Question

Use part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. h(x)=∫ arctan 5t dt on the interval (4, 1/x). I got -5arctan(1/x)/x^2 but it was wrong

Answer 1

well, when you take a derivative of an integral it should just be the function being integrated (arctan(5t)) evaluated at the upper bound (4)

Answer 2

but what about 1/x?

Answer 3

From what I've been taught, the lower bound is insignificant when differentiating an integral.

Answer 4

im sorry you 1/x is the upper bound isn't it. so it should be evaluated at that

Answer 5

so it should be arctan(5/x)

Answer 6

and do you put the 5 in front of that too?

Answer 7

honestly, im a little unsure about this problem. In calc 4, we seem to treat the bottom bound as insignificant, but i did this problem on wolfram alpha and it gave: -arctan(5/x)/(x^2). So maybe im not the best one to be answering this. Sorry.

Answer 8

actually, you are the best one to be answering it because it was correct! Thank you!!

Answer 9

ok, great. like i said i just just used wolfram alpha. So im not really sure how to get to the answer the "correct" way. But maybe you can figure it out.

Answer 10

well regardless thank you. I didn't know about the bottom bound being insignificant, so i'll keep that in mind!

Answer 11

im not sure on this. but that may only be true if its a constant, not if its a function like x or something.

Answer 12

ohh that would make sense

Answer 13

you can check out http://mathmistakes.info/facts/CalculusFacts/learn/doi/doib.html
for an example of how the bottom being a function affects things

Answer 14

Thank you! That would be helpful

Answer 15

No problem.