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8.7: The Fundamental Theorem of Calculus

Difficulty Level: At Grade Created by: CK-12
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Learning objectives

  • Learn the rules of finding the antiderivatives of power functions.
  • Learn how to use the fundamental theorem of calculus to compute the area under a curve.


If you think that evaluating areas under curves is a tedious process you are probably right. Fortunately, there is an easier method. In this section, we shall give a general method of evaluating definite integrals (area under the curve) by using antiderivatives.

The Antiderivative

Another word for the integral that we developed in the previous lesson is antiderivative:

Definition: The Antiderivative

If F ' (x) = f (x), then F (x) is said to be the antiderivative of f(x).

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus makes the relationship between derivatives and integrals clear. Integration performed on a function can be reversed by differentiation.

The Fundamental Theorem of Calculus

If a function f(x) is defined over the interval [a, b] and if F(x) is the antidervative of f on [a, b], then\begin{align*}\int_{a}^{b} f(x) dx = F(x)|^b_a\end{align*}

\begin{align*}= F(b) - F(a)\end{align*}

We can use the relationship between differentiation and integration outlined in the Fundamental Theorem of Calculus to compute definite integrals more quickly.

Example 1: Evaluate \begin{align*}\int_{1}^{2} x^2 dx.\end{align*}

Solution: This integral tells us to evaluate the area under the curve f(x) = x2, which is a parabola over the interval [1, 2], as shown in the figure below.

To compute the integral according to the Fundamental Theorem of Calculus, we need to find the antiderivative of f(x) = x2. It turns out to be F(x) = (1/3)x3 + C, where C is a constant of integration. How can we get this? Think about the functions that will have derivatives of x2. Take the derivative of F(x) to check that we have found such a function. (For more specific rules, see the box after this example). Substituting into the fundamental theorem,

\begin{align*}\int_{a}^{b} f(x) dx\end{align*} \begin{align*}= F(x)|^b_a\end{align*}
\begin{align*}\int_{1}^{2} x^2 dx\end{align*} \begin{align*}= \left [\frac{1} {3} x^3 + C\right ]^2_1\end{align*}
\begin{align*}= \left [\frac{1} {3} (2)^3 + C\right ] - \left [\frac{1} {3} (1)^3 + C\right ]\end{align*}
\begin{align*}= \left [\frac{8} {3} + C\right ] - \left [\frac{1} {3} + C\right ]\end{align*}
\begin{align*}= \frac{7} {3} + C - C\end{align*}
\begin{align*}= \frac{7} {3}\end{align*}

So the area under the curve is (7/3) units of area.

Antiderivative Rules

There are rules for finding the antiderivatives of simple power functions such as f(x) = x2. As you read through them, try to think about why they make sense, keeping in mind that differentiation reverses integration.

Rules of Finding the Antiderivatives of Power Functions

  • The Power Rule

    \begin{align*}\int x^n dx = \frac{1} {n + 1} x^{n + 1} + C\end{align*}

    where C is constant of integration and n is a rational number not equal to -1.

  • A Constant Multiple of a Function Rule

    \begin{align*}\int k x^n dx = k \int x^n dx = k \cdot \frac{1} {n + 1} x^{n + 1} + C\end{align*}

    where k is a constant.

  • Sum and Difference Rule

    \begin{align*}\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx\end{align*}

  • The Constant rule

    \begin{align*}\int k \cdot dx = kx + C\end{align*}

where k is a constant. (Notice that this rule comes as a result of the power rule above.)

You will learn many more rules for finding antiderivatives when you study calculus.

Example 2: Evaluate \begin{align*}\int x^3dx.\end{align*}

Solution: Since \begin{align*}\int x^n dx = \frac{1} {n + 1}x^{n + 1} + C\end{align*}, we have

\begin{align*}\int x^3 dx\end{align*} \begin{align*}= \frac{1} {3 + 1}x^{3 + 1} + C\end{align*}
\begin{align*}= \frac{1} {4}x^4 + C\end{align*}
To check our answer we can take the derivative of \begin{align*} \frac{1} {4}x^4 + C\end{align*} and verify that it is \begin{align*}\,\! x^3\end{align*}, the original function in our integral.

Example 3: Evaluate \begin{align*}\int 5x^2 dx.\end{align*}

Solution: Using the constant multiple of a power rule, the coefficient 5 can be removed outside the integral:

\begin{align*}\int 5x^2 dx = 5 \int x^2 dx\end{align*}
Then we can integrate:
\begin{align*}= 5 \cdot \frac{1} {2 + 1} x^{2 + 1} + C\end{align*}
\begin{align*}= \frac{5} {3} x^3 + C\end{align*}
Again, if we wanted to check our work we could take the derivative of \begin{align*}\frac{5} {3} x^3 + C\end{align*} and verify that we get \begin{align*}\,\! 5x^2\end{align*}.

Example 4: Evaluate \begin{align*}\int (3x^3 - 4x^2 + 2)dx.\end{align*}

Solution: Using the sum and difference rule we can separate our integral into three integrals:

\begin{align*}\int (3x^3 - 4x^2 + 2)dx = 3 \int x^3 dx - 4 \int x^2 dx + \int 2dx\end{align*}
\begin{align*}= 3 \cdot \frac{1} {4} x^4 - 4 \cdot \frac{1} {3}x^3 + 2x + C\end{align*}
\begin{align*}= \frac{3} {4} x^4 - \frac{4} {3} x^3 + 2x + C\end{align*}

Example 5: Evaluate \begin{align*}\int_{2}^{5} \sqrt{x} dx.\end{align*}

Solution: The evaluation of this integral represents calculating the area under the curve \begin{align*}y = \sqrt{x}\end{align*} from x = -2 to x = 3, shown in the figure below.

\begin{align*}\int_{2}^{5} \sqrt{x} dx\end{align*} \begin{align*}= \int_{2}^{5} x^{1/2} dx\end{align*}
\begin{align*}= \left [\frac{1} {\frac{1} {2} + 1} x^{1/2 + 1}\right ]^5_2\end{align*}
\begin{align*}= \left [\frac{1} {3/2} x^{3/2}\right ]^5_2\end{align*}
\begin{align*}= \frac{2} {3} \left [x^{3/2}\right ]^5_2\end{align*}
\begin{align*}= \frac{2} {3}\left [5^{3/2} - 2^{3/2}\right ]\end{align*}
\begin{align*}= 5.57\end{align*}
So the area under the curve is 5.57.

Lesson Summary

At the beginning of this chapter we introduced the two major problems in calculus, that of finding the equation of the tangent line to a graph at a given point and finding the area between a graph and the x-axis. We found that the tangent line problem had to do with differentiation and the area problem had to do with integration. To finish off our introduction to calculus, in this lesson we linked our two original problems together with the Fundamental Theorem of Calculus. We found that differentiation reverses integration. This gave us a way to calculate integrals more easily.

Points to Consider

  1. Given the link between integration and differentiation, and given that derivatives can represent instantaneous rate of change, what might integrals represent?
  2. If you had a function that represented the velocity of an object, how could you find a function that represented the position of the object? What about a function that represented the acceleration of the object?

Review Questions

  1. Evaluate the integral \begin{align*}\int_{0}^{3} 5xdx\end{align*}
  2. Evaluate the integral \begin{align*}\int_{0}^{1} x^4dx\end{align*}
  3. Evaluate the integral \begin{align*}\int_{1}^{4} (x - 3)dx\end{align*}
  4. Evaluate the integral \begin{align*}\int_{-1}^{1} (5x^2 - 3x - 1) dx\end{align*}
  5. Find the integral of x(3x2 + 1) from 0 to 2.
  6. Find the integral of (x + 1)(2x - 3) from -1 to 2.
  7. Find the integral of \begin{align*}\sqrt{x}\end{align*} from 0 to 9.
  8. Sketch y = x3 and y = x on the same coordinate system and then find the area of the region enclosed between them (a) in the first quadrant and (b) in the first and third quadrants.
  9. Evaluate the integral \begin{align*}\int_{-R}^{R} (\pi R^2 - \pi x^2) dx\end{align*} where R is a constant.

Review Answers

  1. \begin{align*}\frac{45} {2}\end{align*}
  2. \begin{align*}\frac{1} {5}\end{align*}
  3. \begin{align*}- \frac{3} {2}\end{align*}
  4. \begin{align*}\frac{4} {3}\end{align*}
  5. 14
  6. \begin{align*}-\frac{9} {2}\end{align*}
  7. 18
  8. (a) \begin{align*}\frac{1} {4}\end{align*} (b) 0
  9. \begin{align*}\frac{4\pi} {3} R^3\end{align*}

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