86811
The area bounded by the parabola \(y^{2}=x\), straight line \(y=4\) and \(Y\)-axis (in sq. units) is
1 \(\frac{16}{3}\)
2 \(\frac{64}{3}\)
3 \(7 \sqrt{2}\)
4 None of these
Explanation:
(B) : Given parabola, \(\mathrm{y}^{2}=\mathrm{x}\) line, \(\mathrm{y}=4\) Now, \(\mathrm{y}^{2}=\mathrm{x} \Rightarrow(4)^{2}=\mathrm{x} \Rightarrow \mathrm{x}=16\) Required area \(\mathrm{A}=\int_{0}^{4} \mathrm{x}\). \(\mathrm{dy}\) \(\mathrm{A}=\int_{0}^{4} \mathrm{y}^{2} \cdot \mathrm{dy}\) \(\mathrm{A}=\left[\frac{\mathrm{y}^{3}}{3}\right]_{0}^{4}=\left[\frac{4^{3}}{3}-0\right]\) \(\Rightarrow \mathrm{A}=\frac{64}{3}\) sq. units
MHT CET-2011
Application of the Integrals
86813
The area bounded by the curves \(y^{2}=4\) ax and \(y\) \(=m x\) is \(\frac{a^{2}}{3}\) then the value of ' \(m\) ' is
1 2
2 3
3 \(1 / 2\)
4 None of these
Explanation:
(A) : Given curves are \(\mathrm{y}^{2}=4 \mathrm{ax}\) and \(\mathrm{y}=\mathrm{mx}\) \(\therefore\) Required area \(\therefore\) Required area \(=\) Area of the shaded region. \(=\int_{0}^{\frac{4 a}{m^{2}}}(\sqrt{4 a x}-m x) d x\) \(=2 \sqrt{a} \cdot\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{\frac{4 a}{m^{2}}}-m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{4 a}{m^{2}}}\) \(=2 \sqrt{\mathrm{a}} \cdot \frac{2}{3}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{3 / 2}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\left(\frac{2 \sqrt{\mathrm{a}}}{\mathrm{m}}\right)^{3}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2} \cdot \frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\) Now, area of the region \(=\frac{\mathrm{a}^{2}}{3}\) (given) \(\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2}\left(\frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\right)=\frac{\mathrm{a}^{2}}{3}\) \(\frac{32}{3} \cdot \frac{\mathrm{a}^{2}}{\mathrm{~m}^{3}}-\frac{8 \mathrm{a}^{2}}{\mathrm{~m}^{3}}=\frac{\mathrm{a}^{2}}{3} \Rightarrow \frac{8}{3 \mathrm{~m}^{3}}=\frac{1}{3} \Rightarrow \mathrm{m}^{3}=8 \Rightarrow \mathrm{m}=2\)
SRM JEE-2011
Application of the Integrals
86814
The area of the figure bounded by the curves \(y^{2}\) \(=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}-\mathbf{1}=\mathbf{0}\) is
1 \(\frac{2}{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 \(\frac{16}{3}\)
Explanation:
(D) : \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}=1, \mathrm{x}=\mathrm{y}+1\) \(\mathrm{y}^{2}=2 \mathrm{x}+1 \Rightarrow \mathrm{y}^{2}=2(1+\mathrm{y})+1\) \(\mathrm{y}^{2}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{y}^{2}-3 \mathrm{y}+\mathrm{y}-3=0\) \((\mathrm{y}-3)(\mathrm{y}+1)=0\) \(\therefore \mathrm{y}=3\) or -1 \(\therefore\) The curves \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(x-y=1\) intersect at \((4,3)\) and \((0,-1)\) and we have to find the area enclosed by the curves \(y^{2}=2 x+1\) and \(x-y=1\) \(\therefore\) Area enclosed by the graph is \(A=\int_{y_{1}}^{y_{2}}\left(\left(x_{2}\right)-\left(x_{1}\right)\right) d y \Rightarrow A=\int_{-1}^{3}(y+1)-\frac{1}{2}\left(y^{2}-1\right) d y\) \(=\int_{-1}^{3}\left(y+1-\frac{y^{2}}{2}+\frac{1}{2}\right) d y=\int_{-1}^{3}\left(y-\frac{y^{2}}{2}+\frac{3}{2}\right) d y\) \(=\left\{\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{6}+\frac{3 \mathrm{y}}{2}\right\}_{-1}^{3}=\frac{9}{2}-\frac{27}{6}+\frac{9}{2}-\left(\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right)\) \(=\frac{32}{6}=\frac{16}{3}\)
86811
The area bounded by the parabola \(y^{2}=x\), straight line \(y=4\) and \(Y\)-axis (in sq. units) is
1 \(\frac{16}{3}\)
2 \(\frac{64}{3}\)
3 \(7 \sqrt{2}\)
4 None of these
Explanation:
(B) : Given parabola, \(\mathrm{y}^{2}=\mathrm{x}\) line, \(\mathrm{y}=4\) Now, \(\mathrm{y}^{2}=\mathrm{x} \Rightarrow(4)^{2}=\mathrm{x} \Rightarrow \mathrm{x}=16\) Required area \(\mathrm{A}=\int_{0}^{4} \mathrm{x}\). \(\mathrm{dy}\) \(\mathrm{A}=\int_{0}^{4} \mathrm{y}^{2} \cdot \mathrm{dy}\) \(\mathrm{A}=\left[\frac{\mathrm{y}^{3}}{3}\right]_{0}^{4}=\left[\frac{4^{3}}{3}-0\right]\) \(\Rightarrow \mathrm{A}=\frac{64}{3}\) sq. units
MHT CET-2011
Application of the Integrals
86813
The area bounded by the curves \(y^{2}=4\) ax and \(y\) \(=m x\) is \(\frac{a^{2}}{3}\) then the value of ' \(m\) ' is
1 2
2 3
3 \(1 / 2\)
4 None of these
Explanation:
(A) : Given curves are \(\mathrm{y}^{2}=4 \mathrm{ax}\) and \(\mathrm{y}=\mathrm{mx}\) \(\therefore\) Required area \(\therefore\) Required area \(=\) Area of the shaded region. \(=\int_{0}^{\frac{4 a}{m^{2}}}(\sqrt{4 a x}-m x) d x\) \(=2 \sqrt{a} \cdot\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{\frac{4 a}{m^{2}}}-m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{4 a}{m^{2}}}\) \(=2 \sqrt{\mathrm{a}} \cdot \frac{2}{3}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{3 / 2}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\left(\frac{2 \sqrt{\mathrm{a}}}{\mathrm{m}}\right)^{3}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2} \cdot \frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\) Now, area of the region \(=\frac{\mathrm{a}^{2}}{3}\) (given) \(\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2}\left(\frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\right)=\frac{\mathrm{a}^{2}}{3}\) \(\frac{32}{3} \cdot \frac{\mathrm{a}^{2}}{\mathrm{~m}^{3}}-\frac{8 \mathrm{a}^{2}}{\mathrm{~m}^{3}}=\frac{\mathrm{a}^{2}}{3} \Rightarrow \frac{8}{3 \mathrm{~m}^{3}}=\frac{1}{3} \Rightarrow \mathrm{m}^{3}=8 \Rightarrow \mathrm{m}=2\)
SRM JEE-2011
Application of the Integrals
86814
The area of the figure bounded by the curves \(y^{2}\) \(=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}-\mathbf{1}=\mathbf{0}\) is
1 \(\frac{2}{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 \(\frac{16}{3}\)
Explanation:
(D) : \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}=1, \mathrm{x}=\mathrm{y}+1\) \(\mathrm{y}^{2}=2 \mathrm{x}+1 \Rightarrow \mathrm{y}^{2}=2(1+\mathrm{y})+1\) \(\mathrm{y}^{2}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{y}^{2}-3 \mathrm{y}+\mathrm{y}-3=0\) \((\mathrm{y}-3)(\mathrm{y}+1)=0\) \(\therefore \mathrm{y}=3\) or -1 \(\therefore\) The curves \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(x-y=1\) intersect at \((4,3)\) and \((0,-1)\) and we have to find the area enclosed by the curves \(y^{2}=2 x+1\) and \(x-y=1\) \(\therefore\) Area enclosed by the graph is \(A=\int_{y_{1}}^{y_{2}}\left(\left(x_{2}\right)-\left(x_{1}\right)\right) d y \Rightarrow A=\int_{-1}^{3}(y+1)-\frac{1}{2}\left(y^{2}-1\right) d y\) \(=\int_{-1}^{3}\left(y+1-\frac{y^{2}}{2}+\frac{1}{2}\right) d y=\int_{-1}^{3}\left(y-\frac{y^{2}}{2}+\frac{3}{2}\right) d y\) \(=\left\{\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{6}+\frac{3 \mathrm{y}}{2}\right\}_{-1}^{3}=\frac{9}{2}-\frac{27}{6}+\frac{9}{2}-\left(\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right)\) \(=\frac{32}{6}=\frac{16}{3}\)
86811
The area bounded by the parabola \(y^{2}=x\), straight line \(y=4\) and \(Y\)-axis (in sq. units) is
1 \(\frac{16}{3}\)
2 \(\frac{64}{3}\)
3 \(7 \sqrt{2}\)
4 None of these
Explanation:
(B) : Given parabola, \(\mathrm{y}^{2}=\mathrm{x}\) line, \(\mathrm{y}=4\) Now, \(\mathrm{y}^{2}=\mathrm{x} \Rightarrow(4)^{2}=\mathrm{x} \Rightarrow \mathrm{x}=16\) Required area \(\mathrm{A}=\int_{0}^{4} \mathrm{x}\). \(\mathrm{dy}\) \(\mathrm{A}=\int_{0}^{4} \mathrm{y}^{2} \cdot \mathrm{dy}\) \(\mathrm{A}=\left[\frac{\mathrm{y}^{3}}{3}\right]_{0}^{4}=\left[\frac{4^{3}}{3}-0\right]\) \(\Rightarrow \mathrm{A}=\frac{64}{3}\) sq. units
MHT CET-2011
Application of the Integrals
86813
The area bounded by the curves \(y^{2}=4\) ax and \(y\) \(=m x\) is \(\frac{a^{2}}{3}\) then the value of ' \(m\) ' is
1 2
2 3
3 \(1 / 2\)
4 None of these
Explanation:
(A) : Given curves are \(\mathrm{y}^{2}=4 \mathrm{ax}\) and \(\mathrm{y}=\mathrm{mx}\) \(\therefore\) Required area \(\therefore\) Required area \(=\) Area of the shaded region. \(=\int_{0}^{\frac{4 a}{m^{2}}}(\sqrt{4 a x}-m x) d x\) \(=2 \sqrt{a} \cdot\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{\frac{4 a}{m^{2}}}-m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{4 a}{m^{2}}}\) \(=2 \sqrt{\mathrm{a}} \cdot \frac{2}{3}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{3 / 2}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\left(\frac{2 \sqrt{\mathrm{a}}}{\mathrm{m}}\right)^{3}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2} \cdot \frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\) Now, area of the region \(=\frac{\mathrm{a}^{2}}{3}\) (given) \(\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2}\left(\frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\right)=\frac{\mathrm{a}^{2}}{3}\) \(\frac{32}{3} \cdot \frac{\mathrm{a}^{2}}{\mathrm{~m}^{3}}-\frac{8 \mathrm{a}^{2}}{\mathrm{~m}^{3}}=\frac{\mathrm{a}^{2}}{3} \Rightarrow \frac{8}{3 \mathrm{~m}^{3}}=\frac{1}{3} \Rightarrow \mathrm{m}^{3}=8 \Rightarrow \mathrm{m}=2\)
SRM JEE-2011
Application of the Integrals
86814
The area of the figure bounded by the curves \(y^{2}\) \(=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}-\mathbf{1}=\mathbf{0}\) is
1 \(\frac{2}{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 \(\frac{16}{3}\)
Explanation:
(D) : \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}=1, \mathrm{x}=\mathrm{y}+1\) \(\mathrm{y}^{2}=2 \mathrm{x}+1 \Rightarrow \mathrm{y}^{2}=2(1+\mathrm{y})+1\) \(\mathrm{y}^{2}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{y}^{2}-3 \mathrm{y}+\mathrm{y}-3=0\) \((\mathrm{y}-3)(\mathrm{y}+1)=0\) \(\therefore \mathrm{y}=3\) or -1 \(\therefore\) The curves \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(x-y=1\) intersect at \((4,3)\) and \((0,-1)\) and we have to find the area enclosed by the curves \(y^{2}=2 x+1\) and \(x-y=1\) \(\therefore\) Area enclosed by the graph is \(A=\int_{y_{1}}^{y_{2}}\left(\left(x_{2}\right)-\left(x_{1}\right)\right) d y \Rightarrow A=\int_{-1}^{3}(y+1)-\frac{1}{2}\left(y^{2}-1\right) d y\) \(=\int_{-1}^{3}\left(y+1-\frac{y^{2}}{2}+\frac{1}{2}\right) d y=\int_{-1}^{3}\left(y-\frac{y^{2}}{2}+\frac{3}{2}\right) d y\) \(=\left\{\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{6}+\frac{3 \mathrm{y}}{2}\right\}_{-1}^{3}=\frac{9}{2}-\frac{27}{6}+\frac{9}{2}-\left(\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right)\) \(=\frac{32}{6}=\frac{16}{3}\)
86811
The area bounded by the parabola \(y^{2}=x\), straight line \(y=4\) and \(Y\)-axis (in sq. units) is
1 \(\frac{16}{3}\)
2 \(\frac{64}{3}\)
3 \(7 \sqrt{2}\)
4 None of these
Explanation:
(B) : Given parabola, \(\mathrm{y}^{2}=\mathrm{x}\) line, \(\mathrm{y}=4\) Now, \(\mathrm{y}^{2}=\mathrm{x} \Rightarrow(4)^{2}=\mathrm{x} \Rightarrow \mathrm{x}=16\) Required area \(\mathrm{A}=\int_{0}^{4} \mathrm{x}\). \(\mathrm{dy}\) \(\mathrm{A}=\int_{0}^{4} \mathrm{y}^{2} \cdot \mathrm{dy}\) \(\mathrm{A}=\left[\frac{\mathrm{y}^{3}}{3}\right]_{0}^{4}=\left[\frac{4^{3}}{3}-0\right]\) \(\Rightarrow \mathrm{A}=\frac{64}{3}\) sq. units
MHT CET-2011
Application of the Integrals
86813
The area bounded by the curves \(y^{2}=4\) ax and \(y\) \(=m x\) is \(\frac{a^{2}}{3}\) then the value of ' \(m\) ' is
1 2
2 3
3 \(1 / 2\)
4 None of these
Explanation:
(A) : Given curves are \(\mathrm{y}^{2}=4 \mathrm{ax}\) and \(\mathrm{y}=\mathrm{mx}\) \(\therefore\) Required area \(\therefore\) Required area \(=\) Area of the shaded region. \(=\int_{0}^{\frac{4 a}{m^{2}}}(\sqrt{4 a x}-m x) d x\) \(=2 \sqrt{a} \cdot\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{\frac{4 a}{m^{2}}}-m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{4 a}{m^{2}}}\) \(=2 \sqrt{\mathrm{a}} \cdot \frac{2}{3}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{3 / 2}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\left(\frac{2 \sqrt{\mathrm{a}}}{\mathrm{m}}\right)^{3}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2} \cdot \frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\) Now, area of the region \(=\frac{\mathrm{a}^{2}}{3}\) (given) \(\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2}\left(\frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\right)=\frac{\mathrm{a}^{2}}{3}\) \(\frac{32}{3} \cdot \frac{\mathrm{a}^{2}}{\mathrm{~m}^{3}}-\frac{8 \mathrm{a}^{2}}{\mathrm{~m}^{3}}=\frac{\mathrm{a}^{2}}{3} \Rightarrow \frac{8}{3 \mathrm{~m}^{3}}=\frac{1}{3} \Rightarrow \mathrm{m}^{3}=8 \Rightarrow \mathrm{m}=2\)
SRM JEE-2011
Application of the Integrals
86814
The area of the figure bounded by the curves \(y^{2}\) \(=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}-\mathbf{1}=\mathbf{0}\) is
1 \(\frac{2}{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 \(\frac{16}{3}\)
Explanation:
(D) : \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}=1, \mathrm{x}=\mathrm{y}+1\) \(\mathrm{y}^{2}=2 \mathrm{x}+1 \Rightarrow \mathrm{y}^{2}=2(1+\mathrm{y})+1\) \(\mathrm{y}^{2}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{y}^{2}-3 \mathrm{y}+\mathrm{y}-3=0\) \((\mathrm{y}-3)(\mathrm{y}+1)=0\) \(\therefore \mathrm{y}=3\) or -1 \(\therefore\) The curves \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(x-y=1\) intersect at \((4,3)\) and \((0,-1)\) and we have to find the area enclosed by the curves \(y^{2}=2 x+1\) and \(x-y=1\) \(\therefore\) Area enclosed by the graph is \(A=\int_{y_{1}}^{y_{2}}\left(\left(x_{2}\right)-\left(x_{1}\right)\right) d y \Rightarrow A=\int_{-1}^{3}(y+1)-\frac{1}{2}\left(y^{2}-1\right) d y\) \(=\int_{-1}^{3}\left(y+1-\frac{y^{2}}{2}+\frac{1}{2}\right) d y=\int_{-1}^{3}\left(y-\frac{y^{2}}{2}+\frac{3}{2}\right) d y\) \(=\left\{\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{6}+\frac{3 \mathrm{y}}{2}\right\}_{-1}^{3}=\frac{9}{2}-\frac{27}{6}+\frac{9}{2}-\left(\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right)\) \(=\frac{32}{6}=\frac{16}{3}\)
86811
The area bounded by the parabola \(y^{2}=x\), straight line \(y=4\) and \(Y\)-axis (in sq. units) is
1 \(\frac{16}{3}\)
2 \(\frac{64}{3}\)
3 \(7 \sqrt{2}\)
4 None of these
Explanation:
(B) : Given parabola, \(\mathrm{y}^{2}=\mathrm{x}\) line, \(\mathrm{y}=4\) Now, \(\mathrm{y}^{2}=\mathrm{x} \Rightarrow(4)^{2}=\mathrm{x} \Rightarrow \mathrm{x}=16\) Required area \(\mathrm{A}=\int_{0}^{4} \mathrm{x}\). \(\mathrm{dy}\) \(\mathrm{A}=\int_{0}^{4} \mathrm{y}^{2} \cdot \mathrm{dy}\) \(\mathrm{A}=\left[\frac{\mathrm{y}^{3}}{3}\right]_{0}^{4}=\left[\frac{4^{3}}{3}-0\right]\) \(\Rightarrow \mathrm{A}=\frac{64}{3}\) sq. units
MHT CET-2011
Application of the Integrals
86813
The area bounded by the curves \(y^{2}=4\) ax and \(y\) \(=m x\) is \(\frac{a^{2}}{3}\) then the value of ' \(m\) ' is
1 2
2 3
3 \(1 / 2\)
4 None of these
Explanation:
(A) : Given curves are \(\mathrm{y}^{2}=4 \mathrm{ax}\) and \(\mathrm{y}=\mathrm{mx}\) \(\therefore\) Required area \(\therefore\) Required area \(=\) Area of the shaded region. \(=\int_{0}^{\frac{4 a}{m^{2}}}(\sqrt{4 a x}-m x) d x\) \(=2 \sqrt{a} \cdot\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{\frac{4 a}{m^{2}}}-m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{4 a}{m^{2}}}\) \(=2 \sqrt{\mathrm{a}} \cdot \frac{2}{3}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{3 / 2}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\left(\frac{2 \sqrt{\mathrm{a}}}{\mathrm{m}}\right)^{3}\right]-\frac{\mathrm{m}}{2}\left[\left(\frac{4 \mathrm{a}}{\mathrm{m}^{2}}\right)^{2}\right]\) \(=\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2} \cdot \frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\) Now, area of the region \(=\frac{\mathrm{a}^{2}}{3}\) (given) \(\frac{4 \sqrt{\mathrm{a}}}{3}\left[\frac{8 \mathrm{a} \sqrt{\mathrm{a}}}{\mathrm{m}^{3}}\right]-\frac{\mathrm{m}}{2}\left(\frac{16 \mathrm{a}^{2}}{\mathrm{~m}^{4}}\right)=\frac{\mathrm{a}^{2}}{3}\) \(\frac{32}{3} \cdot \frac{\mathrm{a}^{2}}{\mathrm{~m}^{3}}-\frac{8 \mathrm{a}^{2}}{\mathrm{~m}^{3}}=\frac{\mathrm{a}^{2}}{3} \Rightarrow \frac{8}{3 \mathrm{~m}^{3}}=\frac{1}{3} \Rightarrow \mathrm{m}^{3}=8 \Rightarrow \mathrm{m}=2\)
SRM JEE-2011
Application of the Integrals
86814
The area of the figure bounded by the curves \(y^{2}\) \(=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}-\mathbf{1}=\mathbf{0}\) is
1 \(\frac{2}{3}\)
2 \(\frac{4}{3}\)
3 \(\frac{8}{3}\)
4 \(\frac{16}{3}\)
Explanation:
(D) : \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(\mathrm{x}-\mathrm{y}=1, \mathrm{x}=\mathrm{y}+1\) \(\mathrm{y}^{2}=2 \mathrm{x}+1 \Rightarrow \mathrm{y}^{2}=2(1+\mathrm{y})+1\) \(\mathrm{y}^{2}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{y}^{2}-3 \mathrm{y}+\mathrm{y}-3=0\) \((\mathrm{y}-3)(\mathrm{y}+1)=0\) \(\therefore \mathrm{y}=3\) or -1 \(\therefore\) The curves \(\mathrm{y}^{2}=2 \mathrm{x}+1\) and \(x-y=1\) intersect at \((4,3)\) and \((0,-1)\) and we have to find the area enclosed by the curves \(y^{2}=2 x+1\) and \(x-y=1\) \(\therefore\) Area enclosed by the graph is \(A=\int_{y_{1}}^{y_{2}}\left(\left(x_{2}\right)-\left(x_{1}\right)\right) d y \Rightarrow A=\int_{-1}^{3}(y+1)-\frac{1}{2}\left(y^{2}-1\right) d y\) \(=\int_{-1}^{3}\left(y+1-\frac{y^{2}}{2}+\frac{1}{2}\right) d y=\int_{-1}^{3}\left(y-\frac{y^{2}}{2}+\frac{3}{2}\right) d y\) \(=\left\{\frac{\mathrm{y}^{2}}{2}-\frac{\mathrm{y}^{3}}{6}+\frac{3 \mathrm{y}}{2}\right\}_{-1}^{3}=\frac{9}{2}-\frac{27}{6}+\frac{9}{2}-\left(\frac{1}{2}+\frac{1}{6}-\frac{3}{2}\right)\) \(=\frac{32}{6}=\frac{16}{3}\)
