# Thread: 2nd Fundamental Theorem of Calculus

1. ## 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)

2. 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,
$\displaystyle \int _{x_0}^x v(t-k) dt=S(x)+C$
Now, the function on the left is a composition function,
$\displaystyle f(t)=t-k$ and, $\displaystyle g(t)=\int_{x_0}^x v(t) dt$
Then,
$\displaystyle \int_{x_0}^x v(t-k) dt= g\circ f$
Then the derivative of this is evaluated throw the can rule.
Since $\displaystyle f,g$ are differenciable we have,
$\displaystyle f' \cdot g'\circ f$
Now,
$\displaystyle f'=1$
$\displaystyle g'=v(t)$----> 2nd Fundamental Theorem.
Thus,
$\displaystyle v(t-k)=S'(x)$

3. 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
$\displaystyle \int v(t-k) dt = S(t)+C$

is a sloppy way of writing:

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

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

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

so:

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

RonL

4. 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:

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

so:

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

RonL