Mathematics

# Solve : $\displaystyle\int_{0}^{3}|3x-1|\ dx$

##### SOLUTION
Let $I=\displaystyle \int_{0}^3 |3x-1|dx$

we know $|2x-1|=\begin{cases} 1-3x & x<1/3 \\ 3x-1 & x\ge 1/3 \end{cases}$

$I=\displaystyle \int_{0}^{1/3} 1-3x\ dx +\displaystyle \int_{1/3}^{3} 3x-1\ dx$

$=x -\dfrac {3x^2}{2}|_0^{1/3}+\dfrac {3x^2}{2}-x|_{1/3}^3$

$=\dfrac {1}{3}-\dfrac {1}{6}+\left (\dfrac {27}{2}-2\right)-\left (\dfrac {1}{6}-\dfrac {1}{3}\right)$

$=\dfrac {2}{3}-\dfrac {1}{3}+\dfrac {27}{3}-3=\dfrac {1}{3}+\dfrac {21}{2}$

$=\dfrac {65}{6}$

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

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