Calculus/Fundamental Theorem of Calculus

The fundamental theorem of calculus is a critical portion of calculus because it links the concept of a derivative to that of an integral. As a result, we can use our knowledge of derivatives to find the area under the curve, which is often quicker and simpler than using the definition of the integral.

As an illustrative example see § Template:Calculus/map page for the connection of natural logarithm and 1/x.

Mean Value Theorem for Integration

We will need the following theorem in the discussion of the Fundamental Theorem of Calculus.

Proof of the Mean Value Theorem for Integration

$f(x)$ satisfies the requirements of the Extreme Value Theorem, so it has a minimum $m$ and a maximum $M$ in $[a,b]$ . Since

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^{n}f(x_k^{*})\cdot \frac{b-a}{n}=\underset{n\to \mathrm{\infty }}{\lim }\frac{b-a}{n}\cdot \sum _{k=1}^{n}f(x_k^{*})$

and since

$m\le f(x_k^{*})\le M$ for all $x_k^{*}\in [a,b]$

we have

$\begin{array}{cc}& \underset{n\to \mathrm{\infty }}{\lim }\frac{b-a}{n}\cdot \sum _{k=1}^{n}m\le \underset{n\to \mathrm{\infty }}{\lim }\frac{b-a}{n}\cdot \sum _{k=1}^{n}f(x_k^{*})\le \underset{n\to \mathrm{\infty }}{\lim }\frac{b-a}{n}\cdot \sum _{k=1}^{n}M\\ & \underset{n\to \mathrm{\infty }}{\lim }mn\cdot \frac{b-a}{n}\le \underset{a}{\overset{b}{\int }}f(x)dx\le \underset{n\to \mathrm{\infty }}{\lim }Mn\cdot \frac{b-a}{n}\\ & \underset{n\to \mathrm{\infty }}{\lim }m(b-a)\le \underset{a}{\overset{b}{\int }}f(x)dx\le \underset{n\to \mathrm{\infty }}{\lim }M(b-a)\\ & m(b-a)\le \underset{a}{\overset{b}{\int }}f(x)dx\le M(b-a)\\ & m\le \frac{1}{b-a}\underset{a}{\overset{b}{\int }}f(x)dx\le M\end{array}$

Since $f$ is continuous, by the Intermediate Value Theorem there is some $f(c)$ with $c\in [a,b]$ such that

$\frac{1}{b-a}\underset{a}{\overset{b}{\int }}f(x)dx=f(c)$

Fundamental Theorem of Calculus

Statement of the Fundamental Theorem

Suppose that $f$ is continuous on $[a,b]$ . We can define a function $F$ by

$F(x)=\underset{a}{\overset{x}{\int }}f(t)dt\text{for }x\in [a,b]$

When we have such functions $F$ and $f$ where $F^{\prime }(x)=f(x)$ for every $x$ in some interval $I$ we say that $F$ is the antiderivative of $f$ on $I$.

Figure 1

Note: a minority of mathematicians refer to part one as two and part two as one. All mathematicians refer to what is stated here as part 2 as The Fundamental Theorem of Calculus.

Proofs

Proof of Fundamental Theorem of Calculus Part I

Suppose $x\in (a,b)$ . Pick $\mathrm{\Delta }x$ so that $x+\mathrm{\Delta }x\in (a,b)$ . Then

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

and

$F(x+\mathrm{\Delta }x)=\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt$

Subtracting the two equations gives

$F(x+\mathrm{\Delta }x)-F(x)=\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt-\underset{a}{\overset{x}{\int }}f(t)dt$

Now

$\underset{a}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt=\underset{a}{\overset{x}{\int }}f(t)dt+\underset{x}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt$

so rearranging this we have

$F(x+\mathrm{\Delta }x)-F(x)=\underset{x}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt$

According to the Mean Value Theorem for Integration, there exists a $c\in [x,x+\mathrm{\Delta }x]$ such that

$\underset{x}{\overset{x+\mathrm{\Delta }x}{\int }}f(t)dt=f(c)\cdot \mathrm{\Delta }x$

Notice that $c$ depends on $\mathrm{\Delta }x$ . Anyway what we have shown is that

$F(x+\mathrm{\Delta }x)-F(x)=f(c)\cdot \mathrm{\Delta }x$

and dividing both sides by $\mathrm{\Delta }x$ gives

$\frac{F(x+\mathrm{\Delta }x)-F(x)}{\mathrm{\Delta }x}=f(c)$

Take the limit as $\mathrm{\Delta }x\to 0$ we get the definition of the derivative of $F$ at $x$ so we have

$F^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{F(x+\mathrm{\Delta }x)-F(x)}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }f(c)$

To find the other limit, we will use the squeeze theorem. $c\in [x,x+\mathrm{\Delta }x]$ , so $x\le c\le x+\mathrm{\Delta }x$ . Hence,

$\underset{\mathrm{\Delta }x\to 0}{\lim }[x+\mathrm{\Delta }x]=x\Rightarrow \underset{\mathrm{\Delta }x\to 0}{\lim }c=x$

As $f$ is continuous we have

$F^{\prime }(x)=\underset{\mathrm{\Delta }x\to 0}{\lim }f(c)=f\left(\underset{\mathrm{\Delta }x\to 0}{\lim }c\right)=f(x)$

which completes the proof.

$◼$

Proof of Fundamental Theorem of Calculus Part II

Define $P(x)=\underset{a}{\overset{x}{\int }}f(t)dt$ . Then by the Fundamental Theorem of Calculus part I we know that $P$ is differentiable on $(a,b)$ and for all $x\in (a,b)$

$P^{\prime }(x)=f(x)$

So $P$ is an antiderivative of $f$ . Since we were assuming that $F$ was also an antiderivative for all $x\in (a,b)$ ,

$\begin{array}{cc}& P^{\prime }(x)=F^{\prime }(x)\\ & P^{\prime }(x)-F^{\prime }(x)=0\\ & (P(x)-F(x){)}^{\prime }=0\end{array}$

Let $g(x)=P(x)-F(x)$ . The Mean Value Theorem applied to $g(x)$ on $[a,\xi ]$ with $a<\xi <b$ says that

$\frac{g(\xi )-g(a)}{\xi -a}=g^{\prime }(c)$

for some $c$ in $(a,\xi )$ . But since $g^{\prime }(x)=0$ for all $x$ in $[a,b]$ , $g(\xi )$ must equal $g(a)$ for all $\xi$ in $(a,b)$ , i.e. g(x) is constant on $(a,b)$ .

This implies there is a constant $C=g(a)=P(a)-F(a)=-F(a)$ such that for all $x\in (a,b)$ ,

$P(x)=F(x)+C$

and as $g$ is continuous we see this holds when $x=a$ and $x=b$ as well. And putting $x=b$ gives

$\underset{a}{\overset{b}{\int }}f(t)dx=P(b)=F(b)+C=F(b)-F(a)$

$◼$

Notation for Evaluating Definite Integrals

The second part of the Fundamental Theorem of Calculus gives us a way to calculate definite integrals. Just find an antiderivative of the integrand, and subtract the value of the antiderivative at the lower bound from the value of the antiderivative at the upper bound. That is

$\underset{a}{\overset{b}{\int }}f(x)dx=F(b)-F(a)$

where $F^{\prime }(x)=f(x)$ . As a convenience, we use the notation

$F(x){|}_a^b$

to represent $F(b)-F(a)$

Integration of Polynomials

Using the power rule for differentiation we can find a formula for the integral of a power using the Fundamental Theorem of Calculus. Let $f(x)=x^n$ . We want to find an antiderivative for $f$ . Since the differentiation rule for powers lowers the power by 1 we have that

$\frac{d}{dx}{x}^{n+1}=(n+1)x^n$

As long as $n+1\ne 0$ we can divide by $n+1$ to get

$\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n=f(x)$

So the function $F(x)=\frac{x^{n+1}}{n+1}$ is an antiderivative of $f$ . If $0\notin [a,b]$ then $F$ is continuous on $[a,b]$ and, by applying the Fundamental Theorem of Calculus, we can calculate the integral of $f$ to get the following rule.

Notice that we allow all values of $n$ , even negative or fractional. If $n>0$ then this works even if $[a,b]$ includes $0$ .

Examples

• To find $\underset{1}{\overset{2}{\int }}{x}^{3}dx$ we raise the power by 1 and have to divide by 4. So

$\underset{1}{\overset{2}{\int }}{x}^{3}dx=\frac{x^4}{4}{|}_1^2=\frac{2^4}{4}-\frac{1^4}{4}=\frac{15}{4}$

• The power rule also works for negative powers. For instance

$\underset{1}{\overset{3}{\int }}\frac{dx}{x^3}=\underset{1}{\overset{3}{\int }}{x}^{-3}dx=\frac{x^{-2}}{-2}{|}_1^3=\frac{1}{-2}(3^{-2}-1^{-2})=-\frac{1}{2}(\frac{1}{3^2}-1)=-\frac{1}{2}(\frac{1}{9}-1)=\frac{1}{2}\cdot \frac{8}{9}=\frac{4}{9}$

• We can also use the power rule for fractional powers. For instance

$\underset{0}{\overset{5}{\int }}\sqrt{x}dx=\underset{0}{\overset{5}{\int }}{x}^{\frac{1}{2}}dx=\frac{x\sqrt{x}}{\frac{3}{2}}{|}_0^5=\frac{2}{3}(5^{\frac{3}{2}}-0^{\frac{3}{2}})=\frac{10\sqrt{5}}{3}$

• Using linearity the power rule can also be thought of as applying to constants. For example,

$=\underset{3}{\overset{11}{\int }}7dx=\underset{3}{\overset{11}{\int }}7x^{0}dx=7\underset{3}{\overset{11}{\int }}{x}^{0}dx=7x{|}_3^{11}=7(11-3)=56$

• Using the linearity rule we can now integrate any polynomial. For example

$\underset{0}{\overset{3}{\int }}(3x^2+4x+2)dx=(x^3+2x^2+2x){|}_0^3=3^3+2\cdot 3^2+2\cdot 3-0=27+18+6=51$