Question

Does this work out right?

Answer 1

----------------------> RIGHT.

Answer 2

\[\int\ x^2(x^3+5)^9\ dx=\left[\sum_{n=1}^{10}\left(\begin{array}c9\\n-1\end{array}\right)5^{(n-1)}x^{(32-3n)}\right]+C\]

Answer 3

\[\int\ x^2(x^3+5)^9\ dx =\frac{(x^3+5)^{10}}{30}+C\]

Answer 4

I can understand Integration only.

Answer 5

i forgot to include my denominators in the original

Answer 6

haha^

Answer 7

I did not get the Sigma notation part o.O

Answer 8

\[\int\ x^2(x^3+5)^9\ dx=\left[\sum_{n=1}^{10}\frac{\left(\begin{array}c9\\n-1\end{array}\right)}{33-3n}5^{(n-1)}x^{(33-3n)}\right]+C\]

Answer 9

\[\int x^2(x^3+5)^9\ dx=\int x^2\sum_{n=0}^{9}{9\choose n}(x^3)^n(5)^{9-n}dx\]

\[=\int\sum_{n=0}^{9}{9\choose n}(x^{3n+2})(5)^{9-n}dx\]

\[=\sum_{n=0}^{9}{9\choose n}\left(\frac{x^{3n+3}}{3n+3}\right)(5)^{9-n}+c\]

Answer 10

in..terne....t, so sl.....ow

Answer 11

why are you shifting the index?

Answer 12

just the way it looked when i did it on the paper :)

Answer 13

my math teacher wants us to do these boring indefinites so i want to give her something to wonder about lol

Answer 14

you mean this was the original problem
\[\int\ x^2(x^3+5)^9\ dx =\frac{(x^3+5)^{10}}{30}+C\]??

Answer 15

yep

Answer 16

you must have a lot of time on your hands....

Answer 17

:)

Answer 18

hmm

Answer 19

not that it makes any difference but i think it is easier to start with
\[(a+b)^n=\sum_{k=0}^n\dbinom{n}{k}a^{n-k}b^k\] rather than using n - 1 and and starting at 1