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Definite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals4 Marks
Question
Evaluate the following definite integrals:
$\int\limits_{0}^{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
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Answer

We have,
$\int_{0}^\limits{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
$=\int_{0}^\limits{1}\text{xe}^{\text{x}}\text{ dx}+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
Applying by parts in 1st integral we get,
$=\text{x}\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}-\int_{0}^\limits{1}\big(\int\text{e}^{\text{x}}\text{ dx}\big)+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
$=\big[\text{xe}^{\text{x}}\big]^1_0-\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}+\bigg[\frac{\sin\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\bigg]^1_0$
$=\big[\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}-0\Big]$
$=\big[\text{e}^{\text{x}}(\text{x}-1)\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}\Big]$
$=0+1+\frac{4}{\pi\sqrt{2}}$
$=1+\frac{2\sqrt{2}}{\pi}$

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