Question

Find the area between the curves:
 
 
 y=2x+2
y=3x^2+1

Answer 1

Integrate. You can do it.

Answer 2

Do you know the limits of integration? I suspect if you do limit from -infinity to infinity the whole integral blows up.

Answer 3

I forgot how to set up the integral..
They intercept at 1 and -1/3

Answer 4

The integrand will be the greater function minus the lesser function.

Answer 5

Looks like \(2x+2\) is the greater function. between the intercepts.

Answer 6

Yes. So integrate. You don't need really to worry about greater minus lesser. Just take the absolute value of them all.

Answer 7

I got a graph:

Answer 8

So really, what is stopping you?

Answer 9

You know how to integrate?
Integrate
A=\[\int\limits_{-1/3}^{1} ((2x+2)-(3x^2+1)) dx\]

Answer 10

\[
A= \int\limits_{-\frac 13}^1(2x+2)-(3x^2+1)\;dx
\]

Answer 11

Okay so i end up with:
\[\LARGE (x^2+2x)-(x^3+x)\]
Right?

Answer 12

With the limits of -1/3 to 1 ofc

Answer 13

Finish it.

Answer 14

Finishing it up I get:
\[\LARGE -x^3+x^2+x\]
\[\LARGE -(1)^3+1^2+1=1\]
\[\LARGE -(-\frac{1}{3})^3+(-\frac{1}{3})^2+(-\frac{1}{3})\]
So the final answer will be from
\[\LARGE (1)-[ -(-\frac{1}{3})^3+(-\frac{1}{3})^2+(-\frac{1}{3})]\]
I think..

Answer 15

That can be simplified dude.