86919
The area of the region (in square unit) bounded by the curve \(x^{2}=4 y\), line \(x=2\) and \(x-\) axis is
1 1
2 \(\frac{2}{3}\)
3 \(\frac{4}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given, Parabola \(\mathrm{x}^{2}=4 \mathrm{y}\) and straight line \(\mathrm{x}=2\) \(4=4 y\), \(y=1\) A \((2,1)\) is the point of intersection of the curve as d straight line Area of shaded reion \(=\int_{0}^{2} \mathrm{ydx}\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{2}}{4}\right) \mathrm{dx}=\left[\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=\frac{8}{12}-0=\frac{2}{3}\) square unit.
Jamia Millia Islamia-2012
Application of the Integrals
86905
The area of the region bounded by the curves \(y\) \(=|\mathbf{x}-\mathbf{2}|, \mathbf{x}=\mathbf{1}, \mathbf{x}=\mathbf{3}\) and \(\mathbf{y}=\mathbf{0}\) is
1 4
2 12
3 3
4 14 (e) 1
Explanation:
(E) : Given, \(\mathrm{y}=|\mathrm{x}-2|, \mathrm{x}=1, \mathrm{x}=3\) and \(\mathrm{y}=0\) According to question. Required area \(=\int_{1}^{3} y d x=\int_{1}^{3}|x-2| d x=\int_{1}^{2}-(x-2) d x+\int_{2}^{3}(x-2) d x\) \(=\int_{1}^{2}(2-x) d x+\int_{2}^{3}(x-2) d x\) \(=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{3}=2-\frac{3}{2}-\frac{3}{2}+2=1\) square unit
Kerala CEE-2019
Application of the Integrals
86790
The area of the region bounded to the curve \(y^{2}\) \(=8 x\) and the line \(y=2 x\) is
86792
Area of the region bounded by the curve \(y=\) \(\cos x, x=0\) and \(x=\pi\) is
1 2 sq. units
2 3 sq. units
3 4 sq. units
4 1 sq. units
Explanation:
(A) : Given, \(\mathrm{y}=\cos \mathrm{x}\) And, \(\quad \mathrm{x}=0, \mathrm{x}=\pi\) At, \(\quad \mathrm{x}=0, \mathrm{y}=\cos 0=1\) And \(\quad \mathrm{x}=\pi, \mathrm{y}=\cos \pi=-1\) Required area, \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\int_0^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^\pi y d x A=\int_0^{\frac{\pi}{2}} \cos x \cdot d x+\int_{\frac{\pi}{2}}^\pi \cos x \cdot d x\) \(\mathrm{A}=[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}+[\sin \mathrm{x}]_{\frac{\pi}{2}}^{\pi}\) \(\mathrm{A}=\left|\sin \frac{\pi}{2}-\sin 0\right|+\left|\sin \pi-\sin \frac{\pi}{2}\right|\) \(A=|1-0|+|0-1|=1+1 \Rightarrow A=2\) sq. units
86919
The area of the region (in square unit) bounded by the curve \(x^{2}=4 y\), line \(x=2\) and \(x-\) axis is
1 1
2 \(\frac{2}{3}\)
3 \(\frac{4}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given, Parabola \(\mathrm{x}^{2}=4 \mathrm{y}\) and straight line \(\mathrm{x}=2\) \(4=4 y\), \(y=1\) A \((2,1)\) is the point of intersection of the curve as d straight line Area of shaded reion \(=\int_{0}^{2} \mathrm{ydx}\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{2}}{4}\right) \mathrm{dx}=\left[\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=\frac{8}{12}-0=\frac{2}{3}\) square unit.
Jamia Millia Islamia-2012
Application of the Integrals
86905
The area of the region bounded by the curves \(y\) \(=|\mathbf{x}-\mathbf{2}|, \mathbf{x}=\mathbf{1}, \mathbf{x}=\mathbf{3}\) and \(\mathbf{y}=\mathbf{0}\) is
1 4
2 12
3 3
4 14 (e) 1
Explanation:
(E) : Given, \(\mathrm{y}=|\mathrm{x}-2|, \mathrm{x}=1, \mathrm{x}=3\) and \(\mathrm{y}=0\) According to question. Required area \(=\int_{1}^{3} y d x=\int_{1}^{3}|x-2| d x=\int_{1}^{2}-(x-2) d x+\int_{2}^{3}(x-2) d x\) \(=\int_{1}^{2}(2-x) d x+\int_{2}^{3}(x-2) d x\) \(=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{3}=2-\frac{3}{2}-\frac{3}{2}+2=1\) square unit
Kerala CEE-2019
Application of the Integrals
86790
The area of the region bounded to the curve \(y^{2}\) \(=8 x\) and the line \(y=2 x\) is
86792
Area of the region bounded by the curve \(y=\) \(\cos x, x=0\) and \(x=\pi\) is
1 2 sq. units
2 3 sq. units
3 4 sq. units
4 1 sq. units
Explanation:
(A) : Given, \(\mathrm{y}=\cos \mathrm{x}\) And, \(\quad \mathrm{x}=0, \mathrm{x}=\pi\) At, \(\quad \mathrm{x}=0, \mathrm{y}=\cos 0=1\) And \(\quad \mathrm{x}=\pi, \mathrm{y}=\cos \pi=-1\) Required area, \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\int_0^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^\pi y d x A=\int_0^{\frac{\pi}{2}} \cos x \cdot d x+\int_{\frac{\pi}{2}}^\pi \cos x \cdot d x\) \(\mathrm{A}=[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}+[\sin \mathrm{x}]_{\frac{\pi}{2}}^{\pi}\) \(\mathrm{A}=\left|\sin \frac{\pi}{2}-\sin 0\right|+\left|\sin \pi-\sin \frac{\pi}{2}\right|\) \(A=|1-0|+|0-1|=1+1 \Rightarrow A=2\) sq. units
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86919
The area of the region (in square unit) bounded by the curve \(x^{2}=4 y\), line \(x=2\) and \(x-\) axis is
1 1
2 \(\frac{2}{3}\)
3 \(\frac{4}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given, Parabola \(\mathrm{x}^{2}=4 \mathrm{y}\) and straight line \(\mathrm{x}=2\) \(4=4 y\), \(y=1\) A \((2,1)\) is the point of intersection of the curve as d straight line Area of shaded reion \(=\int_{0}^{2} \mathrm{ydx}\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{2}}{4}\right) \mathrm{dx}=\left[\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=\frac{8}{12}-0=\frac{2}{3}\) square unit.
Jamia Millia Islamia-2012
Application of the Integrals
86905
The area of the region bounded by the curves \(y\) \(=|\mathbf{x}-\mathbf{2}|, \mathbf{x}=\mathbf{1}, \mathbf{x}=\mathbf{3}\) and \(\mathbf{y}=\mathbf{0}\) is
1 4
2 12
3 3
4 14 (e) 1
Explanation:
(E) : Given, \(\mathrm{y}=|\mathrm{x}-2|, \mathrm{x}=1, \mathrm{x}=3\) and \(\mathrm{y}=0\) According to question. Required area \(=\int_{1}^{3} y d x=\int_{1}^{3}|x-2| d x=\int_{1}^{2}-(x-2) d x+\int_{2}^{3}(x-2) d x\) \(=\int_{1}^{2}(2-x) d x+\int_{2}^{3}(x-2) d x\) \(=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{3}=2-\frac{3}{2}-\frac{3}{2}+2=1\) square unit
Kerala CEE-2019
Application of the Integrals
86790
The area of the region bounded to the curve \(y^{2}\) \(=8 x\) and the line \(y=2 x\) is
86792
Area of the region bounded by the curve \(y=\) \(\cos x, x=0\) and \(x=\pi\) is
1 2 sq. units
2 3 sq. units
3 4 sq. units
4 1 sq. units
Explanation:
(A) : Given, \(\mathrm{y}=\cos \mathrm{x}\) And, \(\quad \mathrm{x}=0, \mathrm{x}=\pi\) At, \(\quad \mathrm{x}=0, \mathrm{y}=\cos 0=1\) And \(\quad \mathrm{x}=\pi, \mathrm{y}=\cos \pi=-1\) Required area, \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\int_0^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^\pi y d x A=\int_0^{\frac{\pi}{2}} \cos x \cdot d x+\int_{\frac{\pi}{2}}^\pi \cos x \cdot d x\) \(\mathrm{A}=[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}+[\sin \mathrm{x}]_{\frac{\pi}{2}}^{\pi}\) \(\mathrm{A}=\left|\sin \frac{\pi}{2}-\sin 0\right|+\left|\sin \pi-\sin \frac{\pi}{2}\right|\) \(A=|1-0|+|0-1|=1+1 \Rightarrow A=2\) sq. units
86919
The area of the region (in square unit) bounded by the curve \(x^{2}=4 y\), line \(x=2\) and \(x-\) axis is
1 1
2 \(\frac{2}{3}\)
3 \(\frac{4}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given, Parabola \(\mathrm{x}^{2}=4 \mathrm{y}\) and straight line \(\mathrm{x}=2\) \(4=4 y\), \(y=1\) A \((2,1)\) is the point of intersection of the curve as d straight line Area of shaded reion \(=\int_{0}^{2} \mathrm{ydx}\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{2}}{4}\right) \mathrm{dx}=\left[\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=\frac{8}{12}-0=\frac{2}{3}\) square unit.
Jamia Millia Islamia-2012
Application of the Integrals
86905
The area of the region bounded by the curves \(y\) \(=|\mathbf{x}-\mathbf{2}|, \mathbf{x}=\mathbf{1}, \mathbf{x}=\mathbf{3}\) and \(\mathbf{y}=\mathbf{0}\) is
1 4
2 12
3 3
4 14 (e) 1
Explanation:
(E) : Given, \(\mathrm{y}=|\mathrm{x}-2|, \mathrm{x}=1, \mathrm{x}=3\) and \(\mathrm{y}=0\) According to question. Required area \(=\int_{1}^{3} y d x=\int_{1}^{3}|x-2| d x=\int_{1}^{2}-(x-2) d x+\int_{2}^{3}(x-2) d x\) \(=\int_{1}^{2}(2-x) d x+\int_{2}^{3}(x-2) d x\) \(=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{3}=2-\frac{3}{2}-\frac{3}{2}+2=1\) square unit
Kerala CEE-2019
Application of the Integrals
86790
The area of the region bounded to the curve \(y^{2}\) \(=8 x\) and the line \(y=2 x\) is
86792
Area of the region bounded by the curve \(y=\) \(\cos x, x=0\) and \(x=\pi\) is
1 2 sq. units
2 3 sq. units
3 4 sq. units
4 1 sq. units
Explanation:
(A) : Given, \(\mathrm{y}=\cos \mathrm{x}\) And, \(\quad \mathrm{x}=0, \mathrm{x}=\pi\) At, \(\quad \mathrm{x}=0, \mathrm{y}=\cos 0=1\) And \(\quad \mathrm{x}=\pi, \mathrm{y}=\cos \pi=-1\) Required area, \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\int_0^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^\pi y d x A=\int_0^{\frac{\pi}{2}} \cos x \cdot d x+\int_{\frac{\pi}{2}}^\pi \cos x \cdot d x\) \(\mathrm{A}=[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}+[\sin \mathrm{x}]_{\frac{\pi}{2}}^{\pi}\) \(\mathrm{A}=\left|\sin \frac{\pi}{2}-\sin 0\right|+\left|\sin \pi-\sin \frac{\pi}{2}\right|\) \(A=|1-0|+|0-1|=1+1 \Rightarrow A=2\) sq. units
86919
The area of the region (in square unit) bounded by the curve \(x^{2}=4 y\), line \(x=2\) and \(x-\) axis is
1 1
2 \(\frac{2}{3}\)
3 \(\frac{4}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given, Parabola \(\mathrm{x}^{2}=4 \mathrm{y}\) and straight line \(\mathrm{x}=2\) \(4=4 y\), \(y=1\) A \((2,1)\) is the point of intersection of the curve as d straight line Area of shaded reion \(=\int_{0}^{2} \mathrm{ydx}\) \(=\int_{0}^{2}\left(\frac{\mathrm{x}^{2}}{4}\right) \mathrm{dx}=\left[\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=\frac{8}{12}-0=\frac{2}{3}\) square unit.
Jamia Millia Islamia-2012
Application of the Integrals
86905
The area of the region bounded by the curves \(y\) \(=|\mathbf{x}-\mathbf{2}|, \mathbf{x}=\mathbf{1}, \mathbf{x}=\mathbf{3}\) and \(\mathbf{y}=\mathbf{0}\) is
1 4
2 12
3 3
4 14 (e) 1
Explanation:
(E) : Given, \(\mathrm{y}=|\mathrm{x}-2|, \mathrm{x}=1, \mathrm{x}=3\) and \(\mathrm{y}=0\) According to question. Required area \(=\int_{1}^{3} y d x=\int_{1}^{3}|x-2| d x=\int_{1}^{2}-(x-2) d x+\int_{2}^{3}(x-2) d x\) \(=\int_{1}^{2}(2-x) d x+\int_{2}^{3}(x-2) d x\) \(=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{3}=2-\frac{3}{2}-\frac{3}{2}+2=1\) square unit
Kerala CEE-2019
Application of the Integrals
86790
The area of the region bounded to the curve \(y^{2}\) \(=8 x\) and the line \(y=2 x\) is
86792
Area of the region bounded by the curve \(y=\) \(\cos x, x=0\) and \(x=\pi\) is
1 2 sq. units
2 3 sq. units
3 4 sq. units
4 1 sq. units
Explanation:
(A) : Given, \(\mathrm{y}=\cos \mathrm{x}\) And, \(\quad \mathrm{x}=0, \mathrm{x}=\pi\) At, \(\quad \mathrm{x}=0, \mathrm{y}=\cos 0=1\) And \(\quad \mathrm{x}=\pi, \mathrm{y}=\cos \pi=-1\) Required area, \(\mathrm{A}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(A=\int_0^{\frac{\pi}{2}} y d x+\int_{\frac{\pi}{2}}^\pi y d x A=\int_0^{\frac{\pi}{2}} \cos x \cdot d x+\int_{\frac{\pi}{2}}^\pi \cos x \cdot d x\) \(\mathrm{A}=[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}+[\sin \mathrm{x}]_{\frac{\pi}{2}}^{\pi}\) \(\mathrm{A}=\left|\sin \frac{\pi}{2}-\sin 0\right|+\left|\sin \pi-\sin \frac{\pi}{2}\right|\) \(A=|1-0|+|0-1|=1+1 \Rightarrow A=2\) sq. units
