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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Definite Integration

NDA 2017 · Mathematics · Definite Integration

What is ∫_ ^ -1 ^ e^2 | x x | dx equal to?

What is ∫\mathrme−1e2∣ln⁡xx∣dx\int_{\mathrme^{-1}}^{\mathrm{e}^2}\left|\frac{\ln \mathrm{x}}{\mathrm{x}}\right| \mathrm{dx} equal to?

Options

  1. A3/2
  2. B5/2
  3. C3
  4. D4

Correct answer

B. 5/2

Step-by-step solution

∫e−1e2∣log⁡xx∣dx=∫e−1e0−log⁡xxdx+∫e0e2log⁡xxdx=−12[(log⁡x)2]e−1e0+12[(log⁡x)2]e0e2=−12[0−(log⁡e−1)2]+12[(log⁡e2)2−0]=−12(−1)+12(2)2=12+42=52\begin{aligned} & \int_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right| d x=\int_{e^{-1}}^{e^0} \frac{-\log x}{x} d x+\int_{e^0}^{e^2} \frac{\log x}{x} d x \\ & =\frac{-1}{2}\left[(\log x)^2\right]_{e^{-1}}^{e^0}+\frac{1}{2}\left[(\log x)^2\right]_{e^0}^{e^2} \\ & =\frac{-1}{2}\left[0-\left(\log e^{-1}\right)^2\right]+\frac{1}{2}\left[\left(\log e^2\right)^2-0\right] \\ & =\frac{-1}{2}(-1)+\frac{1}{2}(2)^2=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}\end{aligned}
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