Section7.4Integration by Change of Variables or Substitution
At the end of the last section, we warned that the symbolic integration techniques we have developed only work for problems that exactly fit our formulas. When we tried integrating an exponential function where the exponent was a constant times \(t\text{,}\) we had to change the base to get a function with only \(t\) in the exponent. We want to develop one more technique of integration, called change of variables or substitution, to handle integrals that are pretty close to our stated rules. This technique is often called \(u\)-substitution and is related to the chain rule for differentiation.
We could do this problem by rewriting the integrand as an explicit seventh degree polynomial and then using the power and sum rules, but that is too much work. Instead, I will notice the integrand looks almost like a power, and thus guess an answer of \(\frac{1}{8} (3x+5)^8+C\text{.}\) I then check by differentiating. Using the chain rule,
As we did in the previous example, we first guess the antiderivative to be \(\frac{1}{n+1} (ax+b)^{n+1}+C\text{.}\) We then take the derivative of that expression and obtain \(a(ax+b)^n\text{.}\) This misses our integrand by a factor of \(a\text{.}\) We adjust by that factor and find the antiderivative is \(\frac{1}{a} \frac{1}{(n+1)} (ax+b)^{n+1}+C\text{.}\)
As we did in the last example, our first guess uses the basic rule without worrying about the linear term, so we guess \(e^{ax+b}+C\text{.}\) We then take the derivative of that expression and obtain \(ae^{ax+b}\text{.}\) This misses our integrand by a factor of \(a\text{.}\) We adjust by that factor and find the antiderivative is \(\frac{1}{a} e^{ax+b}+C\text{.}\)
We run into a problem if we try to extend this method with quadratic terms. If we start with \((x^2+5)^3\) and guess an antiderivative of \(\frac{1}{4} (x^2+5)^4\text{,}\) when we differentiate we get \((x^2+5)^3 2x\) and are off by a factor of \(8x\text{.}\) However, when we divide by that factor to get \(\frac{(x^2+5)^4}{8x}\) as a proposed antiderivative, and then differentiate again, we get
To use this method with \(u\) replacing something more complicated than a linear term, we need to have \(du\) available, with the possible addition of multiplying by a scalar constant.
Subsection7.4.2Change of variables for definite integrals
In the definite integral, we understand that \(a\) and \(b\) are the \(x\)-values of the ends of the integral. We could be more explicit and write \(x=a\) and \(x=b\text{.}\) The last step in solving a definite integral is to substitute the endpoints back into the antiderivative we have found. We can either change the variables for the endpoints as well, or we can convert the antiderivative back to the original variables before substituting. Consider the following example.
Solution 1: Convert everything to \(u\text{.}\) The obvious candidate for \(u\) is \(2x+5\text{.}\) Then \(du=2\, dx\text{.}\) For the lower endpoint, \(x=1\) becomes \(u=2(1)+5=7\text{.}\) For the upper endpoint \(x=3\) becomes \(u=2(3)+5=11\text{.}\) Substituting,
\begin{align*}
\int_1^3 e^{2x+5}\, dx \amp =\left({\frac{1}{2}}\right) (2)\int_1^3 e^{2x+5} \,dx)\quad \text{ (Add needed factors.)}\\
\amp =\frac{1}{2} \int_1^3 e^{2x+5} (2\,dx) \quad \text{ (Make } u \text{ and } du \text{ explicit.)}\\
\amp =\frac{1}{2} \int_{u=7}^{u=11}e^u \,du \quad \text{ (Do the substitution.)}\\
\amp =\left.\frac{1}{2} e^u\right|_7^{11} \quad \text{ (Find the antiderivative.)}\\
\amp =\frac{1}{2} e^{11}-\frac{1}{2} e^7. \quad \text{ (Evaluate.)}\text{.}
\end{align*}
Solution 2: Keeping, but labeling, the endpoints. We have the same \(u\) and \(du\text{,}\) but do not convert the endpoints. To reduce confusion we make sure to label the variable when we are using both \(x\) and \(u\text{.}\) Thus,
\begin{align*}
\int_1^3 e^{2x+5} \, dx \amp =\frac{1}{2} \int_1^3 e^{2x+5} (2\,dx) \quad \text{ (Make } u \text{ and } du \text{ explicit.)}\\
\amp =\frac{1}{2} \int_{x=1}^{x=3}e^u \,du \quad \text{ (Do the substitution.)}\\
\amp =\left.\frac{1}{2} e^u\right|_{x=1}^{x=3} \quad \text{ (Find the antiderivative.)}\\
\amp =\left.\frac{1}{2} e^{2x+3}\right|_{x=1}^{x=3} \quad \text{ (Convert back.)}\\
\amp =\frac{1}{2} e^{11}-\frac{1}{2} e^7. \quad \text{ (Evaluate.)}\text{.}
\end{align*}
It should be noted that when we change variables we may find ourselves looking at an integral from \(a\) to \(b\) where the \(b \lt a\text{.}\) We do not change the order of the endpoints.
(Convert everything to \(u\text{.}\)) The obvious candidate for \(u\) is \(x^2\text{.}\) Then \(du=2x\, dx\text{.}\) For the lower endpoint, \(x=-2\) becomes \(u=(-2)^2=4\text{.}\) For the upper endpoint \(x=1\) becomes \(u=1^2=1\text{.}\) Substituting,
\begin{align*}
\int_{-2}^1 x e^{(x^2)}\,dx \amp =\frac{1}{2} \int_{-2}^1 e^{(x^2)}(2x\, dx) \quad \text{(Make } u \text{ and } du \text{explicit.)}\\
\amp =\frac{1}{2} \int_{4}^{1}e^u\, du \quad \text{ (Do the substitution.)}\\
\amp =\left.\frac{1}{2} e^u\right|_4^{1} \quad \text{ (Find the antiderivative.)}\\
\amp =\frac{1}{2} (1-e^4).\amp \quad \text{ (Evaluate.)}\text{.}
\end{align*}
1.Reading check, Integration by Change of Variable or Substitution.
This question checks your reading comprehension of the material is section 7.4, Integration by Change of Variable or Substitution, of Business Calculus with Excel. Based on your reading, select all statements that are correct. There may be more than one correct answer. The statements may appear in what seems to be a random order.
An investment stream pays out at a rate of $10,000 per year. In computing present value, I assume an investment return rate of 5% compounded continuously. What is the present value of the first 10 years of the payout?
My gas well is returning a payout of $10,000. The well output is expected to decay exponentially with half as much output in 7 years. How much do I make over the next 10 years?
