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Integral (Definite)

We define the definite integral of the function f(x) with respect of x from a to b to be

where F(x) is the anti-derivative of f(x). We call a and b the lower and upper limits of integration respectively. The function being integrated, f(x) is called the Integrand.

Note that the integration constants are not written in definite integrals since they always cancel in them:

Definition of Definite Integral using Riemann Sums:
Given a function f(x) that is continuous on the interval [a,b] we divide the interval into n subintervals of equal width, Δx and from each interval choose a point, . Then the definite integral of f(x) from a to b is

Properties of Definite Integral:
1. We can interchange the limits on any definite integral; all that we need to do is tack a minus sign onto the integral when we do.

2. If the upper and lower limits are the same then there is no work to do, the integral is zero.

3. , where c is any number. So, as with limits, derivatives, and indefinite integrals we can factor out a constant.

4. We can break up definite integrals across a sum or difference.

5.  where c is any number. This property is more important than we might realize at first. One of the main uses of this property is to tell us how we can integrate a function over the adjacent intervals, [a,c] and [c,b]. Note however that c doesn’t need to be between a and b.

6. The point of this property is to notice that as long as the function and limits are the same the variable of integration that we use in the definite integral won’t affect the answer.

7. c is any number.

8. If f(x) ≥ 0 and a ≤ x ≤ b then

9. If f(x) ≥ g(x) and a ≤ x ≤ b then

10. If m ≤ f(x) ≤ M for a ≤ x ≤ b then


Example 1: Evaluate
Solution: Using integration by parts with:

which leads to

Example 2: Given that , and  determine the value of .

This example is mostly an example of property 5 although there are a couple of uses of property 1 in the solution as well.

We need to figure out how to correctly break up the integral using property 5 to allow us to use the given pieces of information. First we’ll note that there is an integral that has a “-5” in one of the limits. It’s not the lower limit, but we can use property 1 to correct that eventually. The other limit is 100 so this is the number c that we’ll use in property 5.

We’ll be able to get the value of the first integral, but the second still isn’t in the list of know integrals. However, we do have second limit that has a limit of 100 in it. The other limit for this second integral is -10 and this will be c in this application of property 5.

At this point all that we need to do is use the property 1 on the first and third integral to get the limits to match up with the known integrals. After that we can plug in for the known integrals.

Example 3: Evaluate
Solution: In this case the integral can be found, because the two points of discontinuity t = ± 1/2 are both outside of the interval of integration. The substitution and converted limits in this case are:

The integral is then:

Example 4: Use the limit definition of definite integral to evaluate .
Solution: Divide the interval [0,3] into n equal parts each of length.

for i=1,2,3,……n. Choose the sampling points to be the right-hand endpoints of the subintervals and given by

for i=1,2,3,……n . The function is f(x) = x2 -1.

The definite integral is therefore: © 2024 | Contact us | Terms of Use | Privacy Policy | Yellow Sparks Network
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