# Fundamental Theorem of Calc. Integration

• Sep 8th 2009, 04:38 PM
Casas4
Fundamental Theorem of Calc. Integration
Evaluate the definite integral using the Fundamental Theorem of Calculus. You will need accuracy to at least 4 decimal places for your numerical answer to be accepted. You can also leave your answer as an algebraic expression involving square roots.

I understand the FTC when you have x in the limits but since there is no x I'm confused on how to solve this.
Thanks!
• Sep 8th 2009, 04:44 PM
luobo
Quote:

Originally Posted by Casas4
Evaluate the definite integral using the Fundamental Theorem of Calculus. You will need accuracy to at least 4 decimal places for your numerical answer to be accepted. You can also leave your answer as an algebraic expression involving square roots.

I understand the FTC when you have x in the limits but since there is no x I'm confused on how to solve this.
Thanks!

$I=\sqrt{2+3t^4}+C$
$I=\sqrt {2+3\times 6^4} - \sqrt {2+3\times 5^4}$
• Sep 8th 2009, 04:48 PM
skeeter
Quote:

Originally Posted by Casas4
Evaluate the definite integral using the Fundamental Theorem of Calculus. You will need accuracy to at least 4 decimal places for your numerical answer to be accepted. You can also leave your answer as an algebraic expression involving square roots.

I understand the FTC when you have x in the limits but since there is no x I'm confused on how to solve this.
Thanks!

$\int_5^6 \frac{6t^3}{\sqrt{2+3t^4}} \, dt = \left[\sqrt{2+3t^4}\right]_5^6$