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  • Stuck here, help me understand: - The area( in sq. units) in the first quadrant bounded by the parabola,, the tangent to it at the p - Integral Calculus - JEE Main

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:

  • Option 1)

    \frac{8}{3}

  • Option 2)

    \frac{187}{24}

  • Option 3)

    \frac{14}{3}

  • Option 4)

    \frac{37}{24}

Answers (1)

best_answer

 

Area between two curves -

\\*I\! \! f \: f\left ( x \right )\geqslant g\left ( x \right )\\* in[a,c)\: \: and \: \: g\left ( x \right ) \geqslant f\left ( x \right )\:in(c,b]\\* Then\: area = \\*\\* \int_{a}^{c}\left ( f\left ( x \right )-g\left ( x \right ) \right )dx+\int_{c}^{b}\left ( g\left ( x \right )-f\left ( x \right ) \right )dx

- wherein

Slope of tangent, m=\frac{dy}{dx}=2x

                       at (2,5)     m=4

Equation of tangent ,  (y - 5) = 4 (x - 2)

                          =>  4x - y = 3  or    y = 4x - 3

Area = \int_{0}^{2}(x^{2}+1)dx-\int_{\frac{3}{4}}^{2}(4x-3)dx

       =\left [ (\frac{x^{3}}{3}+x) \right ]_{0}^{2}-\left [ 4x-3 \right ]_{\frac{3}{4}}^{2}

      =\frac{37}{24}

 

 

 


Option 1)

\frac{8}{3}

Option 2)

\frac{187}{24}

Option 3)

\frac{14}{3}

Option 4)

\frac{37}{24}
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