5.3 The fundamental theorem of calculus

Calculus volume 1 Page 4 / 11

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 $P = {x_{i}}, i = 0, 1,…, n$ be a regular partition of $[a, b] .$ Then, we can write

\begin{matrix} F (b) - F (a) & = F (x_{n}) - F (x_{0}) \\ = [F (x_{n}) - F (x_{n - 1})] + [F (x_{n - 1}) - F (x_{n - 2})] + … + [F (x_{1}) - F (x_{0})] \\ = \sum_{i = 1}^{n} [F (x_{i}) - F (x_{i - 1})] . \end{matrix}

Now, we know F is an antiderivative of f over $[a, b],$ so by the Mean Value Theorem (see The Mean Value Theorem ) for $i = 0, 1,…, n$ we can find $c_{i}$ in $[x_{i - 1}, x_{i}]$ such that

F (x_{i}) - F (x_{i - 1}) = F^{'} (c_{i}) (x_{i} - x_{i - 1}) = f (c_{i}) Δ x .

Then, substituting into the previous equation, we have

F (b) - F (a) = \sum_{i = 1}^{n} f (c_{i}) Δ x .

Taking the limit of both sides as $n \to \infty,$ we obtain

\begin{matrix} F (b) - F (a) & = lim_{n \to \infty} \sum_{i = 1}^{n} f (c_{i}) Δ x \\ = \int_{a}^{b} f (x) d x . \end{matrix}

□

Evaluating an integral with the fundamental theorem of calculus

Use [link] to evaluate

\int_{−2}^{2} (t^{2} - 4) d t .

Recall the power rule for Antiderivatives :

If y = x^{n}, \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C .

Use this rule to find the antiderivative of the function and then apply the theorem. We have

\begin{matrix} \int_{−2}^{2} (t^{2} - 4) d t & = \frac{t^{3}}{3} - {4 t |}_{−2}^{2} \\ = [\frac{{(2)}^{3}}{3} - 4 (2)] - [\frac{{(−2)}^{3}}{3} - 4 (−2)] \\ = (\frac{8}{3} - 8) - (- \frac{8}{3} + 8) \\ = \frac{8}{3} - 8 + \frac{8}{3} - 8 \\ = \frac{16}{3} - 16 \\ = - \frac{32}{3} . \end{matrix}

Analysis

Notice that we did not include the “+ C ” term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the antiderivative with $C = 0 .$ If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.

The region of the area we just calculated is depicted in [link] . Note that the region between the curve and the x -axis is all below the x -axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.

The graph of the parabola f(t) = t^2 – 4 over [-4, 4]. The area above the curve and below the x axis over [-2, 2] is shaded. — The evaluation of a definite integral can produce a negative value, even though area is always positive.

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Evaluating a definite integral using the fundamental theorem of calculus, part 2

Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:

\int_{1}^{9} \frac{x - 1}{\sqrt{x}} d x .

First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:

\int_{1}^{9} \frac{x - 1}{x^{1 / 2}} d x = \int_{1}^{9} (\frac{x}{x^{1 / 2}} - \frac{1}{x^{1 / 2}}) d x .

Use the properties of exponents to simplify:

\int_{1}^{9} (\frac{x}{x^{1 / 2}} - \frac{1}{x^{1 / 2}}) d x = \int_{1}^{9} (x^{1 / 2} - x^{−1 / 2}) d x .

Now, integrate using the power rule:

\begin{matrix} \int_{1}^{9} (x^{1 / 2} - x^{- 1 / 2}) d x & = {(\frac{x^{3 / 2}}{\frac{3}{2}} - \frac{x^{1 / 2}}{\frac{1}{2}}) |}_{1}^{9} \\ = [\frac{{(9)}^{3 / 2}}{\frac{3}{2}} - \frac{{(9)}^{1 / 2}}{\frac{1}{2}}] - [\frac{{(1)}^{3 / 2}}{\frac{3}{2}} - \frac{{(1)}^{1 / 2}}{\frac{1}{2}}] \\ = [\frac{2}{3} (27) - 2 (3)] - [\frac{2}{3} (1) - 2 (1)] \\ = 18 - 6 - \frac{2}{3} + 2 \\ = \frac{40}{3} . \end{matrix}

See [link] .

The graph of the function f(x) = (x-1) / sqrt(x) over [0,9]. The area under the graph over [1,9] is shaded. — The area under the curve from $x = 1$ to $x = 9$ can be calculated by evaluating a definite integral.

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Use [link] to evaluate $\int_{1}^{2} x^{−4} d x .$

$\frac{7}{24}$

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A roller-skating race

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

We need to integrate both functions over the interval $[0, 5]$ and see which value is bigger. For James, we want to calculate

\int_{0}^{5} (5 + 2 t) d t .

Using the power rule, we have

\begin{matrix} \int_{0}^{5} (5 + 2 t) d t & = {(5 t + t^{2}) |}_{0}^{5} \\ = (25 + 25) = 50. \end{matrix}

Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate

\int_{0}^{5} 10 + cos (\frac{π}{2} t) d t .

We know $sin t$ is an antiderivative of $cos t,$ so it is reasonable to expect that an antiderivative of $cos (\frac{π}{2} t)$ would involve $sin (\frac{π}{2} t) .$ However, when we differentiate $sin (\frac{π}{2} t),$ we get $\frac{π}{2} cos (\frac{π}{2} t)$ as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain

\begin{matrix} \int_{0}^{5} 10 + cos (\frac{π}{2} t) d t & = {(10 t + \frac{2}{π} sin (\frac{π}{2} t)) |}_{0}^{5} \\ = (50 + \frac{2}{π}) - (0 - \frac{2}{π} sin 0) \\ \approx 50.6. \end{matrix}

Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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