The slope of the tangent to the curve $y=\int\limits_0^x \frac{1}{1+t^3} d t$ at the point where x = 1, is
Answer & explanation
Correct answer: option 1
We have,
$y=\int\limits_0^x \frac{1}{1+t^3} d t \Rightarrow \frac{d y}{d x}=\frac{1}{1+x^3} \Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{2}$