Practicing Success
Practicing Success
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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 |
$\frac{1}{2}$ 1 $\frac{1}{4}$ non-existent |
$\frac{1}{2}$ |
We have, $y=\int\limits_0^x \frac{1}{1+t^3} d t \Rightarrow \frac{d y}{d x}=\frac{1}{1+x^3} \Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{2}$ |
