The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

How to solve fundamental theory of calculus question ?

How do I go about it3E1EFCC6-36F2-4A04-987B-B76447B9C2FB.jpeg
Define H(x)=∫0xh(x′) dx′H(x) = \int_0^x h(x')\,dx'. You know the derivative of H by FTC.

Write the original integral as a function of H and g, then use the chain rule.
Reply 2
Avatar for R1C3
R1C3
15
I've no idea what is going on here mathematically, but wolframalpha may be able to help?

Here is the LaTeX: \frac{d}{dx}\int_{0}^{f(x)}h(x')dx'
Original post by DFranklin
Define H(x)=∫0xh(x′) dx′H(x) = \int_0^x h(x')\,dx'. You know the derivative of H by FTC.

Write the original integral as a function of H and g, then use the chain rule.

I don’t get it , do we integrate h and subsituite f(x) and 0?
Reply 4
Avatar for Vinny C
Vinny C
20
God...
Reply 5
Avatar for RichE
RichE
15
Original post by studentzz123
I don’t get it , do we integrate h and subsituite f(x) and 0?

How would you rewrite the integral in your problem in terms of the function H that DFranklin introduced?
Original post by RichE
How would you rewrite the integral in your problem in terms of the function H that DFranklin introduced?

(Fx)^2/2 - 0 ?
Original post by studentzz123
(Fx)^2/2 - 0 ?


Huh? No.

Firstly, if H(x)=∫0xh(x′).dx′H(x) = \displaystyle \int_0^x h(x') .dx', then by FToC we have H′(x)=h(x)H'(x) = h(x).

Secondly, simply replacing xx by f(x)f(x) turns the integral formulation into

H(f(x))=∫0f(x)h(x′).dx′H(f(x)) = \displaystyle \int_0^{f(x)} h(x') .dx'

Which means that ddx∫0f(x)h(x′).dx′\dfrac{d}{dx} \displaystyle \int_0^{f(x)} h(x') .dx' is the same as ddxH(f(x))\dfrac{d}{dx}H(f(x)).

Can you proceed and simplify this down to a function made up of hh and ff ?
EB296E39-F848-46CB-A6C4-4CACD050F7D7.jpeg I’m not sure but I think I got it ?

Original post by RDKGames
Huh? No.

Firstly, if H(x)=∫0xh(x′).dx′H(x) = \displaystyle \int_0^x h(x') .dx', then by FToC we have H′(x)=h(x)H'(x) = h(x).

Secondly, simply replacing xx by f(x)f(x) turns the integral formulation into

H(f(x))=∫0f(x)h(x′).dx′H(f(x)) = \displaystyle \int_0^{f(x)} h(x') .dx'

Which means that ddx∫0f(x)h(x′).dx′\dfrac{d}{dx} \displaystyle \int_0^{f(x)} h(x') .dx' is the same as ddxH(f(x))\dfrac{d}{dx}H(f(x)).

Can you proceed and simplify this down to a function made up of hh and ff ?
Original post by studentzz123
I’m not sure but I think I got it ?


Looks good. It's as simple as that.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you