Question

1. integrate |1-x^2|dx from limit -2 to 2.
2. Integrate |2x+3|dx from limit -2 to 2

Answer 1

The problem (1):-

Since $1-x^{2}>0$ when $-1<x<1$ and then $\left|1-x^{2}\right|=1-x^{2}$

$\begin{array}{l} \text { But } 1-x^{2}<0 \quad \text { when } \quad 1<x \text { or } x<-1, \text { thus } \\ \left|1-x^{2}\right|=x^{2}-1 \end{array}$

Thus

$\begin{array}{l} \int_{-2}^{2}\left|1-x^{2}\right| d x \\ =\int_{-2}^{-1}\left|1-x^{2}\right| d x+\int_{-1}^{1}\left|1-x^{2}\right| d x+\int_{1}^{2}\left|1-x^{2}\right| d x \\ =\int_{-2}^{-1}\left(x^{2}-1\right) d x \\ +\int_{-1}^{1}\left(1-x^{2}\right) d x+\int_{1}^{2}\left(x^{2}-1\right) d x \\ =\left(\frac{x^{3}}{3}-x\right)_{-2}^{-1}+\left(x-\frac{x^{3}}{3}\right)_{-1}^{1}+\left(\frac{x^{3}}{3}-x\right)_{1}^{2} \\ =\left(\frac{-1}{3}+1\right)-\left(\frac{-8}{3}+2\right)+\left(\frac{-1}{3}+1\right) \\ =4 \end{array}$

The problem (2):-

Since $2 x+3>0$ when $x>\frac{-3}{2}$ , then $|2 x+3|=2 x+3$

But $2x+3<0$ when $x<\frac{-3}{2}$ , thus $|2 x+3|=-2 x-3$

Therefore we can find the value of the following integral

$\begin{array}{l} \int_{-2}^{2}|2 x+3| d x=\int_{-2}^{\frac{-3}{2}}|2 x+3| d x +\int_{-3}^{2}|2 x+3| d x \\ =\int_{-2}^{\frac{-3}{2}}(-2 x-3) d x+\int_{\frac{-3}{2}}^{2}(2 x+3) d x \\ =\left(-x^{2}-3 x\right)_{-2}^{\frac{-3}{2}}+\left(x^{2}+3 x\right)_{\frac{-3}{2}}^{2} \\ =\left(-\frac{9}{4}+\frac{9}{2}\right)-(-4+6)+(4+6)-\left(\frac{9}{4}-\frac{9}{2}\right) \\ =\frac{25}{2} \end{array}$