# The Fundamental Theorem of Calculus

# The Fundamental Theorem of Calculus (FTC), Part 1

- This theorem has such a big name, because it is a big theorem. It gives a direct connection between two seemingly unrelated topics:
- derivatives, that is slopes of graphs,
- and definite integrals, that is net areas under graphs.

*Theorem.* Let \(f\) be a continuous function on \([a,b]\). Then the function \(g\) defined by \[
g(x)=\int_a^xf(t)\,\mathrm dt\quad a\le x\le b
\] is continuous on \([a,b]\), differentiable on \((a,b)\), and satisfies \[
g'(x)=f(x).
\]
- That is, we have a formula between definite integrals and antiderivatives!
- Note that this implies that if \(F\) is any antiderivative of \(f\), then we have \[
F(x)=\int_a^xf(t)\,\mathrm dt+C\quad a\le x\le b
\] for some constant \(C\).
*Warning.* You need to make sure that the upper boundary \(x\) is a different variable from the variable of integration \(t\). You get wrong answers if you miss this!
- You can see the proof of this theorem in the textbook.

# Exercises

- Since we started considering functions where the function variable (usually \(x\)) is a boundary of an integral, we want to make sense of integrals \(\int_b^af(x)\,\mathrm dx\) in case \(a<b\).
- We define \(\int_b^af(x)\,\mathrm dx\) as \(-\int_a^bf(x)\,\mathrm dx\).
- The interval splitting formula \(\int_a^bf(x)\,\mathrm dx=\int_a^cf(x)\,\mathrm dx+\int_c^bf(x)\,\mathrm dx\) remains true for \(a,b,c\) not necessarily satisfying \(a\le c\le b\).
- Exercises. 5.3: 8, 12, 14, 18, 60, 62, 64, 65.

# FTC Part 2

*Theorem.* Let \(f\) be a continuous function on \([a,b]\). Suppose that \(F\) is an antiderivative of \(f\). Then we have \[
\int_a^bf(x)\,\mathrm dx=F(b)-F(a)
\]
- Usually this is the way we want to calculate definite integrals, so this is the main reason why knowing how to find antiderivatives of functions is important.
*Notation.* Since because of this we see formulas like \(F(b)-F(a)\) a lot, where \(F(x)\) might itself have a long formula, we use the following notations. \[
F(x)\Big|_{x=a}^b=F(x)\Big|_a^b=F(x)\Big]_a^b=\Big[F(x)\Big]_a^b=F(x=b)-F(x=a).
\]
- Exercises. 5.3: 20, 22, 26, 32, 40, 42, 44, 52, 76, 82