# Area Between Two Curves

1. Aug 21, 2010

### Procrastinate

I am having trouble with this question: (QUESTION AS ATTACHED)

This is what I tried to do:
I found the intercepts of the difference of the two functions which I found to be only 8.

Then I integrated this into the new function:
$$\int{\sqrt{2x}-x+4}$$. with limits of 0 and 8 (I don't know how to Latex this)

This, according to my Graphics Calculator yielded 21.3.

However, the answer is 18, what have I done wrong?

Edit: $$\int_0^8 (\sqrt{2x} -x + 4)\,dx$$

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Last edited: Aug 22, 2010
2. Aug 21, 2010

### rock.freak667

First, find where the line intersects the curve.

Then from the points of intersection, draw a horizontal line to the y-axis, what shape is formed with the straight line and these two new lines you just drew? What is the area of this shape?

EDIT:
I see what you are doing now. Your way will be the same, you will need to subtract the same amount of times.

Last edited: Aug 21, 2010
3. Aug 21, 2010

### cepheid

Staff Emeritus
Click on my LaTeX output image below to find out:

$$\int_0^8 (\sqrt{2x} -x + 4)\,dx$$​

4. Aug 21, 2010

### Procrastinate

I am not sure what you mean here, but is the answer a triangle?

5. Aug 21, 2010

### rock.freak667

Yep. You can get the area of the triangle. You can get the area of the region bounded by the curve and the y-axis and then subtract.

6. Aug 22, 2010

### hunt_mat

I think that this would work best as a double integral. Your y-limits are between 2 and 8 and your x-limits are between y^{2}=2x and y=x-4, this will give you the correct answer.

Mat

7. Aug 23, 2010

### ehild

It is easier to calculate the area enclosed by the lines of the inverse functions, x=y+4 and x=y^2/2.

ehild