First slide

Evaluation of definite integrals

Question

The equation of the tangent to the curve
$y = \int_{x^{2}}^{x^{3}} \frac{d t}{\sqrt{1 + t^{2}}} at x = 1$ is

Easy

A

$\sqrt{2} y + 1 = x$
B

$\sqrt{3} x + 1 = y$
C

$\sqrt{3} x + 1 + \sqrt{3} = y$
D

none of these

Solution

$\begin{array}{l} \frac{d y}{d x} = {\frac{1}{\sqrt{1 + t^{2}}}|}_{t = x^{3}} \frac{d}{d x} (x^{3}) - {\frac{1}{\sqrt{1 + t^{2}}}|}_{t = x^{2}} \frac{d}{d x} (x^{2}) \\ = \frac{3 x^{2}}{\sqrt{1 + x^{6}}} - \frac{2 x}{\sqrt{1 + x^{4}}} \\ {\frac{d y}{d x}|}_{x = 1} = \frac{1}{\sqrt{2}} \end{array}$
Also $y (1) = 0 .$ So the equation of tangent is $y - 0$
$= \frac{1}{\sqrt{2}} (x - 1)$

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