Question

What is y'(x) if y=integral from 0 to ln(3x) of sin(3e^t)dt..I know to consider the fundamental theorem of calculus

Answer 1

this question is weird...

Answer 2

you're looking for y'(x)...but you're given y...

Answer 3

and you're supposed to do an integral...

Answer 4

so derivative of an integral? makes sense

Answer 5

just want to make sure though...this question is correct?

Answer 6

yeah
thats correct, so I figured the equation I wanted was y'(x)=sin(3e^(ln3x)) but not sure

Answer 7

i dont think so

Answer 8

I know its incorrect

Answer 9

you are given \[\large y = \int_0^{\ln (3x)} \sin (3e^t)dt\]
if i were you..i'd do \[\int \sin (3e^t) dt\] first. do you know how to?

Answer 10

not really no

Answer 11

you have discussed integration techniques like integration by parts, trig sub, etc right?

Answer 12

i know its

cos(3e^t)*3e^t

Answer 13

that's the derivative

Answer 14

oh yeah..I can't seem to do this one..does u=3e^t?

Answer 15

du=3e^t dt

Answer 16

like i asked you...have you discussed integration by parts?

Answer 17

somewhat

Answer 18

oh..somewhat's not good...try if you can do this by parts

let u = sin (3e^t)

dv = dt

find du and v

Answer 19

du=cos(3e^t)*3e^t dx

Answer 20

further help please?

Answer 21

right

Answer 22

except it's dt

Answer 23

now what would be v if dv = dt

Answer 24

yes correct typo

Answer 25

im not sure to be honest, v as in another u-sub ?

Answer 26

3e^t?

Answer 27

any one able to give me any feedback?

Answer 28

What is y'(x) if y=integral from 0 to ln(3x) of sin(3e^t)dt..I know to consider the fundamental theorem of calculus

\[y=\int_{a}^{b}f(x)\ dx\]or to write it another way\[y=\int_{a}^{b}f'(x)\ dx=f(b)-f(a)\]
\[\frac{d}{dx}\left[y=f(b)-f(a)\right]\to\ \frac{dy}{dx}=f'(b)-f'(a)\]