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4. The Definite Integral

by M. Bourne

In the last section, we used the expression

to find the area under a curve.

F(x) is the integral of f(x);

F(b) is the value of the integral at the upper limit, x = b; and

F(a) is the value of the integral at the lower limit, x = a

This expression is called a definite integral. Note that it does not involve a constant of integration and it gives us a definite value (a number) at the end of the calculation.

We will use definite integrals to solve many practical problems. First, we see how to calculate definite integrals.

Example 1: Evaluate

Solution:

This question requires us to:

1) Find the integral and then write the upper and lower limits with square brackets, as follows:

The upper and lower limits are written like this to mean they will be substituted into the expression in brackets.

2) Next, substitute 5 (the upper limit) into the integral:

[(5)³ + 2(5)² + 5] = 125 + 50 + 5 = 180

3) Then substitute 1 into the integral:

[(1)³ + 2(1)² + 1] = 1 + 2 + 1 = 4

4) Subtract the result of (3) from the result of (2) for our final answer:

180 − 4 = 176

Normally, we would write our complete solution as follows:

=

= [(5)³ + 2(5)² + 5] − [(1)³ + 2(1)² + 1]

= 180 − 4

= 176

Note that our final answer is a number and does not involve "+ K". This is a definite integral.

Example 2: Evaluate

Answer

Example 3: Evaluate

Answer

Recall

Recall the substitution formula for integration:

(n ≠ -1)

 

Definite Integrals and Substitution

When we substitute, we are changing the variable, so we cannot use the same upper and lower limits. We can either:

• Do the problem as an indefinite integral first, then use upper and lower limits later
• Do the problem throughout using the new variable and the new upper and lower limits
• Show the correct variable for the upper and lower limit during the substitution phase.

We will be using the third of these possibilities.

---

Example: Find

using a substitution.

Answer

Here is the LiveMath answer, for comparison.

LIVEMath

Exercises

Application: Work

Question 1:

Find the work done if a force is acting on an object and moves it from x = 1 to x = 5.

That is, evaluate

Answer

For more on work, see: 7. Work by a Variable Force.

Application: Average Value

The average value of a function f(x) in the region x = a to x = b is given by:

Question 2:

Find the average value of x(3x² - 1)³ from 0 to 1.

Here is the graph of the situation:

In this case, b - a = 1, so we simply need to evaluate:

Answer

Application: Displacement

If we know the expression, v, for velocity in terms of t, the time, we can find the displacement (written s) of a moving object from time t = a to time t = b by integration, as follows:

Question 3:

Find the displacement of an object from t = 2 to t = 3, if the velocity of the object at time t is given by

That is, evaluate:

Answer

See more on: displacement, velocity and acceleration as applications of integration.

NOTE: As you can see from the above applications, the definite integral can be used to find work, average value and displacement, and not only areas under curves.

The definite integral only gives us an area when the whole of the curve is above the x-axis in the region from x = a to x = b. If this is not the case, we have to break it up into individual sections. See more at Area Under a Curve.

We now examine a definite integral that we cannot solve using substitution.

Problem:

Find:

Attempted solution:

We try to put u = x² + 1.

Then we find the differential:

du = 2x dx

But the question does not have "2x dx" (it only has "dx") so we cannot replace anything in the question with "du" properly. This means we can't solve it using any of the integration methods used above. (Note: This question can be done using Trigonometric Substitution, however, but we don't meet trigonometric substitution until later.)

Explanation: Let's think about if the question said this instead:

This time, there is a "2x dx" involved in the question and so we would be able to let:

u = x² + 1

Then we would find the differential:

du = 2x dx

Then we could proceed to find the integral like we did in the examples above, by replacing 2x dx with du and the square root part with √u.

However, my "Problem" does not have an "2x" outside of the square root so I cannot use the "u" substitution.

For now, we need to use numerical approaches to evaluate the integral

(Note: Historically, all definite integrals were approximated using numerical methods before Newton and Leibniz developed the integration methods we have learned so far in this chapter.)

We can use two different numerical methods for evaluating an integral:

• Trapezoidal rule
• Simpson's Rule

We meet these methods in the next 2 sections.

 

Need a break? Play a math game. Well, they all involve math... No, really!

Chapter Contents

• Integration
• 1. The Differential
• 2. Antiderivatives and The Indefinite Integral
• 3. The Area Under a Curve
• 4. The Definite Integral
• 5. Trapezoidal Rule
• 6. Simpson’s Rule
• Riemann Sums
• Integration Mini-lectures
• Integration Problem Solver
• Millionaire Calculus Game

• Comments, Questions?

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