Mathematics

# Find the value of $\displaystyle \int _{0}^{2}|1-x|\ dx$.

##### SOLUTION
Now,
$\displaystyle \int _{0}^{2}|1-x|\ dx$
$=\displaystyle \int _{0}^{1}(x-1)\ dx$$+\displaystyle \int _{1}^{2}(1-x)\ dx =\left[\dfrac{x^2}{2}-x\right]_0^1$$+\left[-\dfrac{x^2}{2}+x\right]_1^2$
$=\dfrac{1}{2}-1-\dfrac{3}{2}+1$
$=-1$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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