Integration by Substitution Dr. Julia Arnold and Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

Recall from the previous section that integration is the opposite of differentiation. Recall the Chain Rule for Derivatives: which is the derivative of the outside multiplied by the derivative of the inside So it stands to reason if integration is the opposite of differentiation, that Since identifying this pattern can be difficult, we use a method called substitution to integrate expressions with this form or to reverse the Chain Rule.

The Method of Substitution and How it Works Consider The more complicated expression in the integral is (2x + 4)5. Using a different variable like u, for instance, we will make a substitution for the inside of the exponential function. Let u = 2x + 4, then (2x + 4)5 becomes u5. Now du may be different from dx, so let’s find du. If u = 2x + 4, then du = 2 dx. You will note that we did not write but wrote the differentials on their respective sides.

Now by substitution we can rewrite the problem from the x variable to the u variable. Integrating the expression in the u variable we get: What would this be in the x variable? The problem is solved!! Checking by taking the derivative:

Be sure to check out the simpler examples in the text. Here are some variations on those examples. Example 1a. Find Solution: Remember to pick the most complicated expression and let the inside be u. Which would you pick for u? X2 +3 (X2 +3)4 4x

Example 1a. Find Solution: Remember to pick the most complicated expression and let the inside be u. Which is more complicated, 4x or ? What is the inside? Try again.

Example 1a. Find Solution: Remember to pick the most complicated expression and let the inside be u. Which would you pick for u? u = x²+3, is the correct choice for u.

Example 1a. Find We need all the variables in our integral to be in terms of u, so the next step is to find du. This time the 4x dx doesn’t exactly match the 2x dx, but it only differs by a factor of 2, a constant. So to adjust for this we solve the above for dx. Or, we could multiply by 2

Example 1a. Find Or Substitute just for dx Substitute for 4xdx As you can see, the end result is the same:

To finish the problem

Example 2a. Find Solution: What did you pick for u? u = 3x + 1 du = 3 dx You must change all variables to u. Substitute: Just like with derivatives, we do a rewrite on the square root.

Example 3a. Find Solution: Pick u. Substitute, simplify and integrate:

Example 4a. Find Solution: Pick u. Substitute, simplify and integrate:

Example 5a. Find Solution: Pick u. Substitute and integrate:

Example 6a. Find Solution: Pick u. Substitute and integrate:

Here’s one for you to try: Sometimes it is difficult to pick u. One hint given previously was to pick the inside of the more complicated expression. Another hint is to pick the expression with the highest power of your variable. Using these two hints, click on the expression you would pick for u? or or

x³ is not the right choice. x³ is not the highest power of the variable. Try again.

is not the right choice. Pick the inside of the complicated expression, or in this case the exponent!! Try again.

Good work!! Substitute into the integral, simplify and integrate.

An Application to Business In 1990 the head of the research and development department of the Soloron Corp. claimed that the cost of producing solar cell panels would drop at the rate of dollars per peak watt for the next t years, with t = 0 corresponding to the beginning of the year 1990. ( A peak watt is the power produced at noon on a sunny day.) In 1990 the panels, which are used for photovoltaic power systems, cost $10 per peak watt. Find an expression giving the cost per peak watt of producing solar cell panels at the beginning of year t. What was the cost at the beginning of 2000?

An Application to Business This tells you the expression is a derivative. In 1990 the head of the research and development department of the Soloron Corp. claimed that the cost of producing solar cell panels would drop at the rate of dollars per peak watt for the next t years, with t = 0 corresponding to the beginning of the year 1990. ( A peak watt is the power produced at noon on a sunny day.) In 1990 the panels, which are used for photovoltaic power systems, cost $10 per peak watt. Find an expression giving the cost per peak watt of producing solar cell panels at the beginning of year t. What was the cost at the beginning of 2000?

Since the expression is a dropping rate in cost, the expression is C’(x) or the derivative of the cost C(x) and it should be negative since it is dropping. Thus: The Cost Function C(t) is

The Cost Function C(t) is Use the initial condition that the cost in 1990 was $10, or when t = 0, C(0)= $10, thus Hence,

Now to find the cost per peak watt at the beginning of the year 2000 which is 10 years from 1990 and would correspond to t = 10. Thus the cost per peak watt of producing solar cell panels at the beginning of 2000 is approximately $.94 per peak watt.