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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}$
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Answer

$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}+\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$For
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{-\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}\big\}^0_{-2}+\int\limits^0_{-2}\text{e}^{-\text{x}}\text{ dx}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}-\text{e}^{-\text{x}}\big\}^0_{-2}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{(-1)-\big(2\text{e}^2-\text{e}^2\big)\big\}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-1-\text{e}^2\big\}$
For
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}\big\}^2_{0}-\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big\}^2_{0}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=2\text{e}^2-\text{e}^2+1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\text{e}^2+1$
Hence answer is,
$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}=-1-\text{e}^2+\text{e}^2+1=0$

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