Calculus II Lesson 7: Improper Integrals

  1. Upcoming
  2. Improper Integrals
    1. Divergence
    2. Example
    3. Limits at infinity / Divergence
    4. Exercise
  3. Discontinuities
    1. Middle of Interval
    2. Multiple?
  4. Areas between curves
    1. Strategy
    2. Exercises
  5. Trig Substitution

Upcoming

Reminders:

  1. Problem presentations next two weeks. 5 or 6 each day. Volunteers?
  2. MyOpenMath HW due tomorrow night.

Improper Integrals

How do we evaluate integrals where one bound goes off to infinity?

  1. $\int\limits_a^{\infty} f(x) dx = \lim\limits_{b \rightarrow \infty} \int\limits_a^b f(x) dx$, if this limit exists.
  2. $\int\limits_{-\infty}^{b} f(x) dx = \lim\limits_{a \rightarrow -\infty} \int\limits_a^b f(x) dx$, if this limit exists.

Definitions:

  1. $\int\limits_a^{\infty} f(x) dx = \lim\limits_{b \rightarrow \infty} \int\limits_a^b f(x) dx$, if this limit exists.
  2. $\int\limits_{-\infty}^{b} f(x) dx = \lim\limits_{a \rightarrow -\infty} \int\limits_a^b f(x) dx$, if this limit exists.

If the limit exists, we say it converges. If not, we say it diverges.

Exercise

Compute the following definite integrals by taking the appropriate limit.

  1. $\int\limits_0^{\infty} xe^{-x^2} dx$
  2. $\int\limits_0^{\infty} e^{-x} dx$

Divergence

How would we find $\int_0^{\infty} (e^{-x} + 1) dx$ without computing?

What do you notice about this region?

Question: If $\lim\limits_{x \rightarrow \infty} f(x) \neq 0$, then can $\int\limits_0^{\infty} f(x)dx$ converge? Why or why not?

Example

Compute the following improper integral: \(\int_0^{\infty} \frac{1}{x} dx\)

Since $\frac{1}{x} \rightarrow 0$, must this integral converge?

Limits at infinity / Divergence

Exercise

  1. $\int_1^{\infty} \frac{1}{x^2} dx$
  2. Does $\int\limits_0^{\infty} e^{-x^2} dx$ converge or diverge?

Hint on (2): compare $e^{-x^2}$ with $e^{-x}$, which we already know converges.

Discontinuities

$\int_0^4 \frac{1}{\sqrt{4-x}} dx$? Ideas?

Solution: Set up a limit:

\[\lim\limits_{b \rightarrow 4^-} \int\limits_0^b \frac{1}{\sqrt{4-x}}dx\]

Re-write the function as $(4-x)^{-\frac{1}{2}}$, and use the power rule: \(\lim\limits_{b \rightarrow 4^-} \left. -2(4-x)^{\frac{1}{2}}\right\vert_0^b = \lim\limits_{b\rightarrow 4^-} -2(4-b)^{1/2} + 2(4)^{1/2}\)

Continuing: \(\lim\limits_{b\rightarrow 4^-} -2(4-b)^{1/2} + 2(4)^{1/2} = \lim\limits_{b\rightarrow 4^-} -2(4-b)^{1/2} + 4\)

As $b \rightarrow 4^{-}$, what happens to $(4-b)^{1/2}$? It approaches 0! So the limit is just 4!

Question: why is it important here that we approached 4 from the left?

Exercise:

\[\int_0^2 x \ln(x) dx\]

(Where is the discontinuity here?)

Middle of Interval

$\int_{-1}^1 \frac{1}{x^2} dx$. Ideas?

Idea: Split into 2 integrals.

$\int_{-1}^0 \frac{1}{x^2} dx + \int_0^1 \frac{1}{x^2} dx$

\[\begin{align} &\lim_{b\rightarrow 0^-} \int_{-1}^b \frac{1}{x^2} dx &+ \lim_{a \rightarrow 0^+} \int \frac{1}{x^2} dx \\ &= \lim_{b\rightarrow 0^-} \left. (-\frac{1}{x}) \right\vert_{-1}^b &+ \lim_{a \rightarrow 0^+} \left.(-\frac{1}{x})\right\vert_a^1 \\ &= \ldots &+ \lim_{a \rightarrow 0^+} -1 + \frac{1}{a} \end{align}\]

which diverges. (The other side diverges too.)

Multiple?

What should we do if we have multiple discontinuities?

Areas between curves

Determine the area of the region bounded by the curves $y = x$ and $y = x^2$.

Ideas? First: Where do these curves intersect? What are the endpoints of the region?

\[\begin{align} x &= x^2 \\ x - x^2 &= 0 \\ x(1 - x) &= 0 \\ x = 0 \text{ or } x &= 1 \end{align}\]

What’s the area under just one of those curves?

\[\int_0^1 x dx = \left.\frac{x^2}{2}\right\vert_0^1 = \frac{1}{2}\] \[\int_0^1 x^2 dx = \left.\frac{x^3}{3}\right\vert_0^1 = \frac{1}{3}\]

Then just subtract those areas! $A = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.

Or: take the integral of the difference!

\[\int_0^1 (x - x^2) dx = \left.\frac{x^2}{2}-\frac{x^3}{3}\right\vert_0^1 - \frac{1}{6}\]

Question: Why $x - x^2$ and not $x^2 - x$?

Strategy

  1. Determine where they intersect
  2. Determine which function is greater in this region
  3. Set up the integral $\int_a^b |f(x) - g(x)| dx$.

(Use (2) to determine whether the integral is of $f(x) - g(x)$ or $g(x) - f(x)$.)

Exercises

  1. Find the area between $f(x) = x$ and $g(x) = x^3$ in the first quadrant.
  2. Find the area between $f(x) = \sin(x)$ and $g(x) = \cos(x)$ between $x = \pi/4$ and $x = 5\pi/4$.

Trig Substitution

How do we integrate functions like $\int_0^2 \sqrt{4 - x^2} dx$?

Geometrically:

It’s a quarter circle with radius 2, so the area should be $\frac{1}{4} \pi r^2 = \frac{1}{4} \cdot 4 \pi = \pi$. But how do we see this?

There isn’t a nice antiderivative here. Instead, we will need to make a substitution. Here if we let $x = 2\sin(\theta)$, then $dx = 2\cos(\theta) d\theta$. Then update the bounds:

\[\int_0^{\pi/2} \sqrt{4 - 4\sin^2(\theta)} \cdot 2 \cos(\theta) d\theta\]

$\sqrt{4 - 4\sin^2(\theta)}$ can be simplified to $\sqrt{4}\sqrt{1 - \sin^2(\theta)}$, which is $2 \sqrt{1 - \sin^2(\theta)}$. And we know that $1 - \sin^2(\theta) = \cos^2(\theta)$. So the entire integral is:

\[\int_0^{\pi/2} (2 \sqrt{\cos^2(\theta)})(2 \cos(\theta)) d\theta\]

From $\theta = 0$ to $\pi/2$, $\cos(\theta) \geq 0$, so $\sqrt{\cos^2(\theta)} = \cos(\theta)$. This simplifies now to:

\[4 \int_0^{\pi/2} \cos^2(\theta) d\theta\]

Now we can use our double angle formula: $\cos^2(\theta) = \frac{1}{2} + \frac{1}{2} \cos(2\theta)$ and get: