Integration by Substitution (Change of Variables)

Once again as always in math, the words say it all. We are going to integrate an ugly looking integrand by substituting new variables or renaming the function to be integrated. We are going to change the variables in the expression to be integrated to simplify the antidifferentiation process. Let's do an example and then we'll discuss the finer points of the process.

Rule 1: ALL parts of the integrand, including the dx, must be changed to the new variable. We cannot mix x's and u's.

Example 1: Integrate ò (x + 3) 12 dx

If we wanted to take 2 weeks to raise x + 3 to the 12th power and then antidifferentiate the resulting polynomial, we wouldn't need integration by substitution. Here's the sane way to do it.

We will substitute

u = x + 3, so du = dx.

Now ò (x + 3) 12 dx = ò u 12 du = 1/13 u 13 + C.

However, we started with x as the variable, so now we unsubstitute

and go back to x, the original variable.

ò (x + 3) 12 dx = 1/13 ( x + 3 ) 13 + C

Note 1:

- What is the dx and why do we substitute for it?

- dx in any integrand is the equivalent of the Dx in the Riemann Sum expression f(x)Dx, which represents the area of a typical rectangle with base = Dx and height = f(x). That's why all integrands are of the form f(x)dx.

Example 2: Integrate Though we have a product, which generally means we should use integration by parts, a substitution will solve this easily.

u = x 2 , therefore du = 2x dx, and, dividing by 2, ½ du = x dx Now one where we have to play with the substitution expression in order to replace all parts of the original integrand.

Example 3: u = x - 3 u + 3 = xdu = dx integrate, unsubstitute 2/3 u3 / 2 + 6 u1 / 2 = 2/3 (x - 3)3 / 2 + 6 (x - 3)1 / 2 + C

Note 2: here, we had to substitute for x as well as dx and the other expression so we used the substitution relation statement u = x - 3 to create a replacement expression for x.

Now a definite integral with substitution. Here, we have to change the values for the limits of integration to make them correspond to the new variable. In other words, if the limits on the original variable x are a = 0 and b = 3, if we substitute u = x - 1, our limits of integration on the integral with the new variable will be au = - 1, bu = 2 since the values of u are one less than the values of x.

Example 4: u = 2x 2 - 5 du = 4x dx¼ du = x dx if x = 0, u = 2(0)2 -5 = -5if x = 3, u = 2(3)2 -5 = 13

Note 3: If we didn't change the limits of integration to values of u, then, once we integrate, we would have to unsubstitute and replace x with the old values of 0 and 3. Since the change of limits is a simple arithmetic calculation, it is best to change the values before integrating so that we don't have to take extra steps to complete the question.

Finally, a watch for these type where we must pay close attention to du.

Example 5: Use change of variables to integrate If we set u = x 2 - 4x + 3, then du = (2x - 4) dx = 2(x - 2) dx, so ½du = (x - 2)dx

The integral becomes ½ òu - 3 du = - ¼ u - 2 = - ¼(x 2 - 4x + 3) - 2 + C

In a case like this, we find a common factor in the binomial expression of du. Practice

1) Solve these integrals using integration by substitution or change of variables.

 a) b) c) d) e) f) 2) Find the area of the region between and the x-axis from x = 0 to x = 3.

3) If , find the area of the region under the graph from x = 1 to x = 2.

. .

Solutions

1) a) u = x 3 - 1 du = 3x 2 dx,1/3 du = x 2 dx b) u = 1 - 2x 2 du = - 4x dx- ¼ du = x dx .

c) u = 4 - 3x 2 - 2x 3 du = - 6(x 2 + x) dx- 1/6 du = (x 2 + x) dx d)    e) u = 3 - 2x du = - 2 dxu0 = 3, -½ du = dxu1 = 1 f) u = ln x du = dx / xu2 = ln 2, u5 = ln 5 .

2) Find the area of the region between and the x-axis from x = 0 to x = 3. Area = 3) If , find the area of the region under the graph from x = 1 to x = 2.

 u = x 2 + 1 du = 2 x dx½du = x dx  