Question

what does this mean, i am going over it in my book, but i cant quite get a grip on the concept

Answer 1

it means take the derivative of the integral. which is just the function

Answer 2

Apply the fundamental theorem of calculus

Answer 3

so in this case should i just first integrate cost which would be -sint right, then take the derivative , but that will lead right bakc to cost

Answer 4

at what point do i evluate this problem

Answer 5

do you have to evaluate cos t?

Answer 6

\[(F(\sqrt{x}))'\]

Answer 7

well, its a two part question, it asks me to find the derivative or the problem i posted, by a)evaluating the integral and differentiating the result and b)differntiating the integral directly

Answer 8

i need some help actually doing the problem i am stuck

Answer 9

F=-sint right

Answer 10

Are you familiar with the fundamental theorem of calculus?

Answer 11

well, i read over it .....not to sure how to apply it though

Answer 12

i know one part says that i can take the antiderivative and evluate at the endpoints and get the area

Answer 13

\[(F(\sqrt{x}))' = F'(\sqrt{x})(\sqrt{x})' = \cos(\sqrt{x})(\sqrt{x})' = \frac{1}{2}\frac{cos(\sqrt{x})}{\sqrt{x}} \]

Answer 14

You are just applying the chain rule and the FTC

Answer 15

the derivative of A is B. when you integrate B, you get A. your problem ask you to take the derivative of the integral of B which is the same as taking the derivative of A, which is B.

Answer 16

now, someone please clear this up: do I have to evaluate the integral at 0 and sqrt of x?

Answer 17

well to evluate the integral

Answer 18

\[\int_{0}^{\sqrt{x}}\cos(t)dt = \sin(\sqrt{x})\]
\[(\sin(\sqrt{x}))' =\frac{1}{2}\frac{cos(\sqrt{x})}{\sqrt{x}}\]

Answer 19

But pay attention to the previous solution which uses the fundamental theorem of calculus.

Answer 20

And understand what is happening

Answer 21

\[(F(\sqrt{x}))' = F'(\sqrt{x})(\sqrt{x})' = \cos(\sqrt{x})(\sqrt{x})' = \frac{1}{2}\frac{cos(\sqrt{x})}{\sqrt{x}}\]

Answer 22

F represents the integral

Answer 23

The integrated right

Answer 24

The integral itself is a function.

Answer 25

Alchemista, it looks like you are talking about the chain rule?

Answer 26

There are two rules being used, the chain rule and the fundamental theorem of calculus.

Answer 27

First part of the FTC states:
\[F(x) = \int_{a}^{x}f(x)dx\]\[F'(x) = f(x)\]