4.3: The Fundamental Theorem of Calculus

Motivation

We derived the formula for the area underneath the curve for $f(x) = x^2$ on $[0, 1]$: \[\int^1_0 x^2 \ dx = \lim_{n\rightarrow \infty} \dfrac{1}{n^3} \sum^n_{i = 1}i^2\] This sum formula is not so easy to deal with.

In this section we find a simple way to find definite integrals by answering two questions:

  1. What is the relationship between the derivative and the integral?
  2. How do we exploit this relationship to easily find definite integrals?

FTC Part 1 (the What)

Consider the function \[g(x) = \int^x_a f(t) \ dt\] where $f(t)$ is a continuous function on $[a, b]$.

Consider $f(x) = x^2$ on $[0, 5]$. Then $g(3)$ is equal to \[g(3) = \int^3_0 x^2 \ dx\] which is the area underneath the curve of $x^2$ on $[0, 3]$.

In other words, this function gives you control over the right bound of integration for function $f$.

We now take the derivative of $g(x)$.

Fundamental Theorem of Calculus, Part 1
If $f$ is continuous on $[a, b]$, then the function \[g(x) = \int^x_a f(t) \ dt \qquad a \leq x \leq b\] is also continuous on $[a, b]$, differentiable on $(a, b)$, and $g'(x) = f(x)$.
Find the derivative of the function \[g(x) = \int^x_0 \sqrt{1 + t^2} \ dt\]
Find the derivative of the function \[g(x) = \int^0_x \sin(t) \ dt\]

Intuition for FTC Part 1

Note that the independent variable for $g$ is $x$. When you take the derivative it is respect to $x$.

So FTC Part 1's conclusion, $g'(x) = f(x)$, is really saying \[g'(x) = \dfrac{d}{dx} \int^x_a f(t) \ dt = f(x) \]

In plain English, this says starting with function $f$, if you integrate first, then take the derivative second, then you end up back with $f(x)$.

This means the integral and derivative are inverse processes: each undoes what the other has done.

In Section 3.9 we saw the opposite process of the derivative is the antiderivative.

Therefore, finding integrals involves taking antiderivatives.

FTC Part 2 (the How)

To find any definite integral using antiderivatives, we have the following theorem:

Fundamental Theorem of Calculus, Part 2
If $f(x)$ is continuous on $[a, b]$, then \[\int^b_a f(x) \ dx = F(b) - F(a)\] where $F(x)$ is any antiderivative of $f(x)$.

Notation: to succinctly represent $F(b) - F(a)$, we use a vertical bar\[F(x) \Bigg|^b_a=F(b) - F(a) \]

Integrate $\displaystyle \int^1_0 x^2 \ dx$.
Integrate $\displaystyle \int^1_0 1 + 4x \ dx$.
Integrate $\displaystyle \int^{\pi/2}_0 x + \cos x \ dx$.
Try integrating using FTC Part 2: \[\int^3_{-1} \dfrac{1}{x^2} \ dx\] What is wrong with this calculation?

The Fundamental Theorem of Calculus is one of the most important mathematical achievements in human history.

Instead of having to calculate a difficult infinite limit sum, we can use antiderivatives instead! It's like magic.