NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86810
The area of the region bounded by the curves \(y^{2}=8 x\) and \(y=x\) (in sq. unit) is
1 \(\frac{64}{3}\)
2 \(\frac{32}{3}\)
3 \(\frac{16}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given curve, \(\mathrm{y}^{2}=8 \mathrm{x}\) and \(\mathrm{y}=\mathrm{x}\) Here, \(\mathrm{y}^{2}=8 \mathrm{x}\) \(\mathrm{x}^{2}=8 \mathrm{x} \quad \because \mathrm{y}=\mathrm{x}\) \(x^{2}-8 x=0\) \(\mathrm{x}(\mathrm{x}-8)=0\) \(\mathrm{x}=0, \mathrm{x}=8\) or \(\mathrm{y}=0, \mathrm{y}=8\) Required area \(=\int_{0}^{8} \sqrt{8} \mathrm{dx}-\int_{0}^{8} \mathrm{x} d \mathrm{x}\) \(=2 \sqrt{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{8}-\left[\frac{x^{2}}{2}\right]_{0}^{8}=\frac{4 \sqrt{2}}{3}(8)^{3 / 2}-\frac{1}{2}(8)^{2}\) \(=\left(\frac{4 \sqrt{2}}{3}\right)(16 \sqrt{2})-32=\frac{32}{3}\) sq. units
MHT CET-2012
Application of the Integrals
86908
The area of the region bounded by \(y^{2}=16-x^{2}\), \(\mathbf{y}=\mathbf{0}, \mathbf{x}=\mathbf{0}\) in the first quadrant is (in square units)
1 \(8 \pi\)
2 \(6 \pi\)
3 \(2 \pi\)
4 \(4 \pi\)
5 \(\frac{\pi}{2}\)
Explanation:
(D) : Given, \(\mathrm{y}^{2}=16-\mathrm{x}^{2}, \mathrm{y}=0, \quad \mathrm{x}=0\) According to question, \(y^{2}=16-x^{2}\) \(x^{2}+y^{2}=16\) From equation (i) is equation of a circle with radius 4, Required Area, \(=\int_{0}^{4} \sqrt{16-\mathrm{x}^{2}} \mathrm{dx}\) \(=\left(\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right)_{0}^{4}=4 \pi\)
Application of the Integrals
86909
The area bounded by \(y=x+2, y=2-x\) and the \(\mathbf{x}\)-axis is (in square units)
1 1
2 2
3 4
4 6
5 8
Explanation:
(C) : Given, \(\mathrm{y}=\mathrm{x}+2, \mathrm{y}=2-\mathrm{x}\) According to question . Required Area, \(=\int_{-2}^{0} y d x+\int_{0}^{2} y d x\) \(=\int_{-2}^{0}(x+2) d x+\int_{0}^{2}(2-x) d x\) \(=\left[\frac{x^{2}}{2}+2 x\right]_{-2}^{0}+\left[2 x-\frac{x^{2}}{2}\right]_{0}^{2}=2+2=4\) square unit.
Kerala CEE-2012
Application of the Integrals
86834
Area of the region bounded by the curves \(y=\) \(\mathbf{2}^{x}, y=2 x-x^{2}, x=0\) and \(x=2\) is given by
1 \(\frac{3}{\log 2}-\frac{4}{3}\)
2 \(\frac{3}{\log 2}+\frac{4}{3}\)
3 \(3 \log 2-\frac{4}{3}\)
4 \(3 \log 2+\frac{4}{3}\)
Explanation:
(A) : Given curves, \(y=2^{x} \Rightarrow y=2 x-x^{2}\) and \(\quad \mathrm{x}=0, \mathrm{x}=2\) At, \(\quad \mathrm{x}=0\), then \(\mathrm{y}=0,1\) and \(\quad \mathrm{x}=2\), then \(\mathrm{y}=0,4\) Area, \(A=\int_{0}^{2}\left(y_{1}-y_{2}\right) \cdot d x \Rightarrow A=\left.\int_{0}^{2}\left(2^{x}-2 x /+\left(0, x^{2}\right)\right) \cdot d\right|_{y=2 x-x^{2}} x\) \(A=\int_{0}^{2} 2^{x} \cdot d x-2 \int_{0}^{2} x \cdot d x+\int_{0}^{2} x^{2} \cdot d x \quad x_{x=0}\) \(A=\left[\frac{2^{x}}{\log \cdot 2}-x^{2}+\frac{x^{3}}{3}\right]_{0}^{2}=\left[\frac{4}{\log \cdot 2}-4+\frac{8}{3}-\frac{1}{\log 2}\right]\) \(A=\frac{3}{\log 2}-\frac{4}{3}\) square units.
86810
The area of the region bounded by the curves \(y^{2}=8 x\) and \(y=x\) (in sq. unit) is
1 \(\frac{64}{3}\)
2 \(\frac{32}{3}\)
3 \(\frac{16}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given curve, \(\mathrm{y}^{2}=8 \mathrm{x}\) and \(\mathrm{y}=\mathrm{x}\) Here, \(\mathrm{y}^{2}=8 \mathrm{x}\) \(\mathrm{x}^{2}=8 \mathrm{x} \quad \because \mathrm{y}=\mathrm{x}\) \(x^{2}-8 x=0\) \(\mathrm{x}(\mathrm{x}-8)=0\) \(\mathrm{x}=0, \mathrm{x}=8\) or \(\mathrm{y}=0, \mathrm{y}=8\) Required area \(=\int_{0}^{8} \sqrt{8} \mathrm{dx}-\int_{0}^{8} \mathrm{x} d \mathrm{x}\) \(=2 \sqrt{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{8}-\left[\frac{x^{2}}{2}\right]_{0}^{8}=\frac{4 \sqrt{2}}{3}(8)^{3 / 2}-\frac{1}{2}(8)^{2}\) \(=\left(\frac{4 \sqrt{2}}{3}\right)(16 \sqrt{2})-32=\frac{32}{3}\) sq. units
MHT CET-2012
Application of the Integrals
86908
The area of the region bounded by \(y^{2}=16-x^{2}\), \(\mathbf{y}=\mathbf{0}, \mathbf{x}=\mathbf{0}\) in the first quadrant is (in square units)
1 \(8 \pi\)
2 \(6 \pi\)
3 \(2 \pi\)
4 \(4 \pi\)
5 \(\frac{\pi}{2}\)
Explanation:
(D) : Given, \(\mathrm{y}^{2}=16-\mathrm{x}^{2}, \mathrm{y}=0, \quad \mathrm{x}=0\) According to question, \(y^{2}=16-x^{2}\) \(x^{2}+y^{2}=16\) From equation (i) is equation of a circle with radius 4, Required Area, \(=\int_{0}^{4} \sqrt{16-\mathrm{x}^{2}} \mathrm{dx}\) \(=\left(\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right)_{0}^{4}=4 \pi\)
Application of the Integrals
86909
The area bounded by \(y=x+2, y=2-x\) and the \(\mathbf{x}\)-axis is (in square units)
1 1
2 2
3 4
4 6
5 8
Explanation:
(C) : Given, \(\mathrm{y}=\mathrm{x}+2, \mathrm{y}=2-\mathrm{x}\) According to question . Required Area, \(=\int_{-2}^{0} y d x+\int_{0}^{2} y d x\) \(=\int_{-2}^{0}(x+2) d x+\int_{0}^{2}(2-x) d x\) \(=\left[\frac{x^{2}}{2}+2 x\right]_{-2}^{0}+\left[2 x-\frac{x^{2}}{2}\right]_{0}^{2}=2+2=4\) square unit.
Kerala CEE-2012
Application of the Integrals
86834
Area of the region bounded by the curves \(y=\) \(\mathbf{2}^{x}, y=2 x-x^{2}, x=0\) and \(x=2\) is given by
1 \(\frac{3}{\log 2}-\frac{4}{3}\)
2 \(\frac{3}{\log 2}+\frac{4}{3}\)
3 \(3 \log 2-\frac{4}{3}\)
4 \(3 \log 2+\frac{4}{3}\)
Explanation:
(A) : Given curves, \(y=2^{x} \Rightarrow y=2 x-x^{2}\) and \(\quad \mathrm{x}=0, \mathrm{x}=2\) At, \(\quad \mathrm{x}=0\), then \(\mathrm{y}=0,1\) and \(\quad \mathrm{x}=2\), then \(\mathrm{y}=0,4\) Area, \(A=\int_{0}^{2}\left(y_{1}-y_{2}\right) \cdot d x \Rightarrow A=\left.\int_{0}^{2}\left(2^{x}-2 x /+\left(0, x^{2}\right)\right) \cdot d\right|_{y=2 x-x^{2}} x\) \(A=\int_{0}^{2} 2^{x} \cdot d x-2 \int_{0}^{2} x \cdot d x+\int_{0}^{2} x^{2} \cdot d x \quad x_{x=0}\) \(A=\left[\frac{2^{x}}{\log \cdot 2}-x^{2}+\frac{x^{3}}{3}\right]_{0}^{2}=\left[\frac{4}{\log \cdot 2}-4+\frac{8}{3}-\frac{1}{\log 2}\right]\) \(A=\frac{3}{\log 2}-\frac{4}{3}\) square units.
86810
The area of the region bounded by the curves \(y^{2}=8 x\) and \(y=x\) (in sq. unit) is
1 \(\frac{64}{3}\)
2 \(\frac{32}{3}\)
3 \(\frac{16}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given curve, \(\mathrm{y}^{2}=8 \mathrm{x}\) and \(\mathrm{y}=\mathrm{x}\) Here, \(\mathrm{y}^{2}=8 \mathrm{x}\) \(\mathrm{x}^{2}=8 \mathrm{x} \quad \because \mathrm{y}=\mathrm{x}\) \(x^{2}-8 x=0\) \(\mathrm{x}(\mathrm{x}-8)=0\) \(\mathrm{x}=0, \mathrm{x}=8\) or \(\mathrm{y}=0, \mathrm{y}=8\) Required area \(=\int_{0}^{8} \sqrt{8} \mathrm{dx}-\int_{0}^{8} \mathrm{x} d \mathrm{x}\) \(=2 \sqrt{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{8}-\left[\frac{x^{2}}{2}\right]_{0}^{8}=\frac{4 \sqrt{2}}{3}(8)^{3 / 2}-\frac{1}{2}(8)^{2}\) \(=\left(\frac{4 \sqrt{2}}{3}\right)(16 \sqrt{2})-32=\frac{32}{3}\) sq. units
MHT CET-2012
Application of the Integrals
86908
The area of the region bounded by \(y^{2}=16-x^{2}\), \(\mathbf{y}=\mathbf{0}, \mathbf{x}=\mathbf{0}\) in the first quadrant is (in square units)
1 \(8 \pi\)
2 \(6 \pi\)
3 \(2 \pi\)
4 \(4 \pi\)
5 \(\frac{\pi}{2}\)
Explanation:
(D) : Given, \(\mathrm{y}^{2}=16-\mathrm{x}^{2}, \mathrm{y}=0, \quad \mathrm{x}=0\) According to question, \(y^{2}=16-x^{2}\) \(x^{2}+y^{2}=16\) From equation (i) is equation of a circle with radius 4, Required Area, \(=\int_{0}^{4} \sqrt{16-\mathrm{x}^{2}} \mathrm{dx}\) \(=\left(\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right)_{0}^{4}=4 \pi\)
Application of the Integrals
86909
The area bounded by \(y=x+2, y=2-x\) and the \(\mathbf{x}\)-axis is (in square units)
1 1
2 2
3 4
4 6
5 8
Explanation:
(C) : Given, \(\mathrm{y}=\mathrm{x}+2, \mathrm{y}=2-\mathrm{x}\) According to question . Required Area, \(=\int_{-2}^{0} y d x+\int_{0}^{2} y d x\) \(=\int_{-2}^{0}(x+2) d x+\int_{0}^{2}(2-x) d x\) \(=\left[\frac{x^{2}}{2}+2 x\right]_{-2}^{0}+\left[2 x-\frac{x^{2}}{2}\right]_{0}^{2}=2+2=4\) square unit.
Kerala CEE-2012
Application of the Integrals
86834
Area of the region bounded by the curves \(y=\) \(\mathbf{2}^{x}, y=2 x-x^{2}, x=0\) and \(x=2\) is given by
1 \(\frac{3}{\log 2}-\frac{4}{3}\)
2 \(\frac{3}{\log 2}+\frac{4}{3}\)
3 \(3 \log 2-\frac{4}{3}\)
4 \(3 \log 2+\frac{4}{3}\)
Explanation:
(A) : Given curves, \(y=2^{x} \Rightarrow y=2 x-x^{2}\) and \(\quad \mathrm{x}=0, \mathrm{x}=2\) At, \(\quad \mathrm{x}=0\), then \(\mathrm{y}=0,1\) and \(\quad \mathrm{x}=2\), then \(\mathrm{y}=0,4\) Area, \(A=\int_{0}^{2}\left(y_{1}-y_{2}\right) \cdot d x \Rightarrow A=\left.\int_{0}^{2}\left(2^{x}-2 x /+\left(0, x^{2}\right)\right) \cdot d\right|_{y=2 x-x^{2}} x\) \(A=\int_{0}^{2} 2^{x} \cdot d x-2 \int_{0}^{2} x \cdot d x+\int_{0}^{2} x^{2} \cdot d x \quad x_{x=0}\) \(A=\left[\frac{2^{x}}{\log \cdot 2}-x^{2}+\frac{x^{3}}{3}\right]_{0}^{2}=\left[\frac{4}{\log \cdot 2}-4+\frac{8}{3}-\frac{1}{\log 2}\right]\) \(A=\frac{3}{\log 2}-\frac{4}{3}\) square units.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of the Integrals
86810
The area of the region bounded by the curves \(y^{2}=8 x\) and \(y=x\) (in sq. unit) is
1 \(\frac{64}{3}\)
2 \(\frac{32}{3}\)
3 \(\frac{16}{3}\)
4 \(\frac{8}{3}\)
Explanation:
(B) : Given curve, \(\mathrm{y}^{2}=8 \mathrm{x}\) and \(\mathrm{y}=\mathrm{x}\) Here, \(\mathrm{y}^{2}=8 \mathrm{x}\) \(\mathrm{x}^{2}=8 \mathrm{x} \quad \because \mathrm{y}=\mathrm{x}\) \(x^{2}-8 x=0\) \(\mathrm{x}(\mathrm{x}-8)=0\) \(\mathrm{x}=0, \mathrm{x}=8\) or \(\mathrm{y}=0, \mathrm{y}=8\) Required area \(=\int_{0}^{8} \sqrt{8} \mathrm{dx}-\int_{0}^{8} \mathrm{x} d \mathrm{x}\) \(=2 \sqrt{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{8}-\left[\frac{x^{2}}{2}\right]_{0}^{8}=\frac{4 \sqrt{2}}{3}(8)^{3 / 2}-\frac{1}{2}(8)^{2}\) \(=\left(\frac{4 \sqrt{2}}{3}\right)(16 \sqrt{2})-32=\frac{32}{3}\) sq. units
MHT CET-2012
Application of the Integrals
86908
The area of the region bounded by \(y^{2}=16-x^{2}\), \(\mathbf{y}=\mathbf{0}, \mathbf{x}=\mathbf{0}\) in the first quadrant is (in square units)
1 \(8 \pi\)
2 \(6 \pi\)
3 \(2 \pi\)
4 \(4 \pi\)
5 \(\frac{\pi}{2}\)
Explanation:
(D) : Given, \(\mathrm{y}^{2}=16-\mathrm{x}^{2}, \mathrm{y}=0, \quad \mathrm{x}=0\) According to question, \(y^{2}=16-x^{2}\) \(x^{2}+y^{2}=16\) From equation (i) is equation of a circle with radius 4, Required Area, \(=\int_{0}^{4} \sqrt{16-\mathrm{x}^{2}} \mathrm{dx}\) \(=\left(\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right)_{0}^{4}=4 \pi\)
Application of the Integrals
86909
The area bounded by \(y=x+2, y=2-x\) and the \(\mathbf{x}\)-axis is (in square units)
1 1
2 2
3 4
4 6
5 8
Explanation:
(C) : Given, \(\mathrm{y}=\mathrm{x}+2, \mathrm{y}=2-\mathrm{x}\) According to question . Required Area, \(=\int_{-2}^{0} y d x+\int_{0}^{2} y d x\) \(=\int_{-2}^{0}(x+2) d x+\int_{0}^{2}(2-x) d x\) \(=\left[\frac{x^{2}}{2}+2 x\right]_{-2}^{0}+\left[2 x-\frac{x^{2}}{2}\right]_{0}^{2}=2+2=4\) square unit.
Kerala CEE-2012
Application of the Integrals
86834
Area of the region bounded by the curves \(y=\) \(\mathbf{2}^{x}, y=2 x-x^{2}, x=0\) and \(x=2\) is given by
1 \(\frac{3}{\log 2}-\frac{4}{3}\)
2 \(\frac{3}{\log 2}+\frac{4}{3}\)
3 \(3 \log 2-\frac{4}{3}\)
4 \(3 \log 2+\frac{4}{3}\)
Explanation:
(A) : Given curves, \(y=2^{x} \Rightarrow y=2 x-x^{2}\) and \(\quad \mathrm{x}=0, \mathrm{x}=2\) At, \(\quad \mathrm{x}=0\), then \(\mathrm{y}=0,1\) and \(\quad \mathrm{x}=2\), then \(\mathrm{y}=0,4\) Area, \(A=\int_{0}^{2}\left(y_{1}-y_{2}\right) \cdot d x \Rightarrow A=\left.\int_{0}^{2}\left(2^{x}-2 x /+\left(0, x^{2}\right)\right) \cdot d\right|_{y=2 x-x^{2}} x\) \(A=\int_{0}^{2} 2^{x} \cdot d x-2 \int_{0}^{2} x \cdot d x+\int_{0}^{2} x^{2} \cdot d x \quad x_{x=0}\) \(A=\left[\frac{2^{x}}{\log \cdot 2}-x^{2}+\frac{x^{3}}{3}\right]_{0}^{2}=\left[\frac{4}{\log \cdot 2}-4+\frac{8}{3}-\frac{1}{\log 2}\right]\) \(A=\frac{3}{\log 2}-\frac{4}{3}\) square units.
