Fundamental Theorem of Calculus Part 1

Definition: Fundamental Theorem of Calculus Part 1 (FTC1)

If you have a continuous function, f , on a closed x-interval [ a , b ], then the antiderivative $F\left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)\mathit{dt}$ is continuous on [ a , b ]and differentiable on the open interval ( a , b ), and its derivative is $f\left(x\right)$, then

$F\prime \left(x\right)=\frac{d}{\mathit{dx}}\underset{a}{\overset{x}{\int }}f\left(t\right)\mathit{dt}\mathit{=}f\left(x\right)$

Note : The format for applying the FTC1 is very specific. The lower bound, a , on your integral must be a constant , and the upper bound must be an equation, x .

Meaning:

What this rule is saying is that if you have some equation, $F\left(x\right)$, that is equal to an integral that fits the specific format requirements,

$F\left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)\mathit{dt}$

then when you take the derivative, $\frac{d}{\mathit{dx}}$,on both sides of the equals (just like you were doing an implicit differentiation problem),

$\frac{d}{\mathit{dx}}F\left(x\right)=\frac{d}{\mathit{dx}}\underset{a}{\overset{x}{\int }}f\left(t\right)\mathit{dt}$

your result will be of the form,

$F\prime \left(x\right)=\frac{d}{\mathit{dx}}\underset{a}{\overset{x}{\int }}f\left(t\right)\mathit{dt}\mathit{=}f\left(x\right)$

Which is found by plugging your upper bound, x , into the equation, $f\left(t\right)$, inside you integral.

Essentially a derivative, $\frac{d}{\mathit{dx}}$, undoes an antiderivative, $\underset{a}{\overset{x}{\int }}$, and the result would be what is inside the integral, $f\left(t\right)$, with the upper bound, x , plugged in, $f\left(x\right)$.

Simplifying that final step, you see $F\prime \left(x\right)\mathit{=}f\left(x\right)$. This derivative connection is how you will be asked to connect the FTC1 back to local max, local min, intervals of increasing and decreasing.

Note: If your upper bound is more than just an x , you will need do an additional step beyond plugging in the upper bound, you will also need to multiply the $f\left(x\right)$ by the derivative of the upper bound as well. Just like when you learned derivatives , if that is anything more than just an x , then you are looking at a chain rule derivative , and must multiply by the derivative of the inside piece, DIN .