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

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• Dec 5th 2011, 05:59 AM
Kristenx2
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 :)
• Dec 5th 2011, 07:09 AM
Amer
Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.
You should solve your test alone, am I right ?
• Dec 5th 2011, 07:17 AM
Kristenx2
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? :)
• Dec 5th 2011, 07:30 AM
Amer
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

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

in general

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

indefinite integral

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

definite integral

$\int_1^2 2x^2 -x \; dx = \left(\frac{2(2)^3}{3} - \frac{2^2}{2}\right) - \left(\frac{2}{3} - \frac{1}{2}\right)$
• Dec 5th 2011, 08:15 AM
Kristenx2
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!
• Dec 5th 2011, 05:21 PM
mr fantastic
Re: Definite Integrals - Same Problem, 1st use Definite Integral, then FTC.
Quote:

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.