86865
The area of the region lying between the curve \(y=x^{2}\) and the line \(y=x+2\) in the first quadrant is
1 \(10 / 3\)
2 \(10 / 6\)
3 \(10 / 2\)
4 \(9 / 2\)
Explanation:
Exp:(a) Given the curves \(y=x^2\) and, the line \(y=x+2\), Solving, \(y=x^2 \text { and } y=x+2\) \(x^2=x+2\) \(x^2-x-2=0\) \(x(x-2)+1(x-2)=0\) \(\therefore \quad(\mathrm{x}+1)(\mathrm{x}-2)=0\) But in the first quadrant, \(x=2\) \(\therefore\) Area included \(=\int_{0}^{2}\left[(\mathrm{x}+2)-\mathrm{x}^{2}\right] \mathrm{dx}\) \(=\int_{0}^{2} x d x+2 \int_{0}^{2} d x-\int_{0}^{2} x^{2} d x=\left[\frac{x^{2}}{2}\right]_{0}^{2}+2[x]_{0}^{2}-\left[\frac{x^{3}}{3}\right]_{0}^{2}\) \(=\frac{1}{2}[4-0]+2[2-0]-\frac{1}{3}[8-0]=2+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}\)
AMU-2012
Application of the Integrals
86866
The area of the region bounded by the curves \(y=x^{2}\) and \(x=y^{2}\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 4\)
4 3
Explanation:
(A) : Given, the curves are \(y=x^{2}\) and \(x=y^{2}\) which is the form of parabola The point of intersection are \(-x=\left(x^{2}\right)^{2}=x^{4}{ }_{B}{ }^{y=x}\) \(\therefore \mathrm{x}\left(1-\mathrm{x}^{3}\right)=0 \Rightarrow \mathrm{x}=0,1\) \(\therefore\) Area enclosed \(=\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx}\) \(=\int_{0}^{1} \sqrt{x} \cdot d x-\int_{0}^{1} x^{2} d x=\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{2}{3} \times 1-\frac{1}{3}(1-0)\) \(=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
WB JEE-2014
Application of the Integrals
86867
The area bounded by \(y=x+1\) and \(y=\cos x\) and the \(\mathbf{x}\)-axis, is
1 1 sq.unit
2 \(\frac{3}{2}\) sq.unit
3 \(\frac{1}{4}\) sq.unit
4 \(\frac{1}{8}\) sq.unit
Explanation:
(B) : We have, \(y=x+1\) and \(y=\cos x\) \(\therefore\) Required area \(=\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) \(\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+[\sin x]_{0}^{\pi / 2}=\left[(0)-\left(\frac{1}{2}-1\right)\right]_{-1}^{0}+[1-0]\) \(=\frac{1}{2}+1=\frac{3}{2}\) sq unit.
WB JEE-2019
Application of the Integrals
86868
The straight line through the origin which divides the area formed by the curves \(y=2 x-x^{2}, y=0\) and \(x=1\) into two equal halves is
86865
The area of the region lying between the curve \(y=x^{2}\) and the line \(y=x+2\) in the first quadrant is
1 \(10 / 3\)
2 \(10 / 6\)
3 \(10 / 2\)
4 \(9 / 2\)
Explanation:
Exp:(a) Given the curves \(y=x^2\) and, the line \(y=x+2\), Solving, \(y=x^2 \text { and } y=x+2\) \(x^2=x+2\) \(x^2-x-2=0\) \(x(x-2)+1(x-2)=0\) \(\therefore \quad(\mathrm{x}+1)(\mathrm{x}-2)=0\) But in the first quadrant, \(x=2\) \(\therefore\) Area included \(=\int_{0}^{2}\left[(\mathrm{x}+2)-\mathrm{x}^{2}\right] \mathrm{dx}\) \(=\int_{0}^{2} x d x+2 \int_{0}^{2} d x-\int_{0}^{2} x^{2} d x=\left[\frac{x^{2}}{2}\right]_{0}^{2}+2[x]_{0}^{2}-\left[\frac{x^{3}}{3}\right]_{0}^{2}\) \(=\frac{1}{2}[4-0]+2[2-0]-\frac{1}{3}[8-0]=2+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}\)
AMU-2012
Application of the Integrals
86866
The area of the region bounded by the curves \(y=x^{2}\) and \(x=y^{2}\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 4\)
4 3
Explanation:
(A) : Given, the curves are \(y=x^{2}\) and \(x=y^{2}\) which is the form of parabola The point of intersection are \(-x=\left(x^{2}\right)^{2}=x^{4}{ }_{B}{ }^{y=x}\) \(\therefore \mathrm{x}\left(1-\mathrm{x}^{3}\right)=0 \Rightarrow \mathrm{x}=0,1\) \(\therefore\) Area enclosed \(=\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx}\) \(=\int_{0}^{1} \sqrt{x} \cdot d x-\int_{0}^{1} x^{2} d x=\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{2}{3} \times 1-\frac{1}{3}(1-0)\) \(=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
WB JEE-2014
Application of the Integrals
86867
The area bounded by \(y=x+1\) and \(y=\cos x\) and the \(\mathbf{x}\)-axis, is
1 1 sq.unit
2 \(\frac{3}{2}\) sq.unit
3 \(\frac{1}{4}\) sq.unit
4 \(\frac{1}{8}\) sq.unit
Explanation:
(B) : We have, \(y=x+1\) and \(y=\cos x\) \(\therefore\) Required area \(=\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) \(\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+[\sin x]_{0}^{\pi / 2}=\left[(0)-\left(\frac{1}{2}-1\right)\right]_{-1}^{0}+[1-0]\) \(=\frac{1}{2}+1=\frac{3}{2}\) sq unit.
WB JEE-2019
Application of the Integrals
86868
The straight line through the origin which divides the area formed by the curves \(y=2 x-x^{2}, y=0\) and \(x=1\) into two equal halves is
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Application of the Integrals
86865
The area of the region lying between the curve \(y=x^{2}\) and the line \(y=x+2\) in the first quadrant is
1 \(10 / 3\)
2 \(10 / 6\)
3 \(10 / 2\)
4 \(9 / 2\)
Explanation:
Exp:(a) Given the curves \(y=x^2\) and, the line \(y=x+2\), Solving, \(y=x^2 \text { and } y=x+2\) \(x^2=x+2\) \(x^2-x-2=0\) \(x(x-2)+1(x-2)=0\) \(\therefore \quad(\mathrm{x}+1)(\mathrm{x}-2)=0\) But in the first quadrant, \(x=2\) \(\therefore\) Area included \(=\int_{0}^{2}\left[(\mathrm{x}+2)-\mathrm{x}^{2}\right] \mathrm{dx}\) \(=\int_{0}^{2} x d x+2 \int_{0}^{2} d x-\int_{0}^{2} x^{2} d x=\left[\frac{x^{2}}{2}\right]_{0}^{2}+2[x]_{0}^{2}-\left[\frac{x^{3}}{3}\right]_{0}^{2}\) \(=\frac{1}{2}[4-0]+2[2-0]-\frac{1}{3}[8-0]=2+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}\)
AMU-2012
Application of the Integrals
86866
The area of the region bounded by the curves \(y=x^{2}\) and \(x=y^{2}\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 4\)
4 3
Explanation:
(A) : Given, the curves are \(y=x^{2}\) and \(x=y^{2}\) which is the form of parabola The point of intersection are \(-x=\left(x^{2}\right)^{2}=x^{4}{ }_{B}{ }^{y=x}\) \(\therefore \mathrm{x}\left(1-\mathrm{x}^{3}\right)=0 \Rightarrow \mathrm{x}=0,1\) \(\therefore\) Area enclosed \(=\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx}\) \(=\int_{0}^{1} \sqrt{x} \cdot d x-\int_{0}^{1} x^{2} d x=\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{2}{3} \times 1-\frac{1}{3}(1-0)\) \(=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
WB JEE-2014
Application of the Integrals
86867
The area bounded by \(y=x+1\) and \(y=\cos x\) and the \(\mathbf{x}\)-axis, is
1 1 sq.unit
2 \(\frac{3}{2}\) sq.unit
3 \(\frac{1}{4}\) sq.unit
4 \(\frac{1}{8}\) sq.unit
Explanation:
(B) : We have, \(y=x+1\) and \(y=\cos x\) \(\therefore\) Required area \(=\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) \(\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+[\sin x]_{0}^{\pi / 2}=\left[(0)-\left(\frac{1}{2}-1\right)\right]_{-1}^{0}+[1-0]\) \(=\frac{1}{2}+1=\frac{3}{2}\) sq unit.
WB JEE-2019
Application of the Integrals
86868
The straight line through the origin which divides the area formed by the curves \(y=2 x-x^{2}, y=0\) and \(x=1\) into two equal halves is
86865
The area of the region lying between the curve \(y=x^{2}\) and the line \(y=x+2\) in the first quadrant is
1 \(10 / 3\)
2 \(10 / 6\)
3 \(10 / 2\)
4 \(9 / 2\)
Explanation:
Exp:(a) Given the curves \(y=x^2\) and, the line \(y=x+2\), Solving, \(y=x^2 \text { and } y=x+2\) \(x^2=x+2\) \(x^2-x-2=0\) \(x(x-2)+1(x-2)=0\) \(\therefore \quad(\mathrm{x}+1)(\mathrm{x}-2)=0\) But in the first quadrant, \(x=2\) \(\therefore\) Area included \(=\int_{0}^{2}\left[(\mathrm{x}+2)-\mathrm{x}^{2}\right] \mathrm{dx}\) \(=\int_{0}^{2} x d x+2 \int_{0}^{2} d x-\int_{0}^{2} x^{2} d x=\left[\frac{x^{2}}{2}\right]_{0}^{2}+2[x]_{0}^{2}-\left[\frac{x^{3}}{3}\right]_{0}^{2}\) \(=\frac{1}{2}[4-0]+2[2-0]-\frac{1}{3}[8-0]=2+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}\)
AMU-2012
Application of the Integrals
86866
The area of the region bounded by the curves \(y=x^{2}\) and \(x=y^{2}\) is
1 \(1 / 3\)
2 \(1 / 2\)
3 \(1 / 4\)
4 3
Explanation:
(A) : Given, the curves are \(y=x^{2}\) and \(x=y^{2}\) which is the form of parabola The point of intersection are \(-x=\left(x^{2}\right)^{2}=x^{4}{ }_{B}{ }^{y=x}\) \(\therefore \mathrm{x}\left(1-\mathrm{x}^{3}\right)=0 \Rightarrow \mathrm{x}=0,1\) \(\therefore\) Area enclosed \(=\int_{0}^{1}\left(\sqrt{\mathrm{x}}-\mathrm{x}^{2}\right) \mathrm{dx}\) \(=\int_{0}^{1} \sqrt{x} \cdot d x-\int_{0}^{1} x^{2} d x=\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}-\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{2}{3} \times 1-\frac{1}{3}(1-0)\) \(=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) square unit
WB JEE-2014
Application of the Integrals
86867
The area bounded by \(y=x+1\) and \(y=\cos x\) and the \(\mathbf{x}\)-axis, is
1 1 sq.unit
2 \(\frac{3}{2}\) sq.unit
3 \(\frac{1}{4}\) sq.unit
4 \(\frac{1}{8}\) sq.unit
Explanation:
(B) : We have, \(y=x+1\) and \(y=\cos x\) \(\therefore\) Required area \(=\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) \(\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+[\sin x]_{0}^{\pi / 2}=\left[(0)-\left(\frac{1}{2}-1\right)\right]_{-1}^{0}+[1-0]\) \(=\frac{1}{2}+1=\frac{3}{2}\) sq unit.
WB JEE-2019
Application of the Integrals
86868
The straight line through the origin which divides the area formed by the curves \(y=2 x-x^{2}, y=0\) and \(x=1\) into two equal halves is
