Mathematics

Solve : $$\displaystyle\int_{-2}^{2}|2x+3|\ dx$$


SOLUTION
Let $$I=\displaystyle \int_{-2}^{2}|2x+2|dx$$ 

we know
$$|2x+3|=\begin{cases} -\left( 3x+3 \right)  & x<\dfrac { -3 }{ 2 }  \\ 2x+3 & x\ge \dfrac { 3 }{ 2 }  \end{cases}$$

then

$$I=I=\displaystyle \int_{-2}^{-3/2} -2x-3\ dx +I=\displaystyle \int_{-3/2}^{2} 2x+3\ dx$$

$$=-x^2 -3x|_{-2}^{-3/2} +x^2 +3x|_{-3/2}^2$$

$$=\left (-\dfrac {9}{4}+\dfrac {9}{2}\right)-(5+6)+(4+6)-\left (\dfrac {9}{4}-\dfrac {9}{2}\right)$$

$$=\dfrac {9}{2}-2+10+\dfrac {9}{4}=8+\dfrac {9}{2}$$

$$=\dfrac {25}{2}$$
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Subjective Medium Published on 17th 09, 2020
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