The Fundamental Theorem of Calculus: Learn It 1

  • Understand the Mean Value Theorem for Integrals and both components of the Fundamental Theorem of Calculus
  • Use the Fundamental Theorem of Calculus to find derivatives of integral functions and calculate definite integrals
  • Describe how differentiation and integration are interconnected

In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section, we will look at more powerful and useful techniques for evaluating definite integrals.

These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s, and it is codified in what we now call the Fundamental Theorem of Calculus.

Isaac Newton’s contributions to mathematics and physics changed the way we look at the world. The relationships he discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. To learn more, read a brief biography of Newton with multimedia clips.

The Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if [latex]f(x)[/latex] is continuous, a point [latex]c[/latex] exists in an interval [latex]\left[a,b\right][/latex] such that the value of the function at [latex]c[/latex] is equal to the average value of [latex]f(x)[/latex] over [latex]\left[a,b\right].[/latex]

We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

the mean value theorem for integrals

If [latex]f(x)[/latex] is continuous over an interval [latex]\left[a,b\right],[/latex] then there is at least one point [latex]c\in \left[a,b\right][/latex] such that

[latex]f(c)=\frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(x)dx.[/latex]

This formula can also be stated as

[latex]{\displaystyle\int }_{a}^{b}f(x)dx=f(c)(b-a).[/latex]

Proof

Since [latex]f(x)[/latex] is continuous on [latex]\left[a,b\right],[/latex] by the extreme value theorem, it assumes minimum and maximum values—[latex]m[/latex] and M, respectively—on [latex]\left[a,b\right].[/latex] Then, for all [latex]x[/latex] in [latex]\left[a,b\right],[/latex] we have [latex]m\le f(x)\le M.[/latex] Therefore, by the comparison theorem, we have

[latex]m(b-a)\le {\displaystyle\int }_{a}^{b}f(x)dx\le M(b-a).[/latex]

Dividing by [latex]b-a[/latex] gives us

[latex]m\le \frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(x)dx\le M.[/latex]

Since [latex]\frac{1}{b-a}{\displaystyle\int }_{a}^{b}f(x)dx[/latex] is a number between [latex]m[/latex] and M, and since [latex]f(x)[/latex] is continuous and assumes the values [latex]m[/latex] and M over [latex]\left[a,b\right],[/latex] by the Intermediate Value Theorem, there is a number [latex]c[/latex] over [latex]\left[a,b\right][/latex] such that

[latex]f(c)=\dfrac{1}{b-a}{\displaystyle\int}_{a}^{b}f(x)dx,[/latex]

and the proof is complete.

[latex]_\blacksquare[/latex]

Find the average value of the function [latex]f(x)=8-2x[/latex] over the interval [latex]\left[0,4\right][/latex] and find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of the function over [latex]\left[0,4\right].[/latex]

Given [latex]{\displaystyle\int }_{0}^{3}{x}^{2}dx=9,[/latex] find [latex]c[/latex] such that [latex]f(c)[/latex] equals the average value of [latex]f(x)={x}^{2}[/latex] over [latex]\left[0,3\right].[/latex]