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

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Aicharya · Accepted Answer

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

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

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