Question

evaluate the integral of (x^2+x)dx , [-2,0]

Answer 1

\[=\lim_{n \rightarrow \infty} \frac{ 2 }{ n } \sum_{1}^{n} [(-2+\frac{ 2i }{ n })^2 + (-2+\frac{ 2i }{ n })]\]

The answer is 2/3 but I'm getting other numbers.

Answer 2

\[=\lim_{n \rightarrow \infty}\frac{ 2 }{ n }\sum_{1}^{n}(2-\frac{ 6i }{ n }+\frac{ 4i^2 }{ n^2 })\]
\[=\lim_{n \rightarrow \infty}[2n-\frac{ 6 }{ n }(\frac{ n(n+1) }{ 2 })+\frac{ 4 }{ n^2 }(\frac{ n(n+1)(2n+1) }{ 6 })]\]
\[=\lim_{n \rightarrow \infty}\frac{ 2 }{ n }(-n-3+\frac{ 4n^2+4n+1 }{ 3n })\]
\[=\lim_{n \rightarrow \infty}(-2-3+\frac{ 4n }{ 3 }+\frac{ 4 }{ 3 }+\frac{ 1 }{ 3n })\]
\[=\frac{ -11 }{ 3 }\]

Help?

Answer 3

I'm puzzled by your method of doing integrals...

\[\int\limits_{-2}^{0}(x^2+x) dx = \int\limits_{-2}^{0}x^2 dx + \int\limits_{-2}^{0}x dx\]By the power rule,
\[ \int x^2 dx + \int x dx = (\frac{1}{3}x^3+\frac{1}{2}x^2 + C)\]

Now evaluate that at \(x = 0\) and \(x = -2\) and subtract the latter from the former.

Answer 4

Sorry, should've clarified...the problem asks to evaluate the integral using the form of the definition of the integral
\[\int\limits_{-2}^{0}f(x)dx=\lim_{n \rightarrow \infty}\sum_{1}^{n}f(x _{i})\Delta x\]
where
\[\Delta x=\frac{ b-a }{ n } and x _{i}=a+i \Delta x\]

Answer 5

You should also be aware of the fact that the curve crosses the x-axis in [-2,0]. It may or may not be important, but you WILL want to remember that, at times.

Answer 6

I checked on stewart's homework hints and confirmed I'm right up until I get to
\[=\lim_{n \rightarrow \infty}\frac{ 2 }{ n } \sum_{1}^{n}[(-2+\frac{ 2i }{ n })^2+(-2+\frac{ 2i }{ n })]\]
from here on out it should be algebra and summation distribution but it doesn't give me 2/3

Answer 7

Hmm..

Answer 8

check the third step, you should have had -3/n when you distributed n

Answer 9

oh wait, no that's not it

Answer 10

when you factored out 2/n, you had -n, when it should be +n

Answer 11

Woops, 2/n wasn't factored out. It goes after the limit in the step before it - just forgot to write it in.