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.

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