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