86815
The area of the region bounded by the parabola \(y=x^{2}+2\) and the lines \(y=x, x=0\) and \(x=3\) is
1 15
2 6
3 \(\frac{9}{2}\)
4 \(\frac{21}{2}\)
Explanation:
(D) : equation of the parabola is \(y=x^{2}+2\) or \(\mathrm{x}^{2}=\mathrm{y}-2\) its vertex is \((0,2)\) and axis is \(y\)-axis Graphs of the curve and lines have been shown in the figure \(A=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3} x d x=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}\) \(=\left[\left(\frac{27}{3}+6\right)-0\right]-\left(\frac{9}{2}-0\right)=21 / 2\) Sq units.
nda
Application of the Integrals
86817
The area (in sq. units) bounded by the curves \(y=\) \(\sqrt{x}, 2 y-x+3=0, x\)-axis and lying in the first quadrant is
1 36
2 18
3 \(\frac{27}{4}\)
4 9
Explanation:
(D) : Given curves, \(\mathrm{y}=\sqrt{\mathrm{x}}\) and \(2 y-x+3=0\) From equation (i) and (ii), we get - \(2 \sqrt{\mathrm{x}}-(\sqrt{\mathrm{x}})^{2}+3=0\) \((\sqrt{\mathrm{x}})^{2}-2 \sqrt{\mathrm{x}}-3=0\) \((\sqrt{\mathrm{x}}-3)(\sqrt{\mathrm{x}}+1)=0\) \(\sqrt{\mathrm{x}}=3\) and \(\sqrt{\mathrm{x}}=-1\) \(\because \sqrt{\mathrm{x}}=-1\) not possible \(\therefore \mathrm{y}=\sqrt{\mathrm{x}}=3\) Now, required area, \(A=\int_{0}^{3}\left(x_{2}-x_{1}\right) \cdot d x \Rightarrow A=\int_{0}^{3}\left[(2 y+3]-y^{2}\right] . d y\) \(\mathrm{A}=\left[\left(\frac{2 \mathrm{y}^{2}}{2}\right)+(3 \mathrm{y})-\left(\frac{\mathrm{y}^{3}}{3}\right)\right]_{0}^{3}\) \(\mathrm{A}=\left[\left(\frac{2 \times 3^{2}}{2}-0\right)+(3 \times 3)-\left(\frac{3^{3}}{3}\right)\right]=9+9-\frac{27}{3}\) \(A=18-9 \Rightarrow A=9\) sq. units
COMEDK-2013
Application of the Integrals
86818
The area of the region bounded by \(y=\sqrt{16-x^{2}}\) and \(x\)-axis is
1 \(8 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(16 \pi\) sq. units
4 \(256 \pi\) sq. units
Explanation:
(A) : Given, Equation of curve, \(y=\sqrt{16-x^{2}}\) and line \(\mathrm{x}\)-axis - So, \(\mathrm{y}=0\) Here, \(\sqrt{16-\mathrm{x}^{2}}=0\) \(16-x^{2}=0\) \(\mathrm{x}^{2}=16\) \(\mathrm{x}= \pm 4\) Required area, \(\mathrm{A}=\int_{-4}^{4} \mathrm{y} \cdot \mathrm{dx}\) \(A=2 \int_{0}^{4} y d x \Rightarrow A=2 \int_{0}^{4} \sqrt{16-x^{2}} \cdot d x\) \(A=2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{4}\) \(\mathrm{A}=2\left[\left(0+\frac{8 \sin ^{-1} 4}{4}\right)-(0-0)\right]\) \(A=2\left[8 \sin ^{-1}(1)\right]=2 \times 8 \times \frac{\pi}{2} \Rightarrow A=8 \pi\) sq. units
Karnataka CET-2021
Application of the Integrals
86819
The area in square units of the region bounded by \(y^{2}=9 x\) and \(y=3 x\) is
86815
The area of the region bounded by the parabola \(y=x^{2}+2\) and the lines \(y=x, x=0\) and \(x=3\) is
1 15
2 6
3 \(\frac{9}{2}\)
4 \(\frac{21}{2}\)
Explanation:
(D) : equation of the parabola is \(y=x^{2}+2\) or \(\mathrm{x}^{2}=\mathrm{y}-2\) its vertex is \((0,2)\) and axis is \(y\)-axis Graphs of the curve and lines have been shown in the figure \(A=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3} x d x=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}\) \(=\left[\left(\frac{27}{3}+6\right)-0\right]-\left(\frac{9}{2}-0\right)=21 / 2\) Sq units.
nda
Application of the Integrals
86817
The area (in sq. units) bounded by the curves \(y=\) \(\sqrt{x}, 2 y-x+3=0, x\)-axis and lying in the first quadrant is
1 36
2 18
3 \(\frac{27}{4}\)
4 9
Explanation:
(D) : Given curves, \(\mathrm{y}=\sqrt{\mathrm{x}}\) and \(2 y-x+3=0\) From equation (i) and (ii), we get - \(2 \sqrt{\mathrm{x}}-(\sqrt{\mathrm{x}})^{2}+3=0\) \((\sqrt{\mathrm{x}})^{2}-2 \sqrt{\mathrm{x}}-3=0\) \((\sqrt{\mathrm{x}}-3)(\sqrt{\mathrm{x}}+1)=0\) \(\sqrt{\mathrm{x}}=3\) and \(\sqrt{\mathrm{x}}=-1\) \(\because \sqrt{\mathrm{x}}=-1\) not possible \(\therefore \mathrm{y}=\sqrt{\mathrm{x}}=3\) Now, required area, \(A=\int_{0}^{3}\left(x_{2}-x_{1}\right) \cdot d x \Rightarrow A=\int_{0}^{3}\left[(2 y+3]-y^{2}\right] . d y\) \(\mathrm{A}=\left[\left(\frac{2 \mathrm{y}^{2}}{2}\right)+(3 \mathrm{y})-\left(\frac{\mathrm{y}^{3}}{3}\right)\right]_{0}^{3}\) \(\mathrm{A}=\left[\left(\frac{2 \times 3^{2}}{2}-0\right)+(3 \times 3)-\left(\frac{3^{3}}{3}\right)\right]=9+9-\frac{27}{3}\) \(A=18-9 \Rightarrow A=9\) sq. units
COMEDK-2013
Application of the Integrals
86818
The area of the region bounded by \(y=\sqrt{16-x^{2}}\) and \(x\)-axis is
1 \(8 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(16 \pi\) sq. units
4 \(256 \pi\) sq. units
Explanation:
(A) : Given, Equation of curve, \(y=\sqrt{16-x^{2}}\) and line \(\mathrm{x}\)-axis - So, \(\mathrm{y}=0\) Here, \(\sqrt{16-\mathrm{x}^{2}}=0\) \(16-x^{2}=0\) \(\mathrm{x}^{2}=16\) \(\mathrm{x}= \pm 4\) Required area, \(\mathrm{A}=\int_{-4}^{4} \mathrm{y} \cdot \mathrm{dx}\) \(A=2 \int_{0}^{4} y d x \Rightarrow A=2 \int_{0}^{4} \sqrt{16-x^{2}} \cdot d x\) \(A=2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{4}\) \(\mathrm{A}=2\left[\left(0+\frac{8 \sin ^{-1} 4}{4}\right)-(0-0)\right]\) \(A=2\left[8 \sin ^{-1}(1)\right]=2 \times 8 \times \frac{\pi}{2} \Rightarrow A=8 \pi\) sq. units
Karnataka CET-2021
Application of the Integrals
86819
The area in square units of the region bounded by \(y^{2}=9 x\) and \(y=3 x\) is
86815
The area of the region bounded by the parabola \(y=x^{2}+2\) and the lines \(y=x, x=0\) and \(x=3\) is
1 15
2 6
3 \(\frac{9}{2}\)
4 \(\frac{21}{2}\)
Explanation:
(D) : equation of the parabola is \(y=x^{2}+2\) or \(\mathrm{x}^{2}=\mathrm{y}-2\) its vertex is \((0,2)\) and axis is \(y\)-axis Graphs of the curve and lines have been shown in the figure \(A=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3} x d x=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}\) \(=\left[\left(\frac{27}{3}+6\right)-0\right]-\left(\frac{9}{2}-0\right)=21 / 2\) Sq units.
nda
Application of the Integrals
86817
The area (in sq. units) bounded by the curves \(y=\) \(\sqrt{x}, 2 y-x+3=0, x\)-axis and lying in the first quadrant is
1 36
2 18
3 \(\frac{27}{4}\)
4 9
Explanation:
(D) : Given curves, \(\mathrm{y}=\sqrt{\mathrm{x}}\) and \(2 y-x+3=0\) From equation (i) and (ii), we get - \(2 \sqrt{\mathrm{x}}-(\sqrt{\mathrm{x}})^{2}+3=0\) \((\sqrt{\mathrm{x}})^{2}-2 \sqrt{\mathrm{x}}-3=0\) \((\sqrt{\mathrm{x}}-3)(\sqrt{\mathrm{x}}+1)=0\) \(\sqrt{\mathrm{x}}=3\) and \(\sqrt{\mathrm{x}}=-1\) \(\because \sqrt{\mathrm{x}}=-1\) not possible \(\therefore \mathrm{y}=\sqrt{\mathrm{x}}=3\) Now, required area, \(A=\int_{0}^{3}\left(x_{2}-x_{1}\right) \cdot d x \Rightarrow A=\int_{0}^{3}\left[(2 y+3]-y^{2}\right] . d y\) \(\mathrm{A}=\left[\left(\frac{2 \mathrm{y}^{2}}{2}\right)+(3 \mathrm{y})-\left(\frac{\mathrm{y}^{3}}{3}\right)\right]_{0}^{3}\) \(\mathrm{A}=\left[\left(\frac{2 \times 3^{2}}{2}-0\right)+(3 \times 3)-\left(\frac{3^{3}}{3}\right)\right]=9+9-\frac{27}{3}\) \(A=18-9 \Rightarrow A=9\) sq. units
COMEDK-2013
Application of the Integrals
86818
The area of the region bounded by \(y=\sqrt{16-x^{2}}\) and \(x\)-axis is
1 \(8 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(16 \pi\) sq. units
4 \(256 \pi\) sq. units
Explanation:
(A) : Given, Equation of curve, \(y=\sqrt{16-x^{2}}\) and line \(\mathrm{x}\)-axis - So, \(\mathrm{y}=0\) Here, \(\sqrt{16-\mathrm{x}^{2}}=0\) \(16-x^{2}=0\) \(\mathrm{x}^{2}=16\) \(\mathrm{x}= \pm 4\) Required area, \(\mathrm{A}=\int_{-4}^{4} \mathrm{y} \cdot \mathrm{dx}\) \(A=2 \int_{0}^{4} y d x \Rightarrow A=2 \int_{0}^{4} \sqrt{16-x^{2}} \cdot d x\) \(A=2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{4}\) \(\mathrm{A}=2\left[\left(0+\frac{8 \sin ^{-1} 4}{4}\right)-(0-0)\right]\) \(A=2\left[8 \sin ^{-1}(1)\right]=2 \times 8 \times \frac{\pi}{2} \Rightarrow A=8 \pi\) sq. units
Karnataka CET-2021
Application of the Integrals
86819
The area in square units of the region bounded by \(y^{2}=9 x\) and \(y=3 x\) is
86815
The area of the region bounded by the parabola \(y=x^{2}+2\) and the lines \(y=x, x=0\) and \(x=3\) is
1 15
2 6
3 \(\frac{9}{2}\)
4 \(\frac{21}{2}\)
Explanation:
(D) : equation of the parabola is \(y=x^{2}+2\) or \(\mathrm{x}^{2}=\mathrm{y}-2\) its vertex is \((0,2)\) and axis is \(y\)-axis Graphs of the curve and lines have been shown in the figure \(A=\int_{0}^{3}\left(x^{2}+2\right) d x-\int_{0}^{3} x d x=\left[\frac{x^{3}}{3}+2 x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}\) \(=\left[\left(\frac{27}{3}+6\right)-0\right]-\left(\frac{9}{2}-0\right)=21 / 2\) Sq units.
nda
Application of the Integrals
86817
The area (in sq. units) bounded by the curves \(y=\) \(\sqrt{x}, 2 y-x+3=0, x\)-axis and lying in the first quadrant is
1 36
2 18
3 \(\frac{27}{4}\)
4 9
Explanation:
(D) : Given curves, \(\mathrm{y}=\sqrt{\mathrm{x}}\) and \(2 y-x+3=0\) From equation (i) and (ii), we get - \(2 \sqrt{\mathrm{x}}-(\sqrt{\mathrm{x}})^{2}+3=0\) \((\sqrt{\mathrm{x}})^{2}-2 \sqrt{\mathrm{x}}-3=0\) \((\sqrt{\mathrm{x}}-3)(\sqrt{\mathrm{x}}+1)=0\) \(\sqrt{\mathrm{x}}=3\) and \(\sqrt{\mathrm{x}}=-1\) \(\because \sqrt{\mathrm{x}}=-1\) not possible \(\therefore \mathrm{y}=\sqrt{\mathrm{x}}=3\) Now, required area, \(A=\int_{0}^{3}\left(x_{2}-x_{1}\right) \cdot d x \Rightarrow A=\int_{0}^{3}\left[(2 y+3]-y^{2}\right] . d y\) \(\mathrm{A}=\left[\left(\frac{2 \mathrm{y}^{2}}{2}\right)+(3 \mathrm{y})-\left(\frac{\mathrm{y}^{3}}{3}\right)\right]_{0}^{3}\) \(\mathrm{A}=\left[\left(\frac{2 \times 3^{2}}{2}-0\right)+(3 \times 3)-\left(\frac{3^{3}}{3}\right)\right]=9+9-\frac{27}{3}\) \(A=18-9 \Rightarrow A=9\) sq. units
COMEDK-2013
Application of the Integrals
86818
The area of the region bounded by \(y=\sqrt{16-x^{2}}\) and \(x\)-axis is
1 \(8 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(16 \pi\) sq. units
4 \(256 \pi\) sq. units
Explanation:
(A) : Given, Equation of curve, \(y=\sqrt{16-x^{2}}\) and line \(\mathrm{x}\)-axis - So, \(\mathrm{y}=0\) Here, \(\sqrt{16-\mathrm{x}^{2}}=0\) \(16-x^{2}=0\) \(\mathrm{x}^{2}=16\) \(\mathrm{x}= \pm 4\) Required area, \(\mathrm{A}=\int_{-4}^{4} \mathrm{y} \cdot \mathrm{dx}\) \(A=2 \int_{0}^{4} y d x \Rightarrow A=2 \int_{0}^{4} \sqrt{16-x^{2}} \cdot d x\) \(A=2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{0}^{4}\) \(\mathrm{A}=2\left[\left(0+\frac{8 \sin ^{-1} 4}{4}\right)-(0-0)\right]\) \(A=2\left[8 \sin ^{-1}(1)\right]=2 \times 8 \times \frac{\pi}{2} \Rightarrow A=8 \pi\) sq. units
Karnataka CET-2021
Application of the Integrals
86819
The area in square units of the region bounded by \(y^{2}=9 x\) and \(y=3 x\) is
