87003
The area (in sq units) in the first quadrant bounded by the parabola, \(y=x^{2}+1\), the tangent to it at the point \((2,5)\) and the coordinate axes is
1 \(\frac{14}{3}\)
2 \(\frac{187}{24}\)
3 \(\frac{8}{3}\)
4 \(\frac{37}{24}\)
Explanation:
(D) : We have parabola, \(\mathrm{y}=\mathrm{x}^{2}+1\) Tangent at point \((2,5)\) \(y=x^{2}+1\) and the slope \((m)=4\) Therefore, the equation of tangent is \((2,5)\) \(y-5=4(x-2)\) \(y-5=4 x-8\) \(y=4 x-3\) Hence, the required area is - \(\mathrm{A}=\int_{0}^{2}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}-\left(\frac{1}{2} \times \frac{5}{4} \times 5\right) \Rightarrow \mathrm{A}=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{0}^{2}-\frac{25}{8}\) \(\mathrm{A}=\frac{14}{3}-\frac{25}{8}=\frac{37}{24}\) square unit
JEE Main-2019-11.01.2019
Application of the Integrals
87004
Area of the region bounded by the curve \(y^{2}=\) \(4 x, Y\)-axis and the line \(y=3\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
87003
The area (in sq units) in the first quadrant bounded by the parabola, \(y=x^{2}+1\), the tangent to it at the point \((2,5)\) and the coordinate axes is
1 \(\frac{14}{3}\)
2 \(\frac{187}{24}\)
3 \(\frac{8}{3}\)
4 \(\frac{37}{24}\)
Explanation:
(D) : We have parabola, \(\mathrm{y}=\mathrm{x}^{2}+1\) Tangent at point \((2,5)\) \(y=x^{2}+1\) and the slope \((m)=4\) Therefore, the equation of tangent is \((2,5)\) \(y-5=4(x-2)\) \(y-5=4 x-8\) \(y=4 x-3\) Hence, the required area is - \(\mathrm{A}=\int_{0}^{2}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}-\left(\frac{1}{2} \times \frac{5}{4} \times 5\right) \Rightarrow \mathrm{A}=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{0}^{2}-\frac{25}{8}\) \(\mathrm{A}=\frac{14}{3}-\frac{25}{8}=\frac{37}{24}\) square unit
JEE Main-2019-11.01.2019
Application of the Integrals
87004
Area of the region bounded by the curve \(y^{2}=\) \(4 x, Y\)-axis and the line \(y=3\) is
87003
The area (in sq units) in the first quadrant bounded by the parabola, \(y=x^{2}+1\), the tangent to it at the point \((2,5)\) and the coordinate axes is
1 \(\frac{14}{3}\)
2 \(\frac{187}{24}\)
3 \(\frac{8}{3}\)
4 \(\frac{37}{24}\)
Explanation:
(D) : We have parabola, \(\mathrm{y}=\mathrm{x}^{2}+1\) Tangent at point \((2,5)\) \(y=x^{2}+1\) and the slope \((m)=4\) Therefore, the equation of tangent is \((2,5)\) \(y-5=4(x-2)\) \(y-5=4 x-8\) \(y=4 x-3\) Hence, the required area is - \(\mathrm{A}=\int_{0}^{2}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}-\left(\frac{1}{2} \times \frac{5}{4} \times 5\right) \Rightarrow \mathrm{A}=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{0}^{2}-\frac{25}{8}\) \(\mathrm{A}=\frac{14}{3}-\frac{25}{8}=\frac{37}{24}\) square unit
JEE Main-2019-11.01.2019
Application of the Integrals
87004
Area of the region bounded by the curve \(y^{2}=\) \(4 x, Y\)-axis and the line \(y=3\) is
87003
The area (in sq units) in the first quadrant bounded by the parabola, \(y=x^{2}+1\), the tangent to it at the point \((2,5)\) and the coordinate axes is
1 \(\frac{14}{3}\)
2 \(\frac{187}{24}\)
3 \(\frac{8}{3}\)
4 \(\frac{37}{24}\)
Explanation:
(D) : We have parabola, \(\mathrm{y}=\mathrm{x}^{2}+1\) Tangent at point \((2,5)\) \(y=x^{2}+1\) and the slope \((m)=4\) Therefore, the equation of tangent is \((2,5)\) \(y-5=4(x-2)\) \(y-5=4 x-8\) \(y=4 x-3\) Hence, the required area is - \(\mathrm{A}=\int_{0}^{2}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}-\left(\frac{1}{2} \times \frac{5}{4} \times 5\right) \Rightarrow \mathrm{A}=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{0}^{2}-\frac{25}{8}\) \(\mathrm{A}=\frac{14}{3}-\frac{25}{8}=\frac{37}{24}\) square unit
JEE Main-2019-11.01.2019
Application of the Integrals
87004
Area of the region bounded by the curve \(y^{2}=\) \(4 x, Y\)-axis and the line \(y=3\) is
