# 2nd Fundamental Theorem of Calculus

• Nov 6th 2006, 07:29 AM
wwilpi
2nd Fundamental Theorem of Calculus
These problems are found in a calculus text at the end of a section on the 2nd FTC. I have examined these for a couple of hours, and cannot come to a solution.

Given: The indefinite integral[v(t-k)dt] = S(t) + C, where 'k' and 'C' are constants.
Find: S'(t)+c

Given: The indefinite integral[a*f(x)dx] = g(ax) + C, where 'C' is a constant and 'a' is a constant <>0.
Find: g'(x)
• Nov 6th 2006, 08:42 AM
ThePerfectHacker
Quote:

Originally Posted by wwilpi
Given: The indefinite integral[v(t-k)dt] = S(t) + C, where 'k' and 'C' are constants.
Find: S'(t)+c

The problem is these are not so well-defined and lack math rigor. So I am going to change it slightly.

What you should have written is,
$\int _{x_0}^x v(t-k) dt=S(x)+C$
Now, the function on the left is a composition function,
$f(t)=t-k$ and, $g(t)=\int_{x_0}^x v(t) dt$
Then,
$\int_{x_0}^x v(t-k) dt= g\circ f$
Then the derivative of this is evaluated throw the can rule.
Since $f,g$ are differenciable we have,
$f' \cdot g'\circ f$
Now,
$f'=1$
$g'=v(t)$----> 2nd Fundamental Theorem.
Thus,
$v(t-k)=S'(x)$
• Nov 6th 2006, 08:49 AM
CaptainBlack
Quote:

Originally Posted by wwilpi
These problems are found in a calculus text at the end of a section on the 2nd FTC. I have examined these for a couple of hours, and cannot come to a solution.

Given: The indefinite integral[v(t-k)dt] = S(t) + C, where 'k' and 'C' are constants.
Find: S'(t)+c

$
\int v(t-k) dt = S(t)+C
$

is a sloppy way of writing:

$
\int_0^t v(\tau-k) d\tau + S(0)+C = S(t)+C
$

Now differentiate this with respect to $t$ to get:

$
v(t-k) = S'(t)
$

so:

$
S'(t)+c = v(t-k)+c
$

RonL
• Nov 6th 2006, 08:55 AM
CaptainBlack
Quote:

Originally Posted by wwilpi

Given: The indefinite integral[a*f(x)dx] = g(ax) + C, where 'C' is a constant and 'a' is a constant <>0.
Find: g'(x)

Similar to last problem, applying the same argumet we get:

$
a f(x) = \frac{d}{dx} g(ax) =a\ g'(ax)
$

so:

$
g'(x)=f(x/a)
$

RonL