Question

evaluate the antiderivative of x e^x^2 de from 0 to 3

Answer 1

Let u=2x

Answer 2

I'm sorry, u=x^2

Answer 3

ok
du 2x

Answer 4

aswer = 1/2 e^x^2 + c

Answer 5

?

Answer 6

Did you mean to say evaluate integral?

Answer 7

si

Answer 8

si sorry

Answer 9

yes

Answer 10

Your new integral\[\int\limits_{}^{}e ^{u}(du)/2\]

Answer 11

yea

Answer 12

taking the antiderivative of that will be=e^u + c /2

Answer 13

= e^x^2 + c/2

Answer 14

or the same as 1/2 e^x^2 + c

Answer 15

Si, now you have evaluate 0 to 3.

Answer 16

ohhhhhhhhh

Answer 17

So no constant it is a definite integral

Answer 18

:) stuped

Answer 19

what u mean with no constant is a definite integral?

Answer 20

OK LISTEN VERY CAREFULLY, you need to do a thing called a guassian integral for this one, multily the first integral by integral from 0 to 3 of e^y^2 and then turn into polar coordinates and do rdrd(theta) and there you go x^2 +y^2 turns into r^2 and you have the derivitive sitting right there hope this helps and this is actually how you do the bell curve for stats. DONT FORGET TO DO THE SQUAR OF THAT BECAUSE YOU MULTIPLIED BY ANOTHER INTEGRAL

Answer 21

this is like chinesse.. i am in cal 2.. maybe is a easy way before getting there?

Answer 22

never heard about guassian integral yet..

Answer 23

can u show me a normal way?

Answer 24

I use wolfram as a check. They use the same method we do here. http://www.wolframalpha.com/input/?i=integral+x+e^x^2

Answer 25

put u=x^2..du=2xdx...x-->0,u-->0,x-->3,u-->9..put these values in the integral and u will get 1/2(e^x^2)-1/2

Answer 26

you should get \[\int\limits_{0}^{2\pi}\}\{\int\limits_{0}^{3}\]e^r^2(rdrd(theta)

which you should get \[\sqrt{\pi(e^3-1}/2\]

Answer 27

haha calc 2 you have to do the mcclarean and taylor expansion and series for this GOOD LUCK

Answer 28

:(

Answer 29

hahah so sorry i didnt see the xe^x^2 this one you jjust do a simple u sub

Answer 30

.. omg i was going to cry

Answer 31

:) wait till calc 3 to cry you will see this

Answer 32

jijiiji

Answer 33

so can u show me this one... what is left

Answer 34

i already integrate my answer is
I= 1/2 e^x^2 + c
now integarate freom 0 to 3

Answer 35

You make two brackets
[ ] - [ ]
In the first bracket you put in the integral and plug in 3 for x; the second bracket plug in 0 for x.

Answer 36

That is a subtraction sign between the two brackets.

Answer 37

\[\int\limits_{0}^{3} x e^{x^{2}}dx\]\[\int\limits_{}^{} x e^{x^{2}}dx\]Let:
u = x^2
du = 2x ---> (1/2)du = x
\[1/2\int\limits_{}^{} e^{u}du = 1/2[e^u]\]Substitute u = x^2 \[1/2[e^{x^{2}}]\]and evaluate from 0 to 3.
\[1/2(e^{3^{2}}-e^{0^{2}}) = (1/2)(e^9-1)\]

Answer 38

how u got e^0

Answer 39

\[e^{3^{2}} = e^9\]is evaluating at 3.
And
\[e^{0^{2}} = 1\]is evaluate at 0.

Answer 40

ok

Answer 41

if u need to give the exact answer? what should it be?

Answer 42

\[(1/2)(e^{9}-1)\]

Answer 43

thanks

Answer 44

No problem!

Answer 45

Exact answer means keep the e, don't use calculator approximation.

Answer 46

ohhhhh :),, good point i did not know that

Answer 47

thanks guys.. understanding