Answer to Question #115729 in Calculus for Sunita

Question #115729
1. integrate |1-x^2|dx from limit -2 to 2.
2. Integrate |2x+3|dx from limit -2 to 2
1
Expert's answer
2020-05-15T17:26:57-0400

The problem (1):-


Since 1x2>01-x^{2}>0 when 1<x<1-1<x<1 and then 1x2=1x2\left|1-x^{2}\right|=1-x^{2}

 But 1x2<0 when 1<x or x<1, thus 1x2=x21\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

221x2dx=211x2dx+111x2dx+121x2dx=21(x21)dx+11(1x2)dx+12(x21)dx=(x33x)21+(xx33)11+(x33x)12=(13+1)(83+2)+(13+1)=4\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 2x+3>02 x+3>0 when x>32x>\frac{-3}{2} , then 2x+3=2x+3|2 x+3|=2 x+3

But 2x+3<02x+3<0 when x<32x<\frac{-3}{2} , thus 2x+3=2x3|2 x+3|=-2 x-3

Therefore we can find the value of the following integral

222x+3dx=2322x+3dx+322x+3dx=232(2x3)dx+322(2x+3)dx=(x23x)232+(x2+3x)322=(94+92)(4+6)+(4+6)(9492)=252\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}






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