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
this question is weird...
you're looking for y'(x)...but you're given y...
and you're supposed to do an integral...
so derivative of an integral? makes sense
just want to make sure though...this question is correct?
yeah thats correct, so I figured the equation I wanted was y'(x)=sin(3e^(ln3x)) but not sure
i dont think so
I know its incorrect
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?
not really no
you have discussed integration techniques like integration by parts, trig sub, etc right?
i know its cos(3e^t)*3e^t
that's the derivative
oh yeah..I can't seem to do this one..does u=3e^t?
du=3e^t dt
like i asked you...have you discussed integration by parts?
somewhat
oh..somewhat's not good...try if you can do this by parts let u = sin (3e^t) dv = dt find du and v
du=cos(3e^t)*3e^t dx
further help please?
right
except it's dt
now what would be v if dv = dt
yes correct typo
im not sure to be honest, v as in another u-sub ?
3e^t?
any one able to give me any feedback?
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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