# Integrals and Indefinite Integrals

• Nov 8th 2009, 12:11 PM
Open that Hampster!
Integrals and Indefinite Integrals
1) Evaluate the integral: $\displaystyle \int^5_4(y-2)(2y+3) = \int^5_42y^2-y-6$

Antiderivative of the fucntion is f(x) = $\displaystyle \frac{2y^3}{3} - \frac{y^2}{2}-6y$

So the answer is f(5)-f(4) = 40.8333-22.6666.

Have a feeling I messed up on this one.

2) Evaluate the Integral $\displaystyle \int^6_1 \frac{1}{3x} = \int^6_1 (3x)^{-1}$

Antiderivative of the fucntion being integrated is ln(3x) so the answer is ln(18)-ln(3)

3) Evaluate the indefinite integral: http://www.webassign.net/cgi-bin/sym...2A2%5C%29%20dt

$\displaystyle 18t-t^2+3t^3-\frac{t^4}{4}+C$
• Nov 9th 2009, 02:50 PM
Open that Hampster!
Bump.
• Nov 9th 2009, 04:06 PM
Skerven
Quote:

Originally Posted by Open that Hampster!
Have a feeling I messed up on this one.

f should be a function of y
Quote:

Originally Posted by Open that Hampster!
2) Evaluate the Integral $\displaystyle \int^6_1 \frac{1}{3x} = \int^6_1 (3x)^{-1}$

Antiderivative of the fucntion being integrated is ln(3x) so the answer is ln(18)-ln(3)

No way, $\displaystyle \int^a_b\frac{1}{cx} = \frac{\ln a - \ln b}{c}$
• Nov 9th 2009, 04:52 PM
Open that Hampster!
Quote:

Originally Posted by Skerven
f should be a function of y

That was a typo. The answer is the problem.

Thanks for the second one, though.