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Mathematics 15 Online
OpenStudy (anonymous):

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

OpenStudy (lgbasallote):

this question is weird...

OpenStudy (lgbasallote):

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

OpenStudy (lgbasallote):

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

OpenStudy (lgbasallote):

so derivative of an integral? makes sense

OpenStudy (lgbasallote):

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

OpenStudy (anonymous):

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

OpenStudy (lgbasallote):

i dont think so

OpenStudy (anonymous):

I know its incorrect

OpenStudy (lgbasallote):

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?

OpenStudy (anonymous):

not really no

OpenStudy (lgbasallote):

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

OpenStudy (anonymous):

i know its cos(3e^t)*3e^t

OpenStudy (lgbasallote):

that's the derivative

OpenStudy (anonymous):

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

OpenStudy (anonymous):

du=3e^t dt

OpenStudy (lgbasallote):

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

OpenStudy (anonymous):

somewhat

OpenStudy (lgbasallote):

oh..somewhat's not good...try if you can do this by parts let u = sin (3e^t) dv = dt find du and v

OpenStudy (anonymous):

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

OpenStudy (anonymous):

further help please?

OpenStudy (lgbasallote):

right

OpenStudy (lgbasallote):

except it's dt

OpenStudy (lgbasallote):

now what would be v if dv = dt

OpenStudy (anonymous):

yes correct typo

OpenStudy (anonymous):

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

OpenStudy (anonymous):

3e^t?

OpenStudy (anonymous):

any one able to give me any feedback?

OpenStudy (amistre64):

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

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