# Re: integration- finding area under graph

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Posted by Soroban on September 13, 2002 at 13:17:03:

In Reply to: integration- finding area under graph posted by bana on September 13, 2002 at 07:52:55:

: Hi there, I have a problem here with using integration to find the area under the graph because the graph cuts the x axis at x=6. ( i have to find the area where x=0 to x=9)
: the function is y=25-x^2
: i intergrated to 25x-(1/3)x^3+c
: I have find the area of -18. how can i break it up so that i'd get a correct area figure?

You've done well ~ except a typo: it cuts the X-axis at x = 5.

If part of the area is above the X-axis and part below, you can't
integrate from 0 to 9. The region below gives "negative area".

After integrating, you got: 25x - (1/3)x^3 ... correct!

First, evaluate it for x=0 to x = 5.

Then evaluate it for x=5 to x=9.. and take the absolute value.

Finally, add those two area.

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