How can I accurately calculate the area between two curves?

[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

 

[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.

 

[cepheid]

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)

Click on my LaTeX output image below to find out:

\int_0^8 (\sqrt{2x} -x + 4)\,dx​

 

[Procrastinate]

Then from the points of intersection, draw a horizontal line to the y-axis.

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

 

[rock.freak667]

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

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.

 

[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

 

[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