Answer to Question #101491 in Calculus for Mal

Question

1. Calculate the numerical value of the following integral
∫_(-4)^2[(2-x^2 )-(3x+2)] dx

2. Calculate the curves f (x) = 2 - x2 and g (x) = 3x + 2 total area A between [-4,2]. Compare to the previous problem 1. and notice that there will be a different answer to these! Why came a different answer?

Answer 1

1.

$\int_{-4}^2[2-x^2-(3x+2)]dx=\int_{-4}^2(-x^2-3x)dx=(-\frac{x^3}{3}-\frac{3x^2}{2})|_{x=-4}^{x=2}=-6.$

2.

$A=\int_{-4}^{-3}[(3x+2)-(2-x^2)]dx+\int_{-3}^0[2-x^2-(3x+2)]dx+\\+\int_0^2[(3x+2)-(2-x^2)]dx=$

$=\int_{-4}^{-3}(x^2+3x)dx+\int_{-3}^0(-x^2-3x)dx+\int_0^2(x^2+3x)dx=$

$=(\frac{x^3}{3}+\frac{3x^2}{2})|_{x=-4}^{x=-3}+(-\frac{x^3}{3}-\frac{3x^2}{2})|_{x=-3}^{x=0}+(\frac{x^3}{3}+\frac{3x^2}{2})|_{x=0}^{x=2}=\frac{11}{6}+\frac{9}{2}+\frac{26}{3}=15.$

Answers are different because the area between two curves is always positive and in our case

includes three parts:

-4<x<-3 where g(x)>f(x)

-3<x<0 where f(x)>g(x) and

0<x<2 where g(x)>f(x).