# Thread: fundamental theorem of cal

1. ## fundamental theorem of cal

I'm not sure how to begin/approach/do this problem or apply the theorem.

2. Originally Posted by frog09
I'm not sure how to begin/approach/do this problem or apply the theorem.
You must deduce an equation for f(x) from the given graph. It is the graph of sin(x), shifted 1 place along positive x, and shifted 2 placed along positive y, and squished/expanded by a factor of... something! You must then inject that equation into the integral:

$g(x) = \int_{-3}^{2x-1}f(t)dt$

And carry out the integration to get a function in x for g(x).

3. Originally Posted by frog09
I'm not sure how to begin/approach/do this problem or apply the theorem.
The second fundemental theorem is

$\frac{d}{dx} \int_a^x f(t)\, dt = f(x)$
or if

$g(x) = \int_a^x f(t)\, dt$ then $g'(x) = f(x)$

However to use this theorem, it must look exactly of this form. What you have is

$g(x) = \int_{-3}^{2x-1} f(t)\, dt$

so use the chain rule. Let $u = 2x-1$

$\frac{d}{dx} \int_{-3}^{u} f(t)\, dt = \frac{d}{du} \int_{-3}^{u} f(t)\, dt \cdot \frac{d u}{dx}$
Can you continue?

4. would $f'(x)=g(x)$ be true in this case too?

and how do you use this theorem to find domain of g?

5. Originally Posted by frog09
would $f'(x)=g(x)$ be true in this case too?

and how do you use this theorem to find domain of g?
If $g(x)=\int_{u(x)}^{v(x)} f~dx$ then the domain of $g$ would the values of $x$ that make $\int_{u(x)}^{v(x)} f~dx$ defined.

EDIT: Typo

6. ok i think i may have got it...
$f'(x)=g(x)$
so by the fundamental theorem
$f'(x)= f(u) du/dx$
$f'(x)= g(2x-1)* 2$
which would mean
$g(x)= (1/2) \int_{-3}^u u du$

u=2x-1
so when x=1/2, u=-1; and when x=-3, u=-7

changing the limits would change the domain to [-1, -7]?!

7. Originally Posted by frog09
ok i think i may have got it...
$f'(x)=g(x)$
so by the fundamental theorem
$f'(x)= f(u) du/dx$
$f'(x)= g(2x-1)* 2$
which would mean
$g(x)= (1/2) \int_{-3}^u u du$

u=2x-1
so when x=1/2, u=-1; and when x=-3, u=-7

changing the limits would change the domain to [-1, -7]?!
If $g(x) = \int_{-3}^{2x-1} f(t) \, dt$ then $g' (x)= 2f(2x-1)$, not $f'(x)=g(x)$!

Now we will use $g' (x)= 2f(2x-1)$, to get an idea of the domain of $g(x)$, i.e. places where $g$ is defined. From your graph $f(t)$ is only defined on

$-3 \le t \le 5$ so $f(2x-1)$ is only defined on $-3 \le 2x-1 \le 5\;\; \Rightarrow\;\; -1 \le x \le 3$

So in this interval $f(2x-1)$ is defined so $g'(x)$ is defined and since

$g(x) = \int_{-3}^{2x-1} f(t)\,dt$ then so is $g(x)$.

8. just one more question... in order to find where g(x) is a maximum, how would i adjust the f graph in order to look for the max?

i know a maximum at x will be when it changes from positive to negative at x, but on the f graph, there is no such point.

9. Originally Posted by frog09
just one more question... in order to find where g(x) is a maximum, how would i adjust the f graph in order to look for the max?

i know a maximum at x will be when it changes from positive to negative at x, but on the f graph, there is no such point.
The maximum (or minimum) will occur at a critical point, a place where $g' = 0 \; \text{or}\; g' DNE$. So

$g'(x) = 2 f(2x-1)$ and so $g'(x) = 0 \; \text{when}\;2 f(2x-1) = 0$ and from the graph $f(0)=0$ which means that $x = \frac{1}{2}$