Two new strategies today: Substitution with Definite Integrals (a.k.a. “Change of Bounds”) and Substitution with a change of variables (a.k.a. “Back Substitution”).

Change of Bounds:

I actually find this strategy pretty simple. The only thing that is different about these problems is the bounds on the integral. For these problems, you find u and du as usual, but when it comes time to rewrite the integral in terms of u, the bounds must also change. This is because the bounds on the original integral are for x. Therefore, to solve the problem you must find the u-bounds. All you do to find the u-bounds is plug the x-bounds into the value you chose for u. When you have the new bounds, take the antiderivative as usual. However, you don’t substitute x back in becaue the bounds are now u-bounds. So, to get your answer (which will be a number) just solve using u.

Here is an example:

Change of Bounds:

I actually find this strategy pretty simple. The only thing that is different about these problems is the bounds on the integral. For these problems, you find u and du as usual, but when it comes time to rewrite the integral in terms of u, the bounds must also change. This is because the bounds on the original integral are for x. Therefore, to solve the problem you must find the u-bounds. All you do to find the u-bounds is plug the x-bounds into the value you chose for u. When you have the new bounds, take the antiderivative as usual. However, you don’t substitute x back in becaue the bounds are now u-bounds. So, to get your answer (which will be a number) just solve using u.

Here is an example:

Back Substitution:

These problems are easy enough to detect and solve. This strategy should be used when there is an extra x in the equation. For example, if the function is x[sqrt(3x+2)]dx, and the value chosen for u is 3x+2, then du would equal 3dx, thus leaving an extra x. To fix this problem, you have to switch around u=3x+2 to find x in terms of u, enabling you to plug that into x. In this example, [(u-2)/3] would replace x in the function. Then, since all the variables are u, you can take the antiderivative and solve as usual, although there will be much more expanding and simplifying.

Here is an example:

These problems are easy enough to detect and solve. This strategy should be used when there is an extra x in the equation. For example, if the function is x[sqrt(3x+2)]dx, and the value chosen for u is 3x+2, then du would equal 3dx, thus leaving an extra x. To fix this problem, you have to switch around u=3x+2 to find x in terms of u, enabling you to plug that into x. In this example, [(u-2)/3] would replace x in the function. Then, since all the variables are u, you can take the antiderivative and solve as usual, although there will be much more expanding and simplifying.

Here is an example: