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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Definite Integration

NDA 2018 · Mathematics · Definite Integration

What is ∫_0^π ^ x x d x equal to?

What is ∫0π\mathrmexsin⁡xdx\int_0^\pi \mathrme^{\mathrm{x}} \sin \mathrm{x} d \mathrm{x} equal to?

Options

  1. Aeπ+12\frac{\mathrm{e}^\pi+1}{2}
  2. Beπ−12\frac{\mathrm{e}^\pi-1}{2}
  3. Ceπ+1\mathrm{e}^\pi+1
  4. Deπ+14\frac{\mathrm{e}^\pi+1}{4}

Correct answer

A. eπ+12\frac{\mathrm{e}^\pi+1}{2}

Step-by-step solution

∫0πexsin⁡x⋅dx=II=(sin⁡x⋅ex)0π−∫0πcos⁡x⋅ex⋅dxI=(sin⁡π⋅eπ−sin⁡0⋅e0)−{[cos⁡x⋅ex]0π−∫0πsin⁡x⋅ex⋅dx}I=0−{(cos⁡π⋅eπ−cos⁡0⋅e0)−I}I=−[−eπ−1]−I⇒2I=eπ+1⇒I=eπ+12\begin{aligned} & \int_0^\pi e^x \sin x \cdot d x=I \\ & I=\left(\sin x \cdot e^x\right)_0^\pi-\int_0^\pi \cos x \cdot e^x \cdot d x \\ & I=\left(\sin \pi \cdot e^\pi-\sin 0 \cdot e^0\right)-\left\{\left[\cos x \cdot e^x\right]_0^\pi-\int_0^\pi \sin x \cdot e^x \cdot d x\right\} \\ & I=0-\left\{\left(\cos \pi \cdot e^\pi-\cos 0 \cdot e^0\right)-I\right\} \\ & I=-\left[-e^\pi-1\right]-I \\ & \Rightarrow 2 I=e^\pi+1 \Rightarrow I=\frac{e^\pi+1}{2}\end{aligned}
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