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.

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.

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

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 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.

\[\frac{x^2 + 4x^4}{x^3} = \frac{x^2}{x^3} + \frac{4x^4}{x^3} = \frac{1}{x} + 4x\]

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.

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

Example 3

Evaluating a definite integral

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

Step 1: Apply the Fundamental Theorem of Calculus.

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

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 when sales increase from 20 to 50 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.

By the Fundamental Theorem of Calculus, the additional revenue realized when sales increase from 20 to 50 gadgets is given by

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

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 additional revenue realized when sales increase from 20 to 50 gadgets is \(\$1,710.\)