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As stated before, the second part of the Fundamental Theorem of Calculus states the following:

Theorem: Suppose $f(x)$ and $F(x)$ are real-valued functions on $[a,b]$ such that $F(x)$ is differentiable on $[a,b]$ and $F'(x)=f(x)$ for all $x$ in that interval. Then $$\int_a^b f(x)\, dx=F(b)-F(a).$$

However, just for this section, we will also assume that $f(x)$ is continuous on that interval. This difference will become important later, and we discuss the difference here.

First, let's show that Part 1 of the Fundamental Theorem of Calculus can prove this theorem. Before we begin, we need to start with an important lemma, which is stated and proven here. To repeat, it states the following:

Lemma 1: Suppose $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f'(x)=0$ for all $x$ in $(a,b)$, then prove that $f(x)$ is constant.

With it, we move on to a new lemma. Using the previous lemma, try to prove this on your own first!

Lemma 2: Suppose $f(x)$ and $g(x)$ are differentiable functions in $(a,b)$, and that $f'(x)=g'(x)$ for any $x$ in that interval. Then there exists a constant $c$ such that $g(x)=f(x)+c$ for all $x$ in $(a,b)$.

To prove this lemma, let's define a new function: $$h(x)=g(x)-f(x).$$ We differentiate both sides to find that $$h'(x)=[g(x)-f(x)]'=g'(x)-f'(x)=0.$$ Since this holds for all $x$ in the interval, by Lemma 1 it follows that $h(x)$ is a constant function- i.e. there exists a constant $c$ for which $$g(x)-f(x)=h(x)=c,$$ as desired. Now that we have this lemma, we can use it, along with the First Fundamental Theorem of Calculus, to prove the Second Fundamental Theorem of Calculus! Again, try it on your own first. (Hint: If $f(x)$ is an integrable function, define a new function $$G(x)=\int_a^x f(t)\, dt.$$ Use this with the lemmas that we proved above.)

Okay, let's prove this theorem together. Suppose $f(x)$ and $F(x)$ are real-valued functions on the interval $[a,b]$ such that $F'(x)=f(x)$ for all $x$ in $[a,b]$. We wish to use the First Fundamental Theorem of Calculus to prove that, if $f(x)$ is integrable, then $$\int_a^b f(t)\, dt=F(b)-F(a).$$ To do so, define a new function on $[a,b]$: $$G(x)=\int_a^x f(t)\, dt.$$ By the First Fundamental Theorem of Calculus, $$G'(x)=\frac{d}{dx} G(x)=\frac{d}{dx}\int_a^x f(t)\, dt=f(x)=F'(x).$$ Since $G'(x)=F'(x)$ for all $x$ in $[a,b]$, by Lemma 2 we have $$G(x)=F(x)+C,$$ for some constant $C$. Now, we observe that $$G(a)=\int_a^a f(t)\, dt=0.$$ So, it follows that $$0=G(a)=F(a)+C\Longrightarrow C=-F(a).$$ So now, we have $$G(x)=F(x)-F(a).$$ Letting $x=b$, we now get $$G(b)=\int_a^b f(t)\, dt=F(b)-F(a),$$ as desired.

As explained before, this theorem connects the integral of a function to its anti-derivative, thereby giving us a much easier method of integration. As an example, let's find the value of $$\int_0^2 x^2\, dx.$$ By Lemma 2 from above, we know that any anti-derivative of $f(x)=x^2$ is of the form $F(x)=\frac{1}{3}x^3+C$, where $C$ can be any constant. So, by Part 2 of the Fundamental Theorem of Calculus, $$\int_0^2 x^2\, dx=\left(\frac{1}{3}x^3+C\right)\bigg|_0^2=\left(\frac{1}{3}(2)^3+C\right)-\left(\frac{1}{3}(0)^3+C\right)=\frac{8}{3}.$$ As you've probably realized, this is a much easier way to calculate a definite integral than using the definition of integrals!

Now you try!

We note that for the function $f(x)=\cos{x}$, its anti-derivative is $F(x)=\sin{x}+C$ for some constant $C$. Then, by Part 2 of the Fundamental Theorem of Calculus, $$\int_0^{\pi} \cos{x}\, dx=(\sin{x}+C)\bigg|_0^{\pi}=(\sin{\pi}+C)-(\sin{0}+C)=0.$$ As you've probably noticed, this theorem will become very useful for evaluating integrals. So keep this in mind!