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Prove the following
$\int_0^1 x \cdot e^x d x=1$
$\int_0^1 x \cdot e^x d x=1$
Solution:
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Verified Answer
Let $\mathrm{I}=\int_0^1 \mathrm{x} \cdot \mathrm{e}_{\mathrm{II}}^{\mathrm{x}} \cdot \mathrm{dx}$; Integrating by parts
$\begin{aligned}
I &=\int_0^1 x e^x d x=\left[x e^x\right]_0^1-\int_0^1 x \cdot e^x d x \\
&=e-0-\left[e^x\right]_0^1=e-(e-1)=1=\text { R.H.S. }
\end{aligned}$
$\begin{aligned}
I &=\int_0^1 x e^x d x=\left[x e^x\right]_0^1-\int_0^1 x \cdot e^x d x \\
&=e-0-\left[e^x\right]_0^1=e-(e-1)=1=\text { R.H.S. }
\end{aligned}$
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