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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following definite integrals:$\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
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Answer

Let $\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$ Then,$\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{e}^{\text{x}}}{\text{x}}-\frac{\text{e}^{\text{x}}}{\text{x}^2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
Integrating first term by parts,$\text{I}=\bigg\{\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1-\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}\bigg\}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1+\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$

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