Integrals and Indefinite Integrals

**Open that Hampster!** · November 8th 2009, 12:11 PM

1) Evaluate the integral: $\int^5_4(y-2)(2y+3) = \int^5_42y^2-y-6$

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

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 $\int^6_1 \frac{1}{3x} = \int^6_1 (3x)^{-1}$

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

3) Evaluate the indefinite integral:

$18t-t^2+3t^3-\frac{t^4}{4}+C$

**Open that Hampster!** · November 9th 2009, 02:50 PM

Bump.

**Skerven** · November 9th 2009, 04:06 PM

Originally Posted by Open that Hampster!

Have a feeling I messed up on this one.

f should be a function of y

Originally Posted by Open that Hampster!

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

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

No way, $\int^a_b\frac{1}{cx} = \frac{\ln a - \ln b}{c}$

**Open that Hampster!** · November 9th 2009, 04:52 PM

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.

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