# Thread:

1. ## Integrating Area Formed By Three Points

Use integration to find the area of the triangle having given the vertices
(0,0), (a,0),(b,c)

So I drew out my graph and I got 2 lines
y=c/b(x) and y=(c/b-a)(x-a)
and I know you have to integrate c/b(x) from 0 to b and add that to the integral of (c/b-a)(x-a) from b to a
Im just really having trouble integrating and getting the answer [which is .5(c)(a)]

Please Help

2. Rather than adding the integrals, I think you actually need to use an iterated integral since you're concerned with both the x and y directions simultaneously. In this case, it will obviously be a double integral. When setting it up, you should make y the inner integral since the general horizontal line across the region goes between 2 functions. The x part makes a good candidate for the outer integral because you can then just integrate from 0 to the constant b.

This page has a number of examples involving triangles.

3. I already talked to my teacher about this & he wants me to add them since we're just learning integration

4. Whoops, sorry about that.

5. It's much easier to integrate with respect to y from 0 to c. Have you tried this approach?

6. $\int_0^b \frac{c}{b}x \, dx + \int_b^a \frac{c}{b-a}(x-a) \, dx$

$= \frac{c}{b} \int_0^b x \, dx + \frac{c}{b-a} \int_b^a (x-a) \, dx<br />$

$= \frac{c}{b} \left[ \frac{x^2}{2} \right]_0^b + \frac{c}{b-a} \left[ \frac{x^2}{2}-ax \right]_b^a$

Can you follow what I am doing here?

7. yes, ive done this part already, i just get an answer that looks nothing like .5ac

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