Calculus
posted by Kait .
Find the area of the region bounded by the curves y = x^2  1 and y = cos(x).
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^21
This gives sin(x)(x^3/3)x with limits of [1.177,1.177]
Sin(1.177)1.177^3/31.177((sin(1.177)(1.177^3)/3(1.177))
.923....544...1.177+.923....544...1.177

Calculus 
Reiny
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
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