mathrm{x}^{2}-\mathrm{x}-2=0\) \((\mathrm{x}+1)(\mathrm{x}-2)=0\) \(\mathrm{x}=-1,2\) Area \(=\int_{-1}^{2}\left(2+x-x^{2}\right)\) \(=\left[2 x+\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-1}^{2}=\left(4+2-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right)\) \(=6-3+2-\frac{1}{2}=\frac{9}{2}\)
JEE Main-2021-27.07.2021
Application of the Integrals
86880
The area bounded by the curve \(y=\sin ^{-1} x\) and the line \(x=0,|y|=\frac{\pi}{2}\) is
1 1
2 2
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(B) : Given curve is, \(\mathrm{y}=\sin ^{-1} \mathrm{x}\) and \(\mathrm{y}-\) axis between \(\mathrm{y}=0\) and \(\mathrm{y}=\frac{\pi}{2}\) Hence, the required area \(\mathrm{A}=\int_{0}^{\pi / 2} 2 \sin \mathrm{ydy}=2 \int_{0}^{\pi / 2} \sin \mathrm{y} \mathrm{dy}=2[-\cos \mathrm{y}]_{0}^{\pi / 2}\) \(=-2\left[\cos \frac{\pi}{2}-\cos 0\right]=-2[0-1]=2\) square unit
Jamia Millia Islamia-2011
Application of the Integrals
86881
The area between \(y^{2}+4 x-8=0\), the \(x\)-axis and the line \(x=1\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{2}{3}\) squnit
3 \(\frac{1}{3}\) sq unit
4 1 sq unit
Explanation:
(A) : Given, \(\mathrm{y}^{2}+4 \mathrm{x}-8=0\) \(y^{2}=-4(x-2)\) \(\therefore\) Required area \(=2 \int_{1}^{2} y d x=\int_{1}^{2} \sqrt{8-4 x} d x\) \(=\left[\frac{(8-4 x)^{3 / 2}}{\frac{3}{2}(-4)}\right]_{1}^{2}=-\frac{1}{6}(0-8)=\frac{4}{3}\) sq unit
86877
If the area (in sq. units) of the region \(\left\{(x, y): y^{2}\right.\) \(\leq 4 x, x+y \leq 1, x \leq 0, y \geq 0\}\) is \(a \sqrt{2}+b\) then a \(b\) is equal to
1 \(\frac{10}{3}\)
2 6
3 \(\frac{8}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
(B) : Given, Curve, \(C_{1}: y^{2} \leq 4 x\) \(\mathrm{C}_{2}: \mathrm{x}+\mathrm{y} \leq 1\) \(\mathrm{x}, \mathrm{y} \geq 0\) Point of intersection \(P\) these of two curve, \(y^{2}=4 x, x+y=1\) \(y^{2}-4(1-y)=0\) \(y^{2}+4 y-4=0\) \(y=\frac{-4 \pm \sqrt{16-4(-4)}}{2}=-2 \pm 2 \sqrt{2}\) \(P:(3-2 \sqrt{2},-2+2 \sqrt{2}), O:(0,0) R:(1,0)\) Required area \(=\int_{0}^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\int_{3-2 \sqrt{2}}^{1}(1-x) d x\) \(=\frac{4}{3}\left|x^{\frac{3}{2}}\right|_{0}^{3-2 \sqrt{2}}+\left|x-\frac{x^{2}}{2}\right|_{3-2 \sqrt{2}}^{1}=\frac{8 \sqrt{2}}{3}-\frac{10}{3}=a \sqrt{2}+b\) \(\mathrm{a}=\frac{8}{3}, \mathrm{~b}=-\frac{10}{3}\) \(\therefore \quad \mathrm{a}-\mathrm{b}=\frac{8+10}{3}=6\)
JEE Main-2019-12.04.2019
Application of the Integrals
86878
The area (in sq units) of the region bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{7}{8}\)
2 \(\frac{9}{8}\)
3 \(\frac{5}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(B) : Given, Curve, \(x^{2}=4 y\) and, Straight line \(x=4 y-2\) Point of intersection of the given parabola and line are, \(\left(-1, \frac{1}{4}\right) \text { and }(2,1)\) Required area \(=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x\) \(=\int_{-1}^{2}\left(\frac{x}{4}+\frac{1}{2}-\frac{x^{2}}{4}\right) d x\) \(=\frac{\mathrm{x}^{2}}{8}+\frac{\mathrm{x}}{2}-\left.\frac{\mathrm{x}^{3}}{12}\right|_{-1} ^{2}=\left(\frac{4}{8}+\frac{2}{2}-\frac{8}{12}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)\) \(=\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{3-12+2}{24}\right)\) \(=\left(\frac{3}{2}-\frac{2}{3}\right)-\left(\frac{-7}{24}\right)=\left(\frac{9-4}{6}\right)-\left(\frac{-7}{24}\right)\) \(=\frac{5}{6}+\frac{7}{24}=\frac{20+7}{24}=\frac{27}{24}=\frac{9}{8}\) square unit
JEE Main-2020-11.01.2020
Application of the Integrals
86879
The area of the region bounded by \(y-x=2\) and \(x^{2}=y\) is equal to
1 \(\frac{16}{3}\)
2 \(\frac{2}{3}\)
3 \(\frac{9}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(C) : : Given \(\mathrm{y}-\mathrm{x}=2\) and \(\mathrm{x}^{2}=\mathrm{y}\) Now, \(\mathrm{x}^{2}=2+\mathrm{x}\) \(\mathrm{x}^{2}-\mathrm{x}-2=0\) \((\mathrm{x}+1)(\mathrm{x}-2)=0\) \(\mathrm{x}=-1,2\) Area \(=\int_{-1}^{2}\left(2+x-x^{2}\right)\) \(=\left[2 x+\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-1}^{2}=\left(4+2-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right)\) \(=6-3+2-\frac{1}{2}=\frac{9}{2}\)
JEE Main-2021-27.07.2021
Application of the Integrals
86880
The area bounded by the curve \(y=\sin ^{-1} x\) and the line \(x=0,|y|=\frac{\pi}{2}\) is
1 1
2 2
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(B) : Given curve is, \(\mathrm{y}=\sin ^{-1} \mathrm{x}\) and \(\mathrm{y}-\) axis between \(\mathrm{y}=0\) and \(\mathrm{y}=\frac{\pi}{2}\) Hence, the required area \(\mathrm{A}=\int_{0}^{\pi / 2} 2 \sin \mathrm{ydy}=2 \int_{0}^{\pi / 2} \sin \mathrm{y} \mathrm{dy}=2[-\cos \mathrm{y}]_{0}^{\pi / 2}\) \(=-2\left[\cos \frac{\pi}{2}-\cos 0\right]=-2[0-1]=2\) square unit
Jamia Millia Islamia-2011
Application of the Integrals
86881
The area between \(y^{2}+4 x-8=0\), the \(x\)-axis and the line \(x=1\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{2}{3}\) squnit
3 \(\frac{1}{3}\) sq unit
4 1 sq unit
Explanation:
(A) : Given, \(\mathrm{y}^{2}+4 \mathrm{x}-8=0\) \(y^{2}=-4(x-2)\) \(\therefore\) Required area \(=2 \int_{1}^{2} y d x=\int_{1}^{2} \sqrt{8-4 x} d x\) \(=\left[\frac{(8-4 x)^{3 / 2}}{\frac{3}{2}(-4)}\right]_{1}^{2}=-\frac{1}{6}(0-8)=\frac{4}{3}\) sq unit
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86877
If the area (in sq. units) of the region \(\left\{(x, y): y^{2}\right.\) \(\leq 4 x, x+y \leq 1, x \leq 0, y \geq 0\}\) is \(a \sqrt{2}+b\) then a \(b\) is equal to
1 \(\frac{10}{3}\)
2 6
3 \(\frac{8}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
(B) : Given, Curve, \(C_{1}: y^{2} \leq 4 x\) \(\mathrm{C}_{2}: \mathrm{x}+\mathrm{y} \leq 1\) \(\mathrm{x}, \mathrm{y} \geq 0\) Point of intersection \(P\) these of two curve, \(y^{2}=4 x, x+y=1\) \(y^{2}-4(1-y)=0\) \(y^{2}+4 y-4=0\) \(y=\frac{-4 \pm \sqrt{16-4(-4)}}{2}=-2 \pm 2 \sqrt{2}\) \(P:(3-2 \sqrt{2},-2+2 \sqrt{2}), O:(0,0) R:(1,0)\) Required area \(=\int_{0}^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\int_{3-2 \sqrt{2}}^{1}(1-x) d x\) \(=\frac{4}{3}\left|x^{\frac{3}{2}}\right|_{0}^{3-2 \sqrt{2}}+\left|x-\frac{x^{2}}{2}\right|_{3-2 \sqrt{2}}^{1}=\frac{8 \sqrt{2}}{3}-\frac{10}{3}=a \sqrt{2}+b\) \(\mathrm{a}=\frac{8}{3}, \mathrm{~b}=-\frac{10}{3}\) \(\therefore \quad \mathrm{a}-\mathrm{b}=\frac{8+10}{3}=6\)
JEE Main-2019-12.04.2019
Application of the Integrals
86878
The area (in sq units) of the region bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{7}{8}\)
2 \(\frac{9}{8}\)
3 \(\frac{5}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(B) : Given, Curve, \(x^{2}=4 y\) and, Straight line \(x=4 y-2\) Point of intersection of the given parabola and line are, \(\left(-1, \frac{1}{4}\right) \text { and }(2,1)\) Required area \(=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x\) \(=\int_{-1}^{2}\left(\frac{x}{4}+\frac{1}{2}-\frac{x^{2}}{4}\right) d x\) \(=\frac{\mathrm{x}^{2}}{8}+\frac{\mathrm{x}}{2}-\left.\frac{\mathrm{x}^{3}}{12}\right|_{-1} ^{2}=\left(\frac{4}{8}+\frac{2}{2}-\frac{8}{12}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)\) \(=\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{3-12+2}{24}\right)\) \(=\left(\frac{3}{2}-\frac{2}{3}\right)-\left(\frac{-7}{24}\right)=\left(\frac{9-4}{6}\right)-\left(\frac{-7}{24}\right)\) \(=\frac{5}{6}+\frac{7}{24}=\frac{20+7}{24}=\frac{27}{24}=\frac{9}{8}\) square unit
JEE Main-2020-11.01.2020
Application of the Integrals
86879
The area of the region bounded by \(y-x=2\) and \(x^{2}=y\) is equal to
1 \(\frac{16}{3}\)
2 \(\frac{2}{3}\)
3 \(\frac{9}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(C) : : Given \(\mathrm{y}-\mathrm{x}=2\) and \(\mathrm{x}^{2}=\mathrm{y}\) Now, \(\mathrm{x}^{2}=2+\mathrm{x}\) \(\mathrm{x}^{2}-\mathrm{x}-2=0\) \((\mathrm{x}+1)(\mathrm{x}-2)=0\) \(\mathrm{x}=-1,2\) Area \(=\int_{-1}^{2}\left(2+x-x^{2}\right)\) \(=\left[2 x+\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-1}^{2}=\left(4+2-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right)\) \(=6-3+2-\frac{1}{2}=\frac{9}{2}\)
JEE Main-2021-27.07.2021
Application of the Integrals
86880
The area bounded by the curve \(y=\sin ^{-1} x\) and the line \(x=0,|y|=\frac{\pi}{2}\) is
1 1
2 2
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(B) : Given curve is, \(\mathrm{y}=\sin ^{-1} \mathrm{x}\) and \(\mathrm{y}-\) axis between \(\mathrm{y}=0\) and \(\mathrm{y}=\frac{\pi}{2}\) Hence, the required area \(\mathrm{A}=\int_{0}^{\pi / 2} 2 \sin \mathrm{ydy}=2 \int_{0}^{\pi / 2} \sin \mathrm{y} \mathrm{dy}=2[-\cos \mathrm{y}]_{0}^{\pi / 2}\) \(=-2\left[\cos \frac{\pi}{2}-\cos 0\right]=-2[0-1]=2\) square unit
Jamia Millia Islamia-2011
Application of the Integrals
86881
The area between \(y^{2}+4 x-8=0\), the \(x\)-axis and the line \(x=1\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{2}{3}\) squnit
3 \(\frac{1}{3}\) sq unit
4 1 sq unit
Explanation:
(A) : Given, \(\mathrm{y}^{2}+4 \mathrm{x}-8=0\) \(y^{2}=-4(x-2)\) \(\therefore\) Required area \(=2 \int_{1}^{2} y d x=\int_{1}^{2} \sqrt{8-4 x} d x\) \(=\left[\frac{(8-4 x)^{3 / 2}}{\frac{3}{2}(-4)}\right]_{1}^{2}=-\frac{1}{6}(0-8)=\frac{4}{3}\) sq unit
86877
If the area (in sq. units) of the region \(\left\{(x, y): y^{2}\right.\) \(\leq 4 x, x+y \leq 1, x \leq 0, y \geq 0\}\) is \(a \sqrt{2}+b\) then a \(b\) is equal to
1 \(\frac{10}{3}\)
2 6
3 \(\frac{8}{3}\)
4 \(\frac{-2}{3}\)
Explanation:
(B) : Given, Curve, \(C_{1}: y^{2} \leq 4 x\) \(\mathrm{C}_{2}: \mathrm{x}+\mathrm{y} \leq 1\) \(\mathrm{x}, \mathrm{y} \geq 0\) Point of intersection \(P\) these of two curve, \(y^{2}=4 x, x+y=1\) \(y^{2}-4(1-y)=0\) \(y^{2}+4 y-4=0\) \(y=\frac{-4 \pm \sqrt{16-4(-4)}}{2}=-2 \pm 2 \sqrt{2}\) \(P:(3-2 \sqrt{2},-2+2 \sqrt{2}), O:(0,0) R:(1,0)\) Required area \(=\int_{0}^{3-2 \sqrt{2}} 2 \sqrt{x} d x+\int_{3-2 \sqrt{2}}^{1}(1-x) d x\) \(=\frac{4}{3}\left|x^{\frac{3}{2}}\right|_{0}^{3-2 \sqrt{2}}+\left|x-\frac{x^{2}}{2}\right|_{3-2 \sqrt{2}}^{1}=\frac{8 \sqrt{2}}{3}-\frac{10}{3}=a \sqrt{2}+b\) \(\mathrm{a}=\frac{8}{3}, \mathrm{~b}=-\frac{10}{3}\) \(\therefore \quad \mathrm{a}-\mathrm{b}=\frac{8+10}{3}=6\)
JEE Main-2019-12.04.2019
Application of the Integrals
86878
The area (in sq units) of the region bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{7}{8}\)
2 \(\frac{9}{8}\)
3 \(\frac{5}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(B) : Given, Curve, \(x^{2}=4 y\) and, Straight line \(x=4 y-2\) Point of intersection of the given parabola and line are, \(\left(-1, \frac{1}{4}\right) \text { and }(2,1)\) Required area \(=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x\) \(=\int_{-1}^{2}\left(\frac{x}{4}+\frac{1}{2}-\frac{x^{2}}{4}\right) d x\) \(=\frac{\mathrm{x}^{2}}{8}+\frac{\mathrm{x}}{2}-\left.\frac{\mathrm{x}^{3}}{12}\right|_{-1} ^{2}=\left(\frac{4}{8}+\frac{2}{2}-\frac{8}{12}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)\) \(=\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{3-12+2}{24}\right)\) \(=\left(\frac{3}{2}-\frac{2}{3}\right)-\left(\frac{-7}{24}\right)=\left(\frac{9-4}{6}\right)-\left(\frac{-7}{24}\right)\) \(=\frac{5}{6}+\frac{7}{24}=\frac{20+7}{24}=\frac{27}{24}=\frac{9}{8}\) square unit
JEE Main-2020-11.01.2020
Application of the Integrals
86879
The area of the region bounded by \(y-x=2\) and \(x^{2}=y\) is equal to
1 \(\frac{16}{3}\)
2 \(\frac{2}{3}\)
3 \(\frac{9}{2}\)
4 \(\frac{4}{3}\)
Explanation:
(C) : : Given \(\mathrm{y}-\mathrm{x}=2\) and \(\mathrm{x}^{2}=\mathrm{y}\) Now, \(\mathrm{x}^{2}=2+\mathrm{x}\) \(\mathrm{x}^{2}-\mathrm{x}-2=0\) \((\mathrm{x}+1)(\mathrm{x}-2)=0\) \(\mathrm{x}=-1,2\) Area \(=\int_{-1}^{2}\left(2+x-x^{2}\right)\) \(=\left[2 x+\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-1}^{2}=\left(4+2-\frac{8}{3}\right)-\left(-2+\frac{1}{2}+\frac{1}{3}\right)\) \(=6-3+2-\frac{1}{2}=\frac{9}{2}\)
JEE Main-2021-27.07.2021
Application of the Integrals
86880
The area bounded by the curve \(y=\sin ^{-1} x\) and the line \(x=0,|y|=\frac{\pi}{2}\) is
1 1
2 2
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(B) : Given curve is, \(\mathrm{y}=\sin ^{-1} \mathrm{x}\) and \(\mathrm{y}-\) axis between \(\mathrm{y}=0\) and \(\mathrm{y}=\frac{\pi}{2}\) Hence, the required area \(\mathrm{A}=\int_{0}^{\pi / 2} 2 \sin \mathrm{ydy}=2 \int_{0}^{\pi / 2} \sin \mathrm{y} \mathrm{dy}=2[-\cos \mathrm{y}]_{0}^{\pi / 2}\) \(=-2\left[\cos \frac{\pi}{2}-\cos 0\right]=-2[0-1]=2\) square unit
Jamia Millia Islamia-2011
Application of the Integrals
86881
The area between \(y^{2}+4 x-8=0\), the \(x\)-axis and the line \(x=1\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{2}{3}\) squnit
3 \(\frac{1}{3}\) sq unit
4 1 sq unit
Explanation:
(A) : Given, \(\mathrm{y}^{2}+4 \mathrm{x}-8=0\) \(y^{2}=-4(x-2)\) \(\therefore\) Required area \(=2 \int_{1}^{2} y d x=\int_{1}^{2} \sqrt{8-4 x} d x\) \(=\left[\frac{(8-4 x)^{3 / 2}}{\frac{3}{2}(-4)}\right]_{1}^{2}=-\frac{1}{6}(0-8)=\frac{4}{3}\) sq unit
