# Thread: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.

1. ## Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.

Hi, guys. I'm working on a take-home test for my Calculus 1 class.
The first part of the problem was
"Find the area under f(x)=2x^2-x from 1 to 2 using the definite integral (not FTC)."
I ended up with 18/6.
However, for the second part of the problem, "Compute the definite integral from 1 to 2 (using the antiderivative squiggle) (2x^2-x)dx using the Fudamental Theorem of Caluclus," I got 33/6.

I feel like I probably did the FTC part correctly because it is much less complicated.
But, if you think I'm wrong, please let me know, and show me what I should do! I'm attatching files showing the work I did since it's far too complex to type.
I understand it might be a little difficult to read, but I just hope you can get the just of what I did 2. ## Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.

You should solve your test alone, am I right ?

3. ## Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.

No, it's a take home test so we can use our notes, friends, websites... Anything/Anyone but our professor. Too much for you to figure out, too? 4. ## Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.

in the second attachment file you did this integral wrong this how you should do it

$\displaystyle \int_{1}^{2} 2x^2 -x \;dx = \frac{2x^{2+1}}{2+1}- \frac{x^{1+1}}{1+1} \mid_{1}^{2}$

in general

$\displaystyle \int x^n \; dx = \frac{x^{n+1}}{n+1}+C$

indefinite integral

$\displaystyle \int 2x^2 -x \; dx = \frac{2x^3}{3} - \frac{x^2}{2} +C$

definite integral

$\displaystyle \int_1^2 2x^2 -x \; dx = \left(\frac{2(2)^3}{3} - \frac{2^2}{2}\right) - \left(\frac{2}{3} - \frac{1}{2}\right)$

5. ## Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.

Thank you for catching that!
I also noticed I messed up while taking the limits and adding up the fractions for the first part, 19/6 is the correct answer.
I so much appreciate your help in going through that for me, I would have NEVER caught that mistake I made in FTC!

6. ## Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC. Originally Posted by Kristenx2 No, it's a take home test so we can use our notes, friends, websites... Anything/Anyone but our professor. Too much for you to figure out, too? Uh huh. Then what's the point of handing in the work if it's just copied from someone else? Thread closed.