If $\int \mathrm{e}^{x^2}. x^3 \mathrm{~d} x=\mathrm{e}^{x^2} \mathrm{f}(x)+\mathrm{c}$ and…

If $\int \mathrm{e}^{x^2}. x^3 \mathrm{~d} x=\mathrm{e}^{x^2} \mathrm{f}(x)+\mathrm{c}$ and $\mathrm{f}(\mathrm{I})=0$ (where c is a constant of integration), then the value of $\mathrm{f}(x)$ is
  1. $\frac{x-1}{2}$
  2. $\frac{x^2+1}{2}$
  3. $\frac{x+1}{2}$
  4. $\frac{x^2-1}{2}$

Solution

$\begin{aligned} & \text {Let } \mathrm{I}=\int \mathrm{e}^{\mathrm{x}^2} \cdot x^3 \mathrm{~d} x \\ & \text {Put } x^2=\mathrm{t} \\ & \Rightarrow 2 x \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad I=\frac{1}{2} \int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{t} \mathrm{dt} \\ & =\frac{1}{2}\left(t \cdot e^t-\int 1 \cdot e^t\right) \\ & =\frac{1}{2}\left(\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)+\mathrm{c} \\ & =\frac{1}{2} \mathrm{e}^{\mathrm{t}}(\mathrm{t}-1)+\mathrm{c}=\frac{1}{2} \mathrm{e}^{x^2}\left(x^2-1\right)+\mathrm{c} \\ & \therefore \quad \mathrm{f}(x)=\frac{x^2-1}{2} \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 2)

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