The Fundamental Theorem of Calculus

How to Use Antiderivatives to Evaluate Definite Integrals

The Fundamental Theorem of Calculus

Let \(f\) be a continuous function on \([a, b]\). Then

\[ \int_a^b f(x) ~dx = F(b) - F(a) \]

where \(F\) is any antiderivative of \(f\).

Notation

\[F(x) \Big|_a^b = F(b) - F(a)\]

Example 1

Area under a graph

Write the area under the graph of \(y=x^2\) on \([1,3]\) as a definite integral and then use the Fundamental Theorem of Calculus to evaluate it. Compare the result with the approximations found in Example 1 of the previous section.

Step 1:   Write the area under the graph of \(y=x^2\) on \([1,3]\) as a definite integral.

\[\int_1^3 x^2 ~dx\]

Step 2:   Find an antiderivative of \(x^2\).

\[\begin{align*} \int x^2~dx &= \frac{1}{3}x^3 + C && \text{power rule} \end{align*}\]

Therefore, \(x^3/3\) is an antiderivative of \(x^2\), and can be used to evaluate the definite integral from Step 1.

Step 3:   Apply the Fundamental Theorem of Calculus.

\[\begin{align*} \int_1^3 x^2dx &= \frac{1}{3}x^3 \Biggr|_1^3 && \text{since $\frac{1}{3}x^3$ is an antiderivative of $x^2$} \\ &= \frac{1}{3}3^3 - \frac{1}{3}1^3 && \text{plug in the limits of integration} \\ &= \frac{27}{3} - \frac{1}{3} && \text{simplify} \\ &= 26/3 \end{align*}\]

Therefore, the area under the graph of \(y=x^2\) on \([1,3]\) is \(26/3 = 8.\overline{6}\).

Step 4:   Compare the result from Step 3 to the approximations found in Example 1.

The above calculations show that the area of the region is exactly \(8.\overline{6}\). The approximations we calculated in Example 1 of the previous section were \(10.75\) (using right Riemann sum), \(6.75\) (using left Riemann sum), and \(8.625\) (using the Midpoint Rule).

Example 2

Evaluating a definite integral

Evaluate \(\displaystyle \int_1^2 \frac{x^2 + 4x^4}{x^3} ~dx\).

Step 1:   Simplify the integrand by writing it as a sum.

\[\begin{align*} \frac{x^2 + 4x^4}{x^3} &= \frac{x^2}{x^3} + \frac{4x^4}{x^3} \\ &= \frac{1}{x} + 4x \end{align*}\]

Step 2:   Find an antiderivative of \(\dfrac{1}{x} + 4x\).

\[\begin{align*} \int \frac{1}{x} + 4x ~dx &= \ln|x| + \frac{4x^2}{2} + C \\ &= \ln|x| + 2x^2 + C \end{align*}\]

Therefore, \(\ln|x| + 2x^2 \) is an antiderivative of \(\dfrac{1}{x} + 4x\), and can be used to evaluate the given definite integral.

Step 3:   Apply the Fundamental Theorem of Calculus.

\[\begin{align*} \int_1^2 \frac{x^2 + 4x^4}{x^3} ~dx &= \int_1^2 \frac{1}{x} + 4x ~dx \\ &= \ln|x| + 2x^2 \Big|_1^2 && \hbox{since $\ln|x| + 2x^2$ is an antiderivative}\\ &= (\ln|2| + 2(2)^2) - (\ln|1| + 2(1)^2) && \text{plug in limits of integration}\\ &= \ln(2) + 8 - 0 - 2 && \hbox{since $\ln(1) = 0$}\\ &= 6 + \ln(2). \end{align*}\]

Example 3

Evaluating a definite integral

Evaluate \(\displaystyle \int_{0}^{4} e^x ~dx\).

Step 1:   Apply the Fundamental Theorem of Calculus.

\[\begin{align*} \int_{0}^{4} e^x ~dx &= e^x \Big|_0^4 && \hbox{since $e^x$ is an antiderivative of $e^x$}\\ &= e^4 - e^0 && \text{plug in limits of integration}\\ &= e^4 - 1 && \hbox{since $e^0 = 1$}\\ \end{align*}\]

Example 4

Computing total revenue from marginal revenue

The daily marginal revenue function associated with selling \(m\) gadgets is given by \(R'(m)=0.2m + 50\), where \(R'(m)\) is given in dollars per unit.

• Find the daily total revenue realized from the sale of the first 20 gadgets.

• Find the additional revenue realized from the sale of the 21st through 50th gadgets.

Step 1:   Apply the Fundamental Theorem of Calculus for the first case.

By the Fundamental Theorem of Calculus, the total revenue from the sale of the first 20 gadgets is given by

\[\begin{align*} R(20) &= R(20) - R(0) && \hbox{since $R(0) = 0$} \\ &= \int_0^{20} R'(m)~dm && \hbox{Fundamental Theorem of Calculus}\\ &= \int_0^{20} \frac{2}{10}m + 50 ~dm. \end{align*}\]

We evaluate the integral to get

\[\begin{align*} \frac{2}{10}\cdot \frac{m^2}{2} + 50m \Biggr|_0^{20} &= \frac{m^2}{10} + 50m \Biggr|_0^{20} \\ &= \left(\frac{20^2}{10} + 50(20)\right)-\left(\frac{0^2}{10} + 50(0)\right)\\ &= 40+1000 - 0\\ &=1040. \end{align*}\]

Therefore, the total revenue from the sale of the first 20 gadgets is \(\$1,040\).

Step 2:   Apply the Fundamental Theorem of Calculus for the second case.

Note that the additional revenue realized from the sale of the 21st through 50th gadgets is given by \(R(50) - R(20)\). This is because \(R(50)\) corresponds to the revenue generated from the sale of gadgets 1 through 50 and \(R(20)\) corresponds to the revenue generated from the sale of gadgets 1 through 20. When we subtract the two values, what remains corresponds to the revenue associated with gadgets 21 through 50.

By the Fundamental Theorem of Calculus, we have

\[\begin{align*} R(50) - R(20) &= \int_{20}^{50} R'(m)~dm \\ &= \int_{20}^{50} \frac{2}{10}m + 50 ~dm. \end{align*}\]

Evaluating the integral, we get

\[\begin{align*} \frac{2}{10}\cdot \frac{m^2}{2} + 50m \Biggr|_{20}^{50} &= \frac{m^2}{10} + 50m \Biggr|_{20}^{50} \\ &= \left(\frac{50^2}{10} + 50(50)\right) - \left( \frac{20^2}{10} + 50(20)\right)\\ &=(250+2500)-(40+1000)\\ &=2750-1040 \\ &=1710. \end{align*}\]

Therefore, the sale of gadgets 21 through 50 generate an additional revenue of \(\$1,710.\)