Q100 / 160 · AP EAMCET 2015 · Mathematics

The slope of the tangent to the curve $y=\int_0^x \frac{1}{1+t^3} d t$ at the point, where $x=1$ is

  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{3}$
  • Correct answer
    C
    $\frac{1}{2}$
  • D
    $1$

Solution

We have, $y=\int_0^x \frac{1}{1+t^3} d t$ Differentiating w.r.t. $x$, we get $\frac{d y}{d x}=\frac{1}{1+x^3}$ $\begin{aligned} & \text { At } x=1 \\ & \therefore \quad\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{1+1}=\frac{1}{2} \end{aligned}$

Hence, the correct answer is option: 'C'.

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