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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS3 Marks
Question
Evaluate the definite integral in Exercise:
$\int_{0}^{1}(\text{x}\text{e}^{\text{x}}+\sin\frac{\pi\text{x}}{4})\text{dx}$
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Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{1}\bigg(\text{xe}^{\text{x}}+\sin\frac{\pi\text{x}}{4}\bigg)\text{dx}$
$\int\bigg(\text{xe}^{\text{x}}+\sin\frac{\pi\text{x}}{4}\bigg)\text{dx}=\text{x}\int\text{e}^\text{x}\text{dx}-\int\left\{\bigg(\frac{\text{d}}{\text{dx}}\text{x}\bigg)\int\text{e}^\text{x}\text{dx}\right\}\text{dx}+\left\{\frac{-\cos\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\right\}$
$=\text{xe}^\text{x}-\int\text{e}^\text{x}\text{dx}-\frac{4}{\pi}\cos\frac{\pi\text{x}}{4}$
$=\text{xe}^\text{x}-\text{e}^\text{x}-\frac{4}{\pi}\cos\frac{\pi\text{x}}{4}$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(1)-\text{F}(0)$
$=\bigg(1.\text{e}^{1}-\text{e}^{1}-\frac{4}{\pi}\cos\frac{\pi}{4}\bigg)-\bigg(0.\text{e}^{0}-\text{e}^{0}-\frac{4}{\pi}\cos0\bigg)$
$=\text{e}-\text{e}-\frac{4}{\pi}\bigg(\frac{1}{\sqrt{2}}\bigg)+1+\frac{4}{\pi}$
$=1+\frac{4}{\pi}-\frac{2\sqrt{2}}{\pi}$

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