Calculus/Fundamental Theorem of Calculus

Mean Value Theorem for Integration

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

Mean Value Theorem for Integration

Suppose f(x) is continuous on [a,b]. Then $\frac{\int_{a}^{b}f(x)dx}{b-a}=f(c)$ for some c in [a,b].

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
$\int_{a}^{b}f(x)dx=\lim\limits _{n\to\infty}\sum\limits _{i=1}^{n}f(x_{i}^{*})\frac{b-a}{n}=\lim\limits _{n\to\infty}\frac{b-a}{n}\sum\limits _{i=1}^{n}f(x_{i}^{*})$
and since
$m\leq f(x_{i}^{*})\leq M$ for all $x_{i}^{*}$ in [a,b],
we have

$\lim\limits _{n\to\infty}\frac{b-a}{n}\sum\limits _{i=1}^{n}m\leq\lim\limits _{n\to\infty}\frac{b-a}{n}\sum\limits _{i=1}^{n}f(x_{i}^{*})\leq\lim\limits _{n\to\infty}\frac{b-a}{n}\sum\limits _{i=1}^{n}M$

$\lim\limits _{n\to\infty}\frac{b-a}{n}nm\leq\int_{a}^{b}f(x)dx\leq\lim\limits _{n\to\infty}\frac{b-a}{n}nM$

$\lim\limits _{n\to\infty}(b-a)m\leq\int_{a}^{b}f(x)dx\leq\lim\limits _{n\to\infty}(b-a)M$

$(b-a)m\leq\int_{a}^{b}f(x)dx\leq(b-a)M$

$m\leq\frac{\int_{a}^{b}f(x)dx}{(b-a)}\leq M$

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

$\frac{\int_{a}^{b}f(x)dx}{b-a}=f(c)$

Fundamental Theorem of Calculus

Statement of the Fundamental Theorem

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

$F(x)= \int_a^x f(t)\ dt \mbox{ for } x \mbox{ in } [a,b].$

Fundamental Theorem of Calculus Part I Suppose f is continuous on [a,b] and F is defined by

$F(x)= \int_a^x f(t)\ dt.$

Then F is differentiable on (a,b) and for all $x\in(a,b)$,

$F'(x) = f(x).\,$

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.

Fundamental Theorem of Calculus Part II Suppose that f is continuous on [a,b] and that F is any antiderivative of f. Then

$\ \int_{a}^{b} f(x)\ dx=F(b)-F(a).$

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 is in (a,b). Pick Δx so that x + Δx is also in (a, b). Then

$F(x) = \int_{a}^{x} f(t) dt$

and

$F(x+\Delta x) = \int_{a}^{x+\Delta x} f(t) dt$.

Subtracting the two equations gives

$F(x + \Delta x) - F(x) = \int_{a}^{x + \Delta x} f(t) dt - \int_{a}^{x} f(t) dt.$

Now

$\int_{a}^{x + \Delta x} f(t) dt = \int_{a}^{x} f(t) dt + \int_{x}^{x + \Delta x} f(t) dt$

so rearranging this we have

$F(x + \Delta x) - F(x) = \int_{x}^{x + \Delta x} f(t) dt.$

According to the Mean Value Theorem for Integration, there exists a c in [x, x + Δx] such that

$\int_{x}^{x + \Delta x}f(t) dt = f(c) \Delta x$.

Notice that c depends on Δx. Anyway what we have shown is that

$F(x + \Delta x) - F(x) = f(c) \Delta x \,$,

and dividing both sides by Δx gives

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

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

$F'(x) = \lim_{\Delta x \to 0} f(c).$.

To find the other limit, we will use the squeeze theorem. The number c is in the interval [x, x + Δx], so x≤ c ≤ x + Δx. Also, $\lim_{\Delta x \to 0} x = x$ and $\lim_{\Delta x \to 0} x + \Delta x = x$. Therefore, according to the squeeze theorem,

$\lim_{\Delta x \to 0} c = x$.

As f is continuous we have

$F'(x) = \lim_{\Delta x\to 0} f(c)= f\left(\lim_{\Delta x \to 0} c\right) = f(x)$

which completes the proof.

Proof of Fundamental Theorem of Calculus Part II

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

$P'(x) = f(x).\,$

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

P'(x) = F'(x)

P'(x) − F'(x) = 0

(P(x) − F(x))' = 0.

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

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

for some c in (a,ξ). But since g'(x) = 0 for all x in [a,b], g(ξ) must equal g(a) for all ξ 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

$\int_a^b f(t) dx = P(b) = F(b) + C = F(b) - F(a).$

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

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

where F'(x) = f(x). As a convenience, we use the notation

$F(x)\bigr|_a^b$

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) = xⁿ. 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 $n+1\neq 0$ 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 is not in [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.

Power Rule of Integration I

$\int_a^b x^n dx = {x^{n+1} \over {n+1}}\biggr|_a^b = \frac{b^{n+1}-a^{n+1}}{n+1}\;$ as long as $n\neq -1$ and 0 is not in [a,b].

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

Power Rule of Integration II

$\int_a^b x^n dx = {x^{n+1} \over {n+1}}\biggr|_a^b = \frac{b^{n+1}-a^{n+1}}{n+1}$ as long as n > 0.

Examples

• To find $\int_1^2 x^3 dx$ we raise the power by 1 and have to divide by 4. So

$\int_1^2 x^3 dx = \frac{x^4}{4}\biggr|_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{15}{4}.$

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

$\int_1^3 \frac{1}{x^3} dx = \int_1^3 x^{-3} dx = \frac{x^{-2}}{-2}\biggr|_1^3 = \frac{1}{-2}\left(3^{-2}-1^{-2}\right) = -\frac{1}{2}\left(\frac{1}{3^2} - 1\right) =-\frac{1}{2}\left(\frac{1}{9}-1\right) = \frac{1}{2} \cdot \frac{8}{9} = \frac{4}{9}.$

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

$\int_0^5 \sqrt x\, dx = \int_0^5 x^{\frac{1}{2}}\, dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\Biggr|_0^5 = \frac{2}{3}\left(5^{\frac{3}{2}} - 0^{\frac{3}{2}}\right)=\frac{2}{3}\left(5^{\frac{3}{2}}\right)$

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

$=\int_3^{11} 7\,dx=\int_3^{11} 7x^0\,dx=7\int_3^{11} x^0\,dx=7x \bigr|_3^{11}= 7(11 - 3) = 56$.

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

$\int_0^3 (3x^2 + 4x +2) dx = (x^3 + 2x^2 + 2x)\bigr|_0^3 = 3^3 + 2 \cdot 3^2 + 2 \cdot 3-0 = 27+18+6=51.$

Exercises

1. Evaluate $\int_0^1 x^6 dx$. Compare your answer to the answer you got for exercise 1 in section 4.1.

2. Evaluate $\int_1^2 x^6 dx$. Compare your answer to the answer you got for exercise 2 in section 4.1.

3. Evaluate $\int_0^2 x^6 dx$. Compare your answer to the answer you got for exercise 4 in section 4.1.