Question

evaluate the definite integral

Answer 1

\[\int\limits_{0}^{9} e^x \sin(x) dx\]

Answer 2

this is a classic integral that will require a particular trick
how far have you gotten?

Answer 3

hmph...

Answer 4

I'm on integration by parts, so I'm guessing I need that.

Answer 5

For this case I could use either for x and dv...

Answer 6

and have you chosen u and dv, etc. ???

Answer 7

u*

Answer 8

e^x=u
sinxdx=dv
(actually it doesn't even matter)

Answer 9

I could do u = e^x
du = e^xdx

then dv = sin(x)
v = -cos(x)

Answer 10

or the other way around....

Answer 11

either way is fine, you are good so far

Answer 12

\[-\cos(x)e^x + \int\limits \cos(x) e^xdx?\]

Answer 13

good, keep going (integrate by parts again)

Answer 14

I feel like this is going to go on forever...

Answer 15

yes, but the trick will fix that
do it one more time and show me what you get

Answer 16

u = e^x
du = e^x

dv = cos(x)
v = sin(x)

Answer 17

\[-\cos(x)e^x + \sin(x)e^x - \int\limits \sin(x) e^xdx\]

Answer 18

oh is this whee you divide one another or something.

Answer 19

or add one side to the other...

Answer 20

\[\int\limits \sin(x)e^x = -\cos(x)e^x + \sin(x)e^x - \int\limits \sin(x) e^xdx\]

Answer 21

exactly, look at what you have; the original integral again
this may seem like a bad thing, but in fact now we can use simple algebra:
add that integral to both sides, then solve for that integral by dividing by 2

Answer 22

so now it's

Answer 23

\[2\int\limits \sin(x)e^e = -\cos(x)e^x + \sin(x)e^x\]

Answer 24

\[\int\limits \sin(x)e^e = \frac{-\cos(x)e^x + \sin(x)e^x}{2}\]

Answer 25

besides the typo, lack of dx, and the +C (though in the actual problem the integral is definite, so whatever) and the typo of e^e on the left that should be e^x it looks good :D

Answer 26

e^e yes!

Answer 27

that trick works with any form\[\int e^{ax}\sin(bx)dx\]or\[\int e^{ax}\cos(bx)dx\]but it can get messy depending on the values of a and b as you might imagine...

Answer 28

so all you do is double the first integral and add it to the other side huh.

Answer 29

NOW TRHE FUN PART INTEGRAL FROM 0 TO 9!

Answer 30

\[\frac{-\cos(9)e^9 + \sin(9)e^9}{2} - \frac{-\cos(0)e^0 + \sin(0)e^0}{2}\]

Answer 31

Which I'm lazy to do :p

Answer 32

well a lot of the junk on the right is either 0 or 1

Answer 33

-3367.86 is the first part haha

Answer 34

e^0 is 1....

Answer 35

so it ends up becoming -3367.86 - 1/2 :P

Answer 36

\[\frac{-\cos(9)+ \sin(9)}{2}e^9 - \frac{-1 + 0}{2}=\frac{-\cos(9)+ \sin(9)}{2}e^9+\frac12\]

Answer 37

oh yeah it's -- 1/2 :P

Answer 38

sooo tired.

Answer 39

the devil is in those details...

Answer 40

I think she wants a decimal approv.

Answer 41

I just used wolfram for that answer :P

Answer 42

yeah, at this point those numbers are just too ugly to do any other way

Answer 43

haha guess not fawk...

Answer 44

okay that worked :P ty