# Finding area of a triangle with integrals

I am suppose to find the area of a triangle using integrals with vertices 0,0 1,2 and 3,1

This gives me

$y= 2x$

$y=\frac{1}{3}x$

$y= \frac{-1}{2}x+\frac{5}{2}$

for my slopes

I know that I can calculate the area of the first part by finding

$$\int_{0}^{1}2x-\frac{1}{3}x$$

and the second part by

$$\int_{1}^{3}\frac{-1}{2}x+\frac{5}{2}-\frac{1}{3}x$$

The anti derivative of the top is $x^2 - x^2/3$

and the other one is $-x^2/4 + 5x/2 - x^2/6$

I am not sure what I am doing wrong but I do not get the proper answer of 5/2

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Working with what you have: $$\int_{0}^{1}2x-\dfrac{1}{3}x dx$$ $$\int_{0}^{1}\dfrac{5x}{3}dx$$ $$\dfrac{5}{3}\int_0^{1}x dx$$ $$=\dfrac{5}{6}$$

The second integral before being evaluated from $1$ to $3$ is: $$\dfrac{5x}{2} - \dfrac{5x^{2}}{12}$$

When evaluated at $x = 3$, you get $\dfrac{15}{4}$. When evaluated at $x=1$, you get $\dfrac{25}{12}$.

Get a common denominator and subtract them now. $\dfrac{45}{12} - \dfrac{25}{12} = \dfrac{20}{12} = \dfrac{5}{3}$.

Now simply add the two evaluated integrals:

$$\dfrac{5}{6} + \dfrac{5}{3} = \dfrac{5}{2}$$

Edit: When you said, and the other one is $\dfrac{-x^{2}}{4} + \dfrac{5x}{2} - \dfrac{x^{2}}{6}$, you were correct. However, your first antiderivative was incorrect. When you see something in the original integrand that can be simplified (i.e. your terms with just an $x$ in it), as a general rule, DO IT.

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I don't understand how you got 5/3 – user138246 Apr 29 '12 at 1:56
I will edit my post to show you then. – Joe Apr 29 '12 at 2:02
Where are you having difficulties when evaluating the second integral? – Joe Apr 29 '12 at 2:06
I am not sure what went wrong when I did it, I spent about 2 hours on this problem and was just not able to get an answer. – user138246 Apr 29 '12 at 2:07
Take a look at the bottom of my post. Based on your previous questions, it seems you just need to be careful with your algebra. – Joe Apr 29 '12 at 2:12