Thread: Integral Help (Check if I'm doing these right)

Use the form of the definition of the integral given in the Properties of Integrals to evaluate integrals below:

1. $\displaystyle \int\limits_1^4{(x^2+2x-5)dx}$

I then get:

$\displaystyle \int\limits_1^4{(x^2)dx} + \int\limits_1^4{(2x)dx} - \int\limits_1^4{(5)dx}$

I do the calculations and end up with

$\displaystyle \frac{n(n+1)(2n+1)}{n^3} (\frac{27}{6})$

Take up the limit as n -> infinite and get

$\displaystyle \int\limits_1^4{(x^2+2x-5)dx} = \frac{-1}{2}$

2. $\displaystyle \int\limits_0^5{(1+2x^3)}dx$

basically I did the same thing for this problem and end up with

$\displaystyle \int\limits_0^5{(1+2x^3)}dx = 1$


Here's another problem that I quite don't understand.


3. Find the Riemann sum for $\displaystyle f(x) = sinx$ , $\displaystyle 0 \le x \le \frac{3pi}{2}$
with six terms, taking he saple points to be right endpoints. Give your answer correct to six decimal places.

There are several other parts to this particular question but I think I can manage to answer those if I find out how to do this part.

No, I am sorry to say, you are not doing them right.

What exactly are you trying to do?. I see a possible relationship to a Riemann sum there.

$\displaystyle \int_{1}^{4}{(x^{2}+2x-5)}dx=21$

$\displaystyle {\Delta}x=\frac{4-1}{n}=\frac{3}{n}$

By the right endpoint method:

$\displaystyle x_{k}=1+\frac{3k}{n}$

$\displaystyle f(x_{k}){\Delta}=\left[(1+\frac{3k}{n})^{2}+2(1+\frac{3k}{n})-5\right](\frac{3}{n})$

$\displaystyle =\frac{27k^{2}}{n^{3}}+\frac{36k}{n^{2}}-\frac{6}{n}$

$\displaystyle \frac{27}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{36}{n^{2 }}\sum_{k=1}^{n}k-\frac{6}{n}\sum_{k=1}^{n}1$

Now, use the identities you learned. I see you used one in your post(the sum of the squares). Then, take the limit as $\displaystyle n\rightarrow{\infty}$.
You're shooting for 21, not -1/2.

Try the same technique on this one, if that's what you need to do.

$\displaystyle \int_{0}^{5}(1+2x^{3})dx=\frac{635}{2}$

Originally Posted by galactus

No, I am sorry to say, you are not doing them right.

What exactly are you trying to do?. I see a possible relationship to a Riemann sum there.

$\displaystyle \int_{1}^{4}{(x^{2}+2x-5)}dx=21$

$\displaystyle {\Delta}x=\frac{4-1}{n}=\frac{3}{n}$

By the right endpoint method:

$\displaystyle x_{k}=1+\frac{3k}{n}$

$\displaystyle f(x_{k}){\Delta}=\left[(1+\frac{3k}{n})^{2}+2(1+\frac{3k}{n})-5\right](\frac{3}{n})$

$\displaystyle =\frac{27k^{2}}{n^{3}}+\frac{36k}{n^{2}}-\frac{6}{n}$

$\displaystyle \frac{27}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{36}{n^{2 }}\sum_{k=1}^{n}k-\frac{6}{n}\sum_{k=1}^{n}1$

Now, use the identities you learned. I see you used one in your post(the sum of the squares). Then, take the limit as $\displaystyle n\rightarrow{\infty}$.
You're shooting for 21, not -1/2.

Try the same technique on this one, if that's what you need to do.

$\displaystyle \int_{0}^{5}(1+2x^{3})dx=\frac{635}{2}$

Hmm... I see what I have done wrong. Thank you for your help. I still need help with the last question though which I added while you were posting.

when using K, K = any number in that interval right?

Break sin(x) up into 6 partitions over the designated limits

$\displaystyle \frac{\frac{3\pi}{2}-0}{6}=\frac{\pi}{4}$

That's the width of each rectangle. Use that in your sine to find the heighths of the rectangles. The area of a rectangle is width times height.
Add them up.

$\displaystyle sin(\frac{\pi}{4})+sin(\frac{\pi}{2})+......+sin(\ frac{3\pi}{2})$

Then multiply that by your width of $\displaystyle \frac{\pi}{4}$

Here's a graph:

I'm having problems simplifying this:

$\displaystyle f(x_{i}) = \frac{5}{n} [1 + 2(1 + \frac{5i}{n})^3]$

any help?

$\displaystyle f(x_{i}) = \frac{5}{n} [1 + 2(1 + \frac{5i}{n})^3]$

$\displaystyle = \frac{5}{n} + \frac{10}{n}(1 + \frac{5i}{n})^3$

I get to that point but the cube, or shall I say the 1 +, is bothering me. Is there any way to simplify it or something?

This is my solution to the second one.

$\displaystyle {\Delta}{x} = {\frac{5}{n}}$ and $\displaystyle x_i = \frac{5i}{n}$

$\displaystyle f(xi) = [ 1 + 2(\frac{5i}{n})^3] \frac{5}{n} $

I simplify and get

$\displaystyle = [\frac{15}{n} + \frac{625i^3}{n^4}$

Do the stuff with the sigma notation and get

$\displaystyle \frac{15}{n} + \frac{625}{2}(\frac{n(n+1)}{n4})^2$

I take the limit of that and get $\displaystyle \frac{625}{2}$

Looks like you need to add 5.

$\displaystyle {\Delta}x=\frac{5}{n}$

Then you get:

$\displaystyle \left[1+2(\frac{5k}{n})^{3}\right](\frac{5}{n})=\frac{1250k^{3}}{n^{4}}+\frac{5}{n}$

$\displaystyle \frac{1250}{n^{4}}\sum_{k=1}^{n}k^{3}+\frac{5}{n}\ sum_{k=1}^{n}1$

$\displaystyle \frac{1250}{n^{4}}\cdot\frac{n^{2}(n+1)^{2}}{4}+5$

=$\displaystyle \frac{625}{n}+\frac{625}{2n^{2}}+\frac{635}{2}$

Now, see the limit?.

Originally Posted by galactus

Looks like you need to add 5.

$\displaystyle {\Delta}x=\frac{5}{n}$

Then you get:

$\displaystyle \left[1+2(\frac{5k}{n})^{3}\right](\frac{5}{n})=\frac{1250k^{3}}{n^{4}}+\frac{5}{n}$

$\displaystyle \frac{1250}{n^{4}}\sum_{k=1}^{n}k^{3}+\frac{5}{n}\ sum_{k=1}^{n}1$

$\displaystyle \frac{1250}{n^{4}}\cdot\frac{n^{2}(n+1)^{2}}{4}+5$

=$\displaystyle \frac{625}{n}+\frac{625}{2n^{2}}+\frac{635}{2}$

Now, see the limit?.

hmm. yeah I got to go over and make sure I didn't make the same mistake with other problems. Thanks