Find the area of the region bounded by the curves y = x^2 - 1 and y = cos(x).

Question

I've tried doing this and I got -1.54 which doesn't make sense because area should always be positive right? And I put it in an online calculator to check and it gave me 3.11. I think I may have messed up my negatives somewhere.

Here's my work and can you tell me where I went wrong
The integral from [-1.177,1.177] of cos(x)-x^2-1
This gives sin(x)-(x^3/3)-x with limits of [-1.177,1.177]
Sin(1.177)-1.177^3/3-1.177-((sin(-1.177)-(-1.177^3)/3-(-1.177))
.923...-.544...-1.177+.923...-.544...-1.177

Answer 1

look at the graph
http://www.wolframalpha.com/input/?i=y+%3D+x%5E2+-+1+and+y+%3D+cos(x)

and notice the symmetry. So we need the intersection
cosx = x^2 - 1
which you found correctly as ± 1.177 correct to 3 decimals

area = 2[integral] (cosx - (x^2 -1)) dx from 0 to 1.177,
= 2 (sinx - x^3/3 + x | from 0 to 1.177)

I think I found your mistake, you had -x instead of +x