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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Definite Integration

NDA 2015 · Mathematics · Definite Integration

∫_ -1 ^1 x|x| dx is equal to

\int_{-1}^1 \mathrm{x}|\mathrm{x}| \mathrm{dx} \text{ is equal to }

Options

  1. A0
  2. B2/3
  3. C2
  4. D-1

Correct answer

A. 0

Step-by-step solution

∫−11x∣x∣dx=∫−10x∣x∣dx+∫01x∣x∣dx=∫−10x(−x)dx+∫01x⋅xdx=−∫−10x2dx+∫01x2dx=−[x33]−10+[x33]01=−[0−(−1)33]+13[(1)3−(0)3]=−13+13=0\begin{aligned} & \int_{-1}^1 x|x| d x \\ & =\int_{-1}^0 x|x| d x+\int_0^1 x|x| d x \\ & =\int_{-1}^0 x(-x) d x+\int_0^1 x \cdot x d x \\ & =-\int_{-1}^0 x^2 d x+\int_0^1 x^2 d x \\ & =-\left[\frac{x^3}{3}\right]_{-1}^0+\left[\frac{x^3}{3}\right]_0^1 \\ & =-\left[0-\frac{(-1)^3}{3}\right]+\frac{1}{3}\left[(1)^3-(0)^3\right] \\ & =-\frac{1}{3}+\frac{1}{3}=0\end{aligned}
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