Question

Using the fundamental theorem of calculus, what do I do when my lower limit on my integral is sinx?

Answer 1

Evaluate the lower limit of the integral...?? o.o

Answer 2

\[\int\limits_{a}^{b} f(x) dx = F(b) - F(a)\]

Answer 3

how?! d/dx int[sinx,1] sqrt(1+t^2)dt

Answer 4

You plug sin x in.

Answer 5

\[\int\limits_{1}^{\sin x} dt = \sin x - 1\]

Answer 6

It says to use part 1 of the fundamental theorem of calculus... >.<

Answer 7

As an example.

Answer 8

That's not really what I'm asking. For the theorem, the upper limit has to be some factor of x, but my lower limit is the factor of x instead. Can you tell me how to change that?

Answer 9

Err sorry, I must be going crazy. I can't understand what you're trying to portray >.<

Answer 10

Can you write your problem out nicely, preferably with LaTeX lol.

Answer 11

\[d/dx \int\limits_{sinx}^{1} \sqrt{1+t ^{2}}dt\]

Answer 12

oh i see, it doesn't matter if its the lower limit
what this does is make it negative but its still evaluated the same way

Answer 13

\[d/dx (F(1) - F(\sin x))\]
\[= 0 - \cos x *f(\sin x)\]

Answer 14

So you just took the anti-derivative of \[\sqrt{1+t ^{2}}\] and evaluated it at 1 and sinx?

Answer 15

That's not what I was thinking. Man, I would hate to give the answer, but I just can't explain it lol.

Answer 16

I was thinking:
\[\sqrt(2) - \sqrt(1+\sin^2 x) * \cos x\]

Answer 17

http://www.sosmath.com/calculus/integ/integ03/integ03.html
Check out the bottom of this page.

Answer 18

So you're saying F'(1) = F'(sinx)?

Answer 19

Can you just make the integral negative and switch the bounds? As in:
\[-\int\limits_{1}^{sinx}\]