If $\int e^{x^2} \cdot x^3 d x=e^{x^2} f(x)+C$ (where $C$ is a constant of integration) and $f(1)=0$, then value of $f(2)$ will be

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Aicharya · Accepted Answer

$\int e^{x^2 x^3}=\frac{1}{2} \int x^2 e^{x^2} 2 x d x=\frac{1}{2} \int t e^t d t \quad\left[\operatorname{let} x^2=tight]$ Integrating by parts $\begin{aligned} & =\frac{1}{2}\left[t \cdot e^t-e^tight]+C \ & =\frac{1}{2}\left[x^2 e^{x^2}-e^{x^2}ight]+C=e^{x^2} \cdot \frac{1}{2}\left(x^2-1ight)+C \ & \Rightarrow f(x)=\frac{1}{2}\left(x^2-1ight) \Rightarrow f(2)=\frac{3}{2}\end{aligned}$

If $\int e^{x^2} \cdot x^3 d x=e^{x^2} f(x)+C$ (where $C$ is a constant of integration) and $f(1)=0$, then…

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