Question
Prove $\int_{0}^{1} x e^{x} d x=1$
✓
Answer
To Prove: $\int_{0}^{1} x e^{x} d x=1$
Let $\mathrm{I}=\int_{0}^{1} \mathrm{xe}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
because, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} . \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{1} \mathrm{xe}^{\mathrm{x}} \mathrm{dx}=[\mathrm{x}]_{0}^{1} \cdot \int_{0}^{1} \mathrm{e}^{\mathrm{x}} \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{d}}{\mathrm{dx}}x \cdot\left\{\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}\right\} \cdot \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 \cdot e^{x} d x$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{1}-e^{0}\right]$
$\Rightarrow \int_{0}^{1} x e^{x} d x=e-0-e+1$
$\Rightarrow \int_{0}^{1} x e^{x} d x=1$
Hence Proved.
Let $\mathrm{I}=\int_{0}^{1} \mathrm{xe}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
because, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} . \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{1} \mathrm{xe}^{\mathrm{x}} \mathrm{dx}=[\mathrm{x}]_{0}^{1} \cdot \int_{0}^{1} \mathrm{e}^{\mathrm{x}} \mathrm{dx}-\int_{0}^{1} \frac{\mathrm{d}}{\mathrm{dx}}x \cdot\left\{\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}\right\} \cdot \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 \cdot e^{x} d x$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x e^{x} d x=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{1}-e^{0}\right]$
$\Rightarrow \int_{0}^{1} x e^{x} d x=e-0-e+1$
$\Rightarrow \int_{0}^{1} x e^{x} d x=1$
Hence Proved.
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