Area Between Curves: Can someone please check my work?

Question

Now that I get the funny anti-derivative part, (Muchas muchas gracias to Sean and MathGuy for that!), the answer I have to this problem is as follows:

The area between
y = sin(pi times x) and
y = x^2-x, x=2

is (4/pi) + 1 ....right?

Please show your work, if you don't mind! Thanks.

I know that the intersection is at 1, but the problem asks for the integral for 0 to 2 reguardless... I just did it by splitting the integral into two separate problems; one for 0 to 1 and one for 1 to 2. Was that wrong?

Answer 1

F = sin (piX)
G = X^2 - X

A quick graphing of the equations shows that to get the area:
Integrate F - G from 0 to 1
Integrate G - F from 1 to 2
And sum them together.

Integrate: sin (piX) - X^2 + X => -cos(piX)/pi - X^3/3 + X^2/2
Evaluate: 1/pi - 1/3 + 1/2 + 1/pi = 2/pi +1/6

Integrate: X^2 - X - sin (piX) => X^3/3 -X^2/2 + cos(piX)/pi
Evaluate: 8/3 - 2 + 1/pi - 1/3 + 1/2 + 1/pi = 2/pi + 5/6

Area = 2/pi + 1/6 + 2/pi + 5/6 = 4/pi + 1

Yes you are right.

Answer 2

sorry no answer it is merely complicated!!!!!!!!!!!!!!!!!! is college even properly worth it to income all that non sence i belive in amish college in hassle-free terms bypass to college to the eighth grade then artwork after than that...... who cares what's two +2-5 divided by ability of four hundred to locate the sq. root of x to locate the answer of y wtf none sence bunch of bullshit