The Fundamental Theorem of Calculus: Learn It 3

Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.

After finding approximate areas by adding the areas of [latex]n[/latex] rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.

The Fundamental Theorem of Calculus, Part 2

If [latex]f[/latex] is continuous over the interval [latex]\left[a,b\right][/latex] and [latex]F(x)[/latex] is any antiderivative of [latex]f(x),[/latex] then

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

We often see the notation [latex]{F(x)|}_{a}^{b}[/latex] to denote the expression [latex]F(b)-F(a).[/latex]

We use this vertical bar and associated limits [latex]a[/latex] and [latex]b[/latex] to indicate that we should evaluate the function [latex]F(x)[/latex] at the upper limit (in this case, [latex]b[/latex]), and subtract the value of the function [latex]F(x)[/latex] evaluated at the lower limit (in this case, [latex]a[/latex]).

The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.

Proof

Let [latex]P=\left\{{x}_{i}\right\},i=0,1\text{,…,}n[/latex] be a regular partition of [latex]\left[a,b\right].[/latex] Then, we can write

[latex]\begin{array}{cc}F(b)-F(a)\hfill & =F({x}_{n})-F({x}_{0})\hfill \\ & =\left[F({x}_{n})-F({x}_{n-1})\right]+\left[F({x}_{n-1})-F({x}_{n-2})\right]+\text{…}+\left[F({x}_{1})-F({x}_{0})\right]\hfill \\ \\ & =\underset{i=1}{\overset{n}{\text{∑}}}\left[F({x}_{i})-F({x}_{i-1})\right].\hfill \end{array}[/latex]

Now, we know F is an antiderivative of [latex]f[/latex] over [latex]\left[a,b\right],[/latex] so by the Mean Value Theorem for [latex]i=0,1\text{,…,}n[/latex] we can find [latex]{c}_{i}[/latex] in [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] such that

[latex]F({x}_{i})-F({x}_{i-1})={F}^{\prime }({c}_{i})({x}_{i}-{x}_{i-1})=f({c}_{i})\text{Δ}x.[/latex]

Then, substituting into the previous equation, we have

[latex]F(b)-F(a)=\underset{i=1}{\overset{n}{\text{∑}}}f({c}_{i})\text{Δ}x.[/latex]

Taking the limit of both sides as [latex]n\to \infty ,[/latex] we obtain

[latex]\begin{array}{}\\ \\ F(b)-F(a)\hfill & =\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}f({c}_{i})\text{Δ}x\hfill \\ & ={\displaystyle\int }_{a}^{b}f(x)dx.\hfill \end{array}[/latex]

[latex]_\blacksquare[/latex]

Use the second part of the Fundamental Theorem of Calculus to evaluate

[latex]{\displaystyle\int }_{-2}^{2}({t}^{2}-4)dt.[/latex]

Use the second part of the Fundamental Theorem of Calculus to evaluate [latex]{\displaystyle\int }_{1}^{2}{x}^{-4}dx.[/latex]

James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after [latex]5[/latex] sec wins a prize. If James can skate at a velocity of [latex]f(t)=5+2t[/latex] ft/sec and Kathy can skate at a velocity of [latex]g(t)=10+ \cos (\frac{\pi }{2}t)[/latex] ft/sec, who is going to win the race?