First slide

Evaluation of definite integrals

Question

The slope of the tangent to the curve $y = \int_{0}^{x} \frac{d t}{1 + t^{3}}$ at the point where $x = 1$ is:

Difficult

A

1/2
B

1
C

1/4
D

None

Solution

$y = \int_{0}^{x} \frac{d t}{1 + t^{3}}$
$∴ \frac{d y}{d x} = \frac{1}{1 + x^{3}} .1 - 0 \Rightarrow {(\frac{d y}{d x})}_{x = 1} = \frac{1}{2}$ . by Leibnitz rule

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