The Fundamental Theorem of Calculus

Calculus has two big ideas: the derivative and the integral. The derivative comes from slopes. The integral comes from areas. They look like different topics. The Fundamental Theorem of Calculus shows that they are opposite processes, like multiplying and dividing.

The theorem also gives a fast way to compute integrals. No more long limits of sums.

The area function

Let $f$ be continuous on $[a,b]$. For each $x$ in $[a,b]$, define

$\displaystyle g(x)=\int_a^x f(t)\,dt$.

If $f\ge 0$, then $g(x)$ is the area under the graph of $f$ from $a$ to $x$. As $x$ moves to the right, the area changes. So $g$ is a function of $x$. (We use the letter $t$ inside the integral, because $x$ is now a limit of integration.)

The area function $g(x)$ is the light teal area from $a$ to $x$. When $x$ grows by $h$, the area grows by the thin dark strip, which is about $f(x)\cdot h$.

Example 1: Let $f(t)=t$ and $a=0$. Find $g(x)$ and $g'(x)$.

Solution:

The region under $y=t$ from $0$ to $x$ is a triangle with base $x$ and height $x$. So

$\displaystyle g(x)=\int_0^x t\,dt=\dfrac{1}{2}x\cdot x=\dfrac{x^2}{2}$.

Then $g'(x)=x$. This is the original function $f$, with $t$ replaced by $x$.

This is not a coincidence. It is always true.

Part 1: the derivative of an integral

The Fundamental Theorem of Calculus, Part 1: If $f$ is continuous on $[a,b]$, then the function

$\displaystyle g(x)=\int_a^x f(t)\,dt$

is differentiable, and

$g'(x)=f(x)$.

In words: if you integrate $f$ and then differentiate, you get $f$ back.

Why: Look at the figure. When $x$ grows by a small amount $h$, the area grows by a thin strip. The strip is $h$ wide and about $f(x)$ tall. So

$g(x+h)-g(x)\approx f(x)\cdot h$.

Divide by $h$:

$\dfrac{g(x+h)-g(x)}{h}\approx f(x)$.

As $h\to 0$, the strip becomes thinner, and the estimate becomes exact. The left side approaches $g'(x)$. So $g'(x)=f(x)$.

Example 2: Find the derivative of

$\displaystyle g(x)=\int_1^x\sqrt{1+t^3}\,dt$.

Solution:

The integrand $\sqrt{1+t^3}$ is continuous. By Part 1, replace $t$ by $x$:

$g'(x)=\sqrt{1+x^3}$

We did not need to find the integral. (In fact, it has no simple formula.)

Example 3: Find

$\displaystyle\dfrac{d}{dx}\int_1^{x^2}\cos t\,dt$.

Solution:

The upper limit is $x^2$, not $x$. So use the chain rule. Let $u=x^2$. Then the integral is $\displaystyle\int_1^u\cos t\,dt$, and

$\begin{align*}&\dfrac{d}{dx}\int_1^{x^2}\cos t\,dt\\&=\cos u\cdot\dfrac{du}{dx}\\&=\cos(x^2)\cdot 2x\end{align*}$

Part 2: computing integrals

The Fundamental Theorem of Calculus, Part 2: If $f$ is continuous on $[a,b]$, and $F$ is any antiderivative of $f$, then

$\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$.

In words: to integrate $f$ from $a$ to $b$, find an antiderivative, and subtract its values at the two ends.

We often write $F(b)-F(a)$ in a short form:

$\Big[F(x)\Big]_a^b=F(b)-F(a)$

Why: By Part 1, $g(x)=\displaystyle\int_a^x f(t)\,dt$ is an antiderivative of $f$. Two antiderivatives differ by a constant, so $F(x)=g(x)+C$. Then

$\begin{align*}&F(b)-F(a)\\&=[g(b)+C]-[g(a)+C]\\&=g(b)-g(a)\\&=\int_a^b f(t)\,dt-0\end{align*}$

(Here $g(a)=0$, because an integral from $a$ to $a$ is $0$.)

Any antiderivative works, because the constant $C$ cancels. So we use the one with $C=0$.

Example 4: Find

$\displaystyle\int_0^1 x^2\,dx$.

Solution:

An antiderivative of $x^2$ is $\dfrac{x^3}{3}$. So

$\displaystyle\int_0^1 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{3}-0=\dfrac{1}{3}$.

In areas and Riemann sums, we needed sigma notation, a sum formula, and a limit to get this answer. Now it takes one line.

Example 5: Find

$\displaystyle\int_1^3(3x^2-2x+1)\,dx$.

Solution:

An antiderivative is $x^3-x^2+x$.

$\begin{align*}&\int_1^3(3x^2-2x+1)\,dx\\&=\Big[x^3-x^2+x\Big]_1^3\\&=(27-9+3)-(1-1+1)\\&=21-1\\&=20\end{align*}$

Use parentheses around each value, so the minus sign applies to all of $F(a)$.

Example 6: Find the area under one arch of $y=\sin x$, from $0$ to $\pi$.

Solution:

An antiderivative of $\sin x$ is $-\cos x$. Angles are in radians.

$\begin{align*}&\int_0^\pi\sin x\,dx\\&=\Big[-\cos x\Big]_0^\pi\\&=-\cos\pi-(-\cos 0)\\&=1+1\\&=2\end{align*}$

The area is exactly $2$.

Example 7: Find each integral.

(a) $\displaystyle\int_1^e\dfrac{1}{x}\,dx$

(b) $\displaystyle\int_1^4\sqrt{x}\,dx$

Solution:

(a) An antiderivative of $\dfrac{1}{x}$ is $\ln x$ (here $x$ is positive).

$\displaystyle\int_1^e\dfrac{1}{x}\,dx=\ln e-\ln 1=1-0=1$

(b) An antiderivative of $x^{1/2}$ is $\dfrac{2}{3}x^{3/2}$.

$\begin{align*}&\int_1^4\sqrt{x}\,dx\\&=\left[\dfrac{2}{3}x^{3/2}\right]_1^4\\&=\dfrac{2}{3}(8)-\dfrac{2}{3}(1)\\&=\dfrac{14}{3}\end{align*}$

Here $4^{3/2}=(\sqrt{4})^3=8$.

Be careful: $f$ must be continuous

Example 8: What is wrong with this work?

$\begin{align*}&\int_{-1}^{1}\dfrac{1}{x^2}\,dx\\&=\left[-\dfrac{1}{x}\right]_{-1}^{1}\\&=-1-1\\&=-2\end{align*}$

Solution:

The function $1/x^2$ is always positive, so its integral cannot be negative. Something is wrong.

The problem is that $1/x^2$ is not continuous on $[-1,1]$. It is undefined at $x=0$, and it approaches $\infty$ there. So Part 2 of the theorem does not apply.

Summary

Part 1: $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)$. Integrating and then differentiating gives back $f$.
Part 2: $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$, where $F$ is any antiderivative of $f$.
Both parts need $f$ to be continuous on the interval.
Differentiation and integration are opposite processes.

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