Question

How do you use the second fundamental theorem of calculus to find a derivative of the function of an integral from -4 to sin(x)?

Answer 1

\[\int\limits_{-4}^{\sin(x)}[\cos(t^5)+t] dt\]

Answer 2

not sure what the second one is, but i think you are looking for: \[\fracd{dx}\int_{a}^{b}f(t)~dt=f(b)b'-f(a)a'\]

Answer 3

The second part is the actual question that I have. I'll write out what the second fundamental theorem says

Answer 4

if \[f(x)=\int\limits_{a}^{x}g(t)dt \] then \[f'(x)= g(x)\]

Answer 5

I can find the derivative with respect to sin(x) but not just x. I think. What I got was
\[f'(\sin(x))= \frac{ \cos(\sin^5(x))+\sin(x) }{ \cos(x) }\]

Answer 6

\[\frac d{dx}\int\limits_{-4}^{\sin(x)}[\cos(t^5)+t] dt\]\[\hspace{5em}=[~cos(sin^5(x)+sin(x)~]~cos(x) dt\]

Answer 7

why is it multiplied instead of divided?

Answer 8

by cos(x)

Answer 9

chain rule

Answer 10

Yes, but going back a step, wouldn't it be
\[f'(sinx)*\cos(x)=\frac{ d }{ dx }(\int\limits_{-4}^{\sin(x)}(\cos(t^5)+t) dt)\]?

Answer 11

by the definition of integrating
\[\int_{a(x)}^{b(x)} f(t)~dt=F(b(x))-F(a(x))\]

and by the definition of derivatives
\[D_x[F(b(x))-F(a(x))]\]
\[=D_x[F(b(x))]-D_x[F(a(x))]\]
\[=f(b(x))*b'(x)-f(a(x))*a'(x)\]

Answer 12

your a'=0 so that goes away

Answer 13

so you are left with f(b)*b'

Answer 14

b=sinx; b'=cosx

Answer 15

i think you are trying to make it more difficult simply because f' is not very pretty

Answer 16

Ive never seen the definition of a derivative written like that before. I'm having a hard time interpreting. But I still don't understand why the cos(x) is being multiplied instead of divided, and if that is f'(x)? I thought it was f'(sin(x))

Answer 17

does the chain rule pop out a division or a multiplication?

Answer 18

Multiplication, but when it changes sides of the equation, it'd be division

Answer 19

why would it change sides of the equation?

Answer 20

To get f'(x) by itself

Answer 21

I'm having a hard time with this concept because I was only taught through a video due to a strike. The textbook isn't making it easier.

Answer 22

your confusing something .... show me your steps so i can see what square peg your trying to cram into the round hole

Answer 23

Okay.

Answer 24

Since the upper bound is sin(x), I'm getting f(sin(x)) instead of f(x). That's my first issue.
I have it written like this:
\[f'(\sin(x))*\cos(x)=(\cos(\sin^5(x))+\sin(x))\]

Answer 25

Which is why I'm confused when it comes to the cos(x) being multiplied on the other side like you have it

Answer 26

would you agree that:\[F(x)=F(b)-F(a)\]

Answer 27

\[F(x)=F(sin(x))-F(-4)\]

Answer 28

I'm supposed to be using the second fundamental theorem, which I stated near the top of this... THats the first fundamental theorem

Answer 29

there is no appreciable difference between the two of them

Answer 30

the second is simply stating the consequences of the first if the lower limit is a constant

Answer 31

I can vaguely see that... Let me process for a minute.

Answer 32

Okay can you help me solve it using that and then help me translate it to the way that it is written for the Second Fundamental Theorem?

Answer 33

sure, in fact im using the the first to establish that F(x) is, then by deriving F(x) we get to the second thrms by default

Answer 34

so now we just differentiate both sides?

Answer 35

Oh wait.... the derivative.... what... F'(x)=F'(sin(x))??

Answer 36

Oh chain rule

Answer 37

so F is the integral function, right? And so F'(x)=F'(sinx)*cos(x)?

Answer 38

yes,
\[\Large F(x)=\int_{k}^{sinx}f(t)~dt\]
\[\Large F(x)=F(sin(x))-F(k)\]
\[\Large F'(x)=F'(sin(x))(sinx)'-F'(k)k'\]
since k', the derivative of any constant, is 0

\[\Large F'(x)=F'(sin(x))(sinx)'\]
\[\Large F'(x)=F'(sin(x))(cosx)\]

and F' = f

Answer 39

therefore\[F'(x)=f(sinx)~cosx\]

Answer 40

But f(x) is defined by an integral... so I can't type that in... because\[f(x)= \int\limits_{-4}^{sinx}(\cos(t^5)+t)dt\] right?

Answer 41

oh wait no...

Answer 42

no ...

Answer 43

f(x) is the inside of the integral, right?

Answer 44

f(t) is the inside of the integral; and the integral itself become F(x) the conventional use of upper and lower case f might be causing you some grief

Answer 45

Okay that is the right answer... But I am still confused how to approach it from the other direction

Answer 46

the derivative of F is f, and we are looking for the derivative of F

Answer 47

I understand the way that we just did it now.

Answer 48

if i were to try to adapt your setup
\[f(x)=\int\limits_{a}^{x}g(t)dt\]
\[f'(x)=g(x)\]

thats fine, but lets define the limit as a function of x, instead of just x; say u(x) = x .... then we have to play with a chain rule

\[f(x)=\int\limits_{a}^{u(x)}g(t)dt\]
\[f'(x)=g(u(x))*u'(x)\]

Answer 49

doesnt that revert to:

f'(x) = g(x)? since u(x) = x ; and u' = 1

Answer 50

Man, that's ugly.

Answer 51

:)

Answer 52

You've been a huge help. But, the way that mine is set up, why isn't it ending up with f'(u(x))?

Answer 53

becasue we are defining f'(x) in terms of g(u(x)) and u'(x) your setup seems a little better at expressing the point

Answer 54

if we ignore the (x) parts we can prolly read this even better

f' = g(u) * u'

Answer 55

Umm nope, it was making more sense with the 'x's to me.

Answer 56

lol, then leave them in ;)

Answer 57

Haha but even though it makes *more* sense, it doesn't quite click.

Answer 58

Because to me, you have x's in one equation, then you change one x to u(x), why doesnt the other x also change to u(x)?

Answer 59

no, go thru the long way, integrate up, insert the limits, and derive it back down .... that is the best way to understand it.

Answer 60

Argh. I'll ask my prof to help me through in more detail tomorrow. Thank you! Would you mind helping with one that is a definite integral?

Answer 61

\[f(x)=\int\limits_{a}^{u(x)}g(t)dt\]
\[f(x)=G(u(x))-G(k)\]
\[f'(x)=g(u(x))~u'(x)-g(k)\cancel{k'}^0\]
\[f'(x)=g(u(x))~u'(x)\]

Answer 62

ill help

Answer 63

in the first one, we are not replacing x with u(x), its just a consequence of the integration is all

Answer 64

i mean f(x) is not equal to f(u(x))

Answer 65

I think I need to think about that one by myself for a while :P
Okay. I already know the answer, I just cant get to it with these concepts.
\[\int\limits_{0}^{1}t^2dt\]

Answer 66

this ones pretty basic, what do you need help with?

Answer 67

I know that it's basic, I want to understand it so I can apply the same steps to a harder one. I'm supposed to use the set up that I gave you to evaluate it.

Answer 68

Which means I need x's on the boundaries, right? So I can split it into 2 integrals

Answer 69

you dont need xs ... we you trying to determine the derivative of the inegration again? or are you just trying to find the value of the integration itself?

Answer 70

Just the value. Sorry.

Answer 71

do you recall the power rule for derivatives?

Answer 72

And not using integration techniques. I mean I know how to integrate that.

Answer 73

not using integration techniques .... then what shall we try to use?

Answer 74

the limit of reimann sums?

Answer 75

The second fundamental theorem of calculus.

Answer 76

We've used riemann sums to prove this one in class.

Answer 77

I know it's sketchy going to a site introduced to you by a stranger on the internet, but cosketch is nice and clean and easier than typing these things out. It's just drawing... Could we use that?

Answer 78

the 2nd fundamental thrm states:

\[if~~f(x)=\int\limits_{a}^{x}g(t)~dt~~then~~f'(x)=g(x) \]

im not sure how we would use the second thrm to determine the value of the definite integration without actually intgrating it

Answer 79

Neither am I, but we haven't learned the reverse chain rule and the original problem I have would need that to integrate. So I know it can be done.

Answer 80

That's why I said that we needed x on the boundary

Answer 81

Oh! I misread! It says the First! I think I can solve it now.

Answer 82

in any case, you are going to need to integrate it up to a function that derives down to t^2

Answer 83

power functions derive down a degree and integrate up a degree

so there is some function k t^3 that has a derivate t^2

k t^3 derives to 3k t^2; what does k have to be in order for this to equal t^2?

Answer 84

Yeah I dont know how to solve without integrating.... I Know how to integrate it. But I'm not supposed to.

Answer 85

one third.

Answer 86

i think you need to integrate it, but then apply the first thrm to find the value

Answer 87

But the question that I have to solve is not integrable with the rules that we have learned. And we should be able to solve it.

Answer 88

so, F(t) = 1/3 t^3; and the first thrm says that the value is equal to F(b)-F(a)

Answer 89

I have a second, messy one. I just wanted to see if I could get the answer through these methods for an easy one first.

Answer 90

We arent supposed to integrate.

Answer 91

youre going to have to clear that up with your teacher than.

Answer 92

Okay. -sigh- Thanks anyways.

Answer 93

good luck with it :)