Question

Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis

A. Set up a sum of integrals, not containing an absolute value symbol,that can be used to find the area of this region

B. Find the area of the region by using your answer from part A. Don't use calculator

What is a sum of integrals, and can it just be integral [-5,5] x^2+x-12 dx? Thanks!

Answer 1

ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.

`\int\limits^(-5)_(-4) {x^2+x-12} \, dx + \int\limits^(-4)_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx`

B.
sepearte the integrals

`\int\limits^(-5)_(-4) {x^2+x-12} \, dx = [(x^3)/(3)+(x^2)/(2)-12x]^(-5)_(-4)=((-125)/(3)+(25)/(2)+60)-((64)/(3)+8+48)=(23)/(6)`

next one

`\int\limits^(-4)_3 {-x^2-x+12} \, dx=-1[(x^3)/(3)+(x^2)/(2)-12x]^(-4)_(3)=-1((-64/3)+8+48)-(9+(9/2)-36))=(343)/(6)`

the last one you can do yourself, it is
`(50)/(3)`
the sum is
`(23)/(6)+(343)/(6)+(50)/(3)=(233)/(3)`


so the area under the curve is
`(233)/(3)`