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\[(\int\limits_{0}^{x}(3t-1)^{10} dt)/3x\]
as x approaches 0 I'm confused because the question has two variables: x and t. How should I approach this?
0
Evaluate the limit in the top first.
sorry, I didn't see 3x :)
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Use L'Hospital rule
Oh right. L'Hopital's rule is the best choice: \[\frac{d}{dx}\int\limits_0^x(3t-1)^{10}dt=(3x-1)^{10}.\]
The derivative of the bottom is obviously 3. Hence the limit is \(\frac{3(0)-1)^{10}}{3}=\frac{1}{3}\).
This makes sense, but how did you know to use l'hopital's rule?
If we plug x=0, you will get \(\frac{0}{0}\). If you plug x in the top, you get an integral from 0 to 0, which results in a value of 0. I think it's very obvious in the denominator.
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oh okay, thanks a ton!
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