• Use pattern recognition to find an indefinite integral.
  • Use a change of variables to find an indefinite integral.
  • Use the General Power Rule for Integration to find an indefinite integral.
  • Use a change of variables to evaluate a definite integral.
  • Evaluate a definite integral involving an even or odd function.

Assignment

  • Vocabulary and teal boxes
  • p343 1–4, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 59, 63, 67–79 odd, 82, 86 91–95 odd, 117–120

Additional Resources

Differentiating something like $\sin x^2$ requires use of the chain rule, and in this case results in $\cos x^2 \cdot 2x$. You need to differentiate the outer function, then multiply it by the derivative of the inner function.

\[\frac{d}{dx} F(g(x)) = F'(g(x)) \cdot g'(x)\]

Since we are now dealing with integration and antidifferentiation, we will look for this pattern so we can work backwards: a composite function multiplied by the derivative of the inner function.

\[\int (x^2 + 1)^2 \cdot 2x \, dx\]

Our composite function is $(x^2 + 1)^2$, and it’s multiplied by the inner’s derivative, $2x$ in this case. Since the pattern exists, we can move on, and we’re going to use some substitution to make our lives easier. First, we’re going to let $u = x^2 + 1$. Then, we can say $\frac{du}{dx} = 2x$, which we can rewite as $du = 2x \, dx $. Now our integral looks like this.

\[\int (u)^2 \, du\]

After antidifferentiating, we can substitute back in our originals.

\[\int (u)^2 \, du = \frac{1}{3}u^3 + C =\frac{1}{3}(x^2+1)^3 + C\]

Multiplying by a Constant

So, this pattern we are looking for is very specific, and obviously won’t always be there. We can include more candidates by multiplying by a constant. The integral below is close to what we need, but requires $2x$ where it only has an $x$. We can multiply by $2$, and then by $1/2$, to get the $2$ that we need, without technically changing the problem.

\[\begin{align*} \int x(x^2 + 1)^2 \, dx &= \int \frac{1}{2} \cdot 2 \cdot x(x^2 + 1)^2 \, dx \\ &= \frac{1}{2} \int u^2 \, du &&\textcolor{BrickRed}{\text{Let }u=x^2 + 1, du = 2x\, dx} \\ &= \frac{1}{2} \cdot \frac{1}{3}u^3 + C \\ &= \frac{1}{6} (x^2+1)^3 + C &&\textcolor{BrickRed}{\text{Substitute in for } u} \end{align*}\]

Change of Variables for Definite Integrals

So far, we’ve dealt with indefinite integrals. Definite integrals require that we adjust our interval bounds when we change our variable. When you decide what $u$ will equal, you need to run your old interval bounds though that equation. Below, $u=x^2+1$ so our old interval of $[0,1]$ becomes $[1,2]$ because $(0)^2 + 1 = 1$ and $(1)^2 + 1 = 2$.

\[\begin{align*} \int_0^1 x(x^2+1)^3 \, dx &= \frac{1}{2}\int_0^1 2x(x^2+1)^3 \, dx \\ &= \frac{1}{2}\int_1^2 u^3 \, du &&\textcolor{BrickRed}{\text{Let }u=x^2 + 1, du = 2x\, dx}\\ &= \frac{1}{2} \left[\frac{1}{4}u^4\right]_1^2 \\ &= \frac{1}{2} \left(4 - \frac{1}{4}\right) \\ &= \frac{15}{8} \end{align*}\]

Note that we don’t need to substitute our equation back in for $u$ since we already adjusted the interval. Changing the variable to $u$ doesn’t change the result of the integral, as long as the interval is adjusted properly.

Alternate Example 9

The book evaluates example 9 one way, but here is another that is more in line with what we’ve done so far. Both versions introduce what to do if you have extra $x$ terms lying around after you perform your substitution.

\[\begin{align*} \int_1^5 \frac{x}{\sqrt{2x-1}} \, dx &= \frac{1}{2} \int_1^5 2x(2x-1)^{-1/2} \, dx && \textcolor{BrickRed}{\text{In anticipation of }u=2x - 1, du = 2\, dx} \\ &= \frac{1}{2} \int_1^9 \frac{u-1}{2}u^{-1/2} \, du && \textcolor{BrickRed}{\text{Since } u=2x - 1, x=\frac{u-1}{2}} \\ &= \frac{1}{4} \int_1^9 u^{1/2} + u^{-1/2} \, du \\ &= \frac{1}{4} \left[\frac{2}{3}u^{3/2} + 2u^{1/2}\right]_1^9 \\ &= \frac{1}{4}\left(24-\frac{8}{3}\right) \\ &= \frac{16}{3} \end{align*}\]

The General Power Rule for Integration and Integration of Even and Odd Functions

Two parts of this section deal with shortcuts that can make your life easier. The General Power Rule you will likely pick up on as you complete problems involving integrating powers. Use it as you see fit.

The even and odd shortcut has you taking advantage of symmetry in certain functions. Even functions are mirrored over the $y$-axis, so you can just integrate one side and double it. Odd functions have symmetry with respect to the origin, so the areas end up cancelling each other out. Just watch your intervals when you are using these.