Question

more integration.... will need the equation editor

Answer 1

true?

Answer 2

\[\int\limits_{0}^{1} x^2 e^-(x^3) dx\]

looks like I am supposed to use u substitution u=x^3 so du = 3x^2

Answer 3

That looks like a good method to do it.

Answer 4

I get to \[\int\limits_{0}^{1} e^-(u) (x ^{2}/1) (1/3x^2) = 1/3 \int\limits_{?}^{?}e^-u \]

and then I do something wrong

Answer 5

\[\int\limits_{0}^{1} x^2 e^{-x^3} dx\]Let \(u=x^3\). Then \(du=3x^2 \;\;dx\). Hence, we're integrating \[{1 \over 3} \int\limits e^u du={e^u \over 3}\]Now replace for x again to get\[\large {e^{-x^3}\over 3} =\LARGE{|} \large{_0^1}\](Sorry for the formatting there)

Now we just evaluate.

Answer 6

my correction key has an extra negative at the front...I thought it was me. or is the integration of e^-u -e^-u?

Answer 7

The book is correct. The integral of \(e^{-u}\) is \(-e^{-u}\)

Answer 8

So to correct my previous post:

Let \(u=-x^3\), then \(du = -3x^2 \;\; dx\). If we go through the work again, we get \[\large -{e^{-x^3}\over 3} =\LARGE{|} \large{_0^1}\]

Answer 9

ok so then i get ((-1/3) e^-1) + 1/3

but i'm not liking - ((-1/3) e^-0) but that still looks like +1/3 since e^(-0) would be 1

Answer 10

I'm getting\[\large -{e^{-1^3}\over 3} +\large {e^{0^3}\over 3} = -{e^{-1} \over 3}+{e^0 \over 3}\]\[\large =-{{1 \over e} \over 3}+{1 \over 3}={1-{1 \over e } \over 3}={{e-1 \over e} \over3}={{e-1} \over 3e}\]

Answer 11

hmmm cant cut and paste equation things -- that is rough

but your - (1 over e) over 3 Plus 1/3 would jive with my answer....my key is saying...
(-e^(-1) - 1) / 3 and I am really not believing it.

Answer 12

unfortunately - I got a different wrong answer on the test so I won't be able to fight it baaah

Answer 13

You actually can copy/paste by right clicking the equation, going to "Show Math As" and clicking on "TeX command."

Answer 14

now there is a useful tip thanks a ton

Answer 15

You're welcome.

Answer 16

The key to this seems a bit wrong. After checking with Wolfram, it agrees with my answer.

Answer 17

then I am going with wrong answer on the test and right answer at OpenStudy =)

Answer 18

I've done the same.