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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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(where $C$ is a constant of integration)
and $f(1)=0$, then value of $f(2)$ will be
Solution:
2531 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{2}$
$\int e^{x^2 x^3}=\frac{1}{2} \int x^2 e^{x^2} 2 x d x=\frac{1}{2} \int t e^t d t \quad\left[\operatorname{let} x^2=t\right]$
Integrating by parts
$\begin{aligned} & =\frac{1}{2}\left[t \cdot e^t-e^t\right]+C \\ & =\frac{1}{2}\left[x^2 e^{x^2}-e^{x^2}\right]+C=e^{x^2} \cdot \frac{1}{2}\left(x^2-1\right)+C \\ & \Rightarrow f(x)=\frac{1}{2}\left(x^2-1\right) \Rightarrow f(2)=\frac{3}{2}\end{aligned}$
Integrating by parts
$\begin{aligned} & =\frac{1}{2}\left[t \cdot e^t-e^t\right]+C \\ & =\frac{1}{2}\left[x^2 e^{x^2}-e^{x^2}\right]+C=e^{x^2} \cdot \frac{1}{2}\left(x^2-1\right)+C \\ & \Rightarrow f(x)=\frac{1}{2}\left(x^2-1\right) \Rightarrow f(2)=\frac{3}{2}\end{aligned}$
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