4. The Definite Integral

by M. Bourne

In the last section, we used the expression

to find the area under a curve. This expression is called a definite integral. It does not involve the constant of integration and it gives us a definite value (a number) at the end of the calculation.

CARE:The 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.

It is quite possible to have a definite integral having value 0 or negative. There are many more applications of the integral than areas under curves.

Example: Evaluate

Solution:

This requires:

1) Find the integral

2) Substitute 1.6 into the integral

3) Substitute 1.2 into the integral

4) Subtract.

math expression

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.

Here is the LiveMath answer.

LIVEMath

Answer

Exercises

In physics, work is done when a force acting upon an object causes a displacement. (For example, riding a bicycle...).

If the force is not constant, we must use integration to find the work done.

We use

where F(x) is the variable force.

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

Question 1:

Find the work done if the force is and it is acting from x = 1 to x = 5.

That is, evaluate

Answer

Application:

The average value of a function f(x) from a to b is given by:

For more: 9. Average Value using Integration.

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:

If we are given an expression for velocity, we can find the displacement of a moving object by integration:

More on: displacement, velocity and acceleration.

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

Problem:

Find

We put u = x² + 1.

So du = 2x dx

But the question does not have "x dx"so we cannot solve it using any of the integration methods used above. (It can be done using trigonometric substitution, however).

We need to use numerical approaches. When software like Scientific Notebook, Matlab or Mathcad perform definite integrals, it uses numerical methods.

We use two methods: