# Math Help - Area between curves

1. ## Area between curves

I have some wrong answers .. help?

Calculate the area between y-sin(x) and y=cos(x) , -pi<or equal to x <or equal to 2pi.
I have -1.9984.

Calculate the area between y=x^2 and y=3x+4.
I have -18.7.

Find the area of the region enclosed between y=2sin(x) and y=2cos(x) from x=0 and x=0.8pi
I have .6334465.

Can someone tell me what i did wrong?

2. Originally Posted by keemariee
I have some wrong answers .. help?

Calculate the area between y-sin(x) and y=cos(x) , -pi<or equal to x <or equal to 2pi.
I have -1.9984.

Calculate the area between y=x^2 and y=3x+4.
I have -18.7.

Find the area of the region enclosed between y=2sin(x) and y=2cos(x) from x=0 and x=0.8pi
I have .6334465.

Can someone tell me what i did wrong?
We want to compute the area of

$\int_{-\pi}^{2\pi}f(x)-g(x)$ were f(x) and g(x) are to be determined

First off did you realize taht you must split this into four integrals?

3. Originally Posted by keemariee
Calculate the area between y=x^2 and y=3x+4.
I have -18.7.
First, we need to know our boundaries, so we want to know where the lines cross. At these points, they will have the same (x,y) value.

So x^2 = y = 3x+4
this gives us x^2 = 3x+4
x^2-3x-4 = 0
(x-4)(x+1)=0

So at x=4, and x=-1, these two graphs will cross. Then these are our boundaries, so we will integrate from -1 to 4.

Now we need to know which function is above the other, substituting in x=0 (because it is a point inside of our boundaries) we get that y=3x+4 is above.

So we know that we will take the integral of 3x+4 and then subtract the integral of x^2 (if you don't understand why to do this, look at the attached image, it should be rather clear.

so $A = \int_{-1}^4 (3x+4-x^2)~dx$

$= \left[\frac 32x^2+4x-\frac 13x^3 \right]_{-1}^4$

$= \frac 32*16 + 16 - \frac 13*64 - \frac 32 +4 +\frac 13$

$= \frac {125}6$