Question

@DanJS

Answer 1

i have no idea how to solve it @DanJS

Answer 2

here is the graph with the region

Answer 3

okk

Answer 4

Take the integral difference of the top function - bottom function, from the bounds where the two functions intersect.

Answer 5

wait what am i taking the integral of

Answer 6

you can see on the graph where they intersect, those are the bounds [a,b]
\[\int\limits_{a}^{b}\sin(x)dx - \int\limits_{a}^{b}[x^2-1]dx\]

Answer 7

Here is a few examples..
http://www.millersville.edu/~bikenaga/calculus/areacur/areacur.html

Answer 8

When i put the integrals into my calculator , from the approximate decimal bounds where sin(x) = x^2-1

It says the area is about 1.67

Answer 9

[a,b] = [-0.637 , 1.41]

\[\int\limits\limits_{a}^{b}\sin(x)dx - \int\limits\limits_{a}^{b}[x^2-1]dx\]

Answer 10

so my final answer should be 1.67? i was way off... ahhaha

Answer 11

\[[\cos(a)-\cos(b)] - [\frac{ -a^3 }{ 3 }+a+\frac{ b^3 }{ 3 }-b]\]

Answer 12

Those are the integrals evalueated, just plug in the a and b bounds

Answer 13

and that is how you got 1.67?

Answer 14

let me check the numbers again

Answer 15

okk

Answer 16

i think that is right, yea

Answer 17

so the area is 1.67?

Answer 18

I think so, double check my math... plug in those a and b into the evaluated integrals

Answer 19

okkk thanks(:

Answer 20

seems reasonable