Math 2411-Notes for February 12, 2004

Let's review improper integrals before moving on to new material. A definite integral is called an improper integral if either there is an infinite limit(s) of integration or the function has a vertical asymptote in the interval over which you are integrating. For example, each of the definite integrals

$$\displaystyle \int^{\infty}_{2}\displaystyle \frac{1}{x^{3/2}}\;d... ...sqrt{x}}\;dx,\;\;\;\; \displaystyle \int^{4}_{2}\displaystyle \frac{1}{x-3}\;dx$$

are improper.

Up to this point, when evaluating definite integrals we have used the Fundamental Theorem of Calculus (FTC) which says the definite integral can be solved by evaluating the antiderivative at the endpoints of the interval.

$$\displaystyle \int^{b}_{a}f(x)\;dx=F(b)-F(a)\;\;$$where$$\;\;F'(x)=f(x)$$

However, this theorem only holds when both $a$ and $b$ are finite and the function $f(x)$ is continuous on the interval $[a,b]$. So we are unable to use the FTC for improper integrals.

In order to evaluate an improper integral we will need to relate the problem to a ``proper'' integral and then apply a limit. For example,

$$\displaystyle \int^{\infty}_{1}\displaystyle \frac{1}{x^2}\;dx$$

can not be evaluated with the FTC since the upper limit of integration is infinite. But we can certainly evaluate

$$\displaystyle \int^{b}_{1}\displaystyle \frac{1}{x^2}\;dx$$

for any value $b$. Using the FTC we get

$$\displaystyle \int^{b}_{1}\displaystyle \frac{1}{x^2}\;dx=1-\displaystyle \frac{1}{b}$$

So given any interval $[1,b]$ the area under the curve is exactly $1-1/b$. If $b=10$ then the area under the curve is $1-1/10=.9$. If the interval is $[1,1000]$ then the area is $1-1/1000=.999$. If $b=100000$ the area under the curve is exactly $.99999$. It should be clear that the larger the interval we choose, the closer the area under the curve gets to one. So it seems to make sense that

$$\displaystyle \int^{\infty}_{1}\displaystyle \frac{1}{x^2}\;dx=1$$

Lets now look at

$$\displaystyle \int^{\infty}_{1}\displaystyle \frac{1}{x}\;dx$$

Again we can not use the FTC so we start with a problem that we can evaluate. As long as $a$ is finite we get

$$\displaystyle \int^{a}_{1}\displaystyle \frac{1}{x}\;dx=\ln(a)$$

This means that if we chose the interval $[1,a]$ the area under the curve is exactly $ln(a)$. If $a=5$ then the area under the curve is $\ln(5)\approx1.60944$. If we let $a=100$ the area under the curve is $\ln(100)\approx4.60517$. Over the interval $[1,1000000]$ the area under the curve is exactly $\ln(1000000)\approx13.8155$. The amount of area under the curve is growing slowly as we increase the size of our interval, and it will continue to grow infinitely large as we let $a$ get infinitely large. Since we are not approaching a single value as we let $a$ get large it doesn't make sense to define a value for the improper integral. So some improper integrals are defined and some are not. If an improper integral is defined as some value, we say it converges to that value. Otherwise we say it diverges.

If we have an improper integral of the form

$$\displaystyle \int^{\infty}_{a}f(x)\;dx$$

we say it converges if and only if the following limit exists.

$$\displaystyle \lim_{b\to\infty}\displaystyle \int^{b}_{a}f(x)\;dx$$

We will always treat improper integral as limits of ``proper'' integrals. So our first example should be treated as

$$\displaystyle \int^{\infty}_{1}\displaystyle \frac{1}{x^2}\;dx= = \displaystyle \lim_{b\to\infty}\displaystyle \int^{b}_{1}\displaystyle \frac{1}{x^2}\;dx$$

$$= \displaystyle \lim_{b\to\infty}\left[-\displaystyle \frac{1}{x}\displaystyle \right]^{b}_{1}$$

$$= \displaystyle \lim_{b\to\infty}\left[1-\displaystyle \frac{1}{b}\right]$$

$$= 1$$

So the improper integral converges to the value one. However, the improper integral

$$\displaystyle \int^{\infty}_{1}\displaystyle \frac{1}{x}\;dx= = \displaystyle \lim_{b\to\infty}\displaystyle \int^{b}_{1}\displaystyle \frac{1}{x}\;dx$$

$$= \displaystyle \lim_{b\to\infty}\left[\ln\vert x\vert\displaystyle \right]^{b}_{1}$$

$$= \displaystyle \lim_{b\to\infty}\ln(b)$$

$$= \infty$$

diverges since the above limit does not exist.

Although it is difficult to get a handle on the geometric side of things, we might want to think that a convergent improper integral is an infinite length curve bounding a finite ``amount'' of area between it and the $x$-axis. Then a divergent improper integral would have an infinite ``amount'' of area under its infinite length curve and so we can not assign it a real numbered value.

We will limit our decisions about convergence and divergence to the definition given above. So

$$\displaystyle \int^{0}_{-\infty}\cos(x)\;dx = \displaystyle \lim_{a\to-\infty}\displaystyle \int^{0}_{a}\cos(x)\;dx$$

$$= \displaystyle \lim_{a\to-\infty}\left[\;\sin(x)\;\displaystyle \right]^{0}_{a}$$

$$= \displaystyle \lim_{a\to-\infty}\left[\;\sin(0)-\sin(a)\;\right]$$

$$= -\displaystyle \lim_{a\to-\infty}\sin(a)$$

diverges (because the limit does not exists), even though we might think that each segment above the axis cancels out the area of a segment below the axis. Again, we set our intuition aside, and since the limit does not exist we say that the improper integral diverges.

The other type if improper integral is one where the the function has a vertical asymptote in the interval of integration. For example,

$$\displaystyle \int^{0}_{-2}\displaystyle \frac{1}{\sqrt{\vert x+1\vert}}\;dx$$

is improper since $1/\sqrt{\vert x+1\vert}$ has a vertical asymptote at $x=-1$.

We will first break up the improper integral into enough intervals so that each interval has exactly one ``problem''; i.e., a single asymptote in the interval. This means that at each vertical asymptote we make a new interval. Then we will use our limit definition of convergence for each interval. Since here there is only one vertical asymptote at $x=-1$ we get two intervals, one from $-2$ to $-1$ and the other from $-1$ to zero.

$$\displaystyle \int^{0}_{-2}\displaystyle \frac{1}{\sqrt{\vert x+1\vert}}\;dx = \displaystyle \int^{-1}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx+ \displaystyle \int^{0}_{-1}\displaystyle \frac{1}{\sqrt{x+1}}\;dx$$

$$= \displaystyle \lim_{b\to-1^{-}}\displaystyle \int^{b}_{-2}\displa... ...im_{a\to-1^{+}}\displaystyle \int^{0}_{a}\displaystyle \frac{1}{\sqrt{x+1}}\;dx$$

Both limits must exist for the integral to converge so we start by looking at the first piece. Before we evaluate the limit let's put our calculator to use and try some numbers where $b$ is close to $-1$. First try $b=-1.1$.

$$\displaystyle \int^{-1.1}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx=1.36754$$

Lets get a little closer and let $b=-1.01$

$$\displaystyle \int^{-1.01}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx=1.8$$

And a little closer at each step.....

$$\displaystyle \int^{-1.0001}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx=1.98$$

$$\displaystyle \int^{-1.00001}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx=1.99368$$

$$\displaystyle \int^{-1.000001}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx=1.998$$

$$\displaystyle \int^{-1.00000001}_{-2}\displaystyle \frac{1}{\sqrt{-(x+1)}}\;dx=1.9998$$

Any guesses yet? Let's evalate the limit to get the real solution.

$$\displaystyle \lim_{b\to-1^{-}}\displaystyle \int^{b}_{-2}\displaystyle \frac{1}{(-x-1)^{1/2}}\;dx = \displaystyle \lim_{b\to-1^{-}}\left[-2\sqrt{-(x+1)}\displaystyle \right]^{b}_{-2}$$

$$= \displaystyle \lim_{b\to-1^{-}}\left[\left(-2\sqrt{-(b+1)}\right)- \left(-2\sqrt{-(-2+1)}\right)\right]$$

$$= \displaystyle \lim_{b\to-1^{-}}\left[2-2\sqrt{-(b+1)}\right]$$

$$= 2-2\displaystyle \lim_{b\to-1^{-}}\sqrt{-(b+1)}$$

$$= 2$$

We can now argue, using symmetry about the line $x=-1$, that the second piece also converges to 2. So since both limits exist the series converges to the sum of those limits and

$$\displaystyle \int^{0}_{-2}\displaystyle \frac{1}{\sqrt{\vert x+1\vert}}\;dx=4$$

It is important to recognize improper integrals and to decide their convergence or divergence, especially when we use them to evaluate infinite series later in the course. Now we move on the section 6.1.

Finding Area Between Curves:

We are already comfortable with finding area under a curve and interpreting the meaning of this value, but what about finding the area between two curves?

This is an easy problem. Simply determine the area under the upper function and subtract the area from the lower function. Suppose we wanted to know the area between the curves in the figure above on the interval $[0,1]$. We would get

$$Area = \displaystyle \int^{1}_{0}(9-x^2)\;dx-\displaystyle \int^{1}_{0}(x+1)\;dx$$

$$= \displaystyle \int^{1}_{0}8-x^2-x\;dx$$

$$= 8x-\displaystyle \frac{x^3}{3}-\displaystyle \frac{x^2}{2}\displaystyle \Bigg\vert^{1}_{0}$$

$$= \Big(8-\displaystyle \frac{1}{3}-\displaystyle \frac{1}{2}\Big)-(\;0\;)$$

$$= \displaystyle \frac{43}{6}$$

If we wanted to know the area bewteen the curves on the interval $[0,4]$ we would need two integrals since the top function changes where they intersect. Since this intersection occurs at $x=(-1+\sqrt{33})/2\approx2.37228$ we get

$$Area = \Bigg[\displaystyle \int^{2.372}_{0}(9-x^2)\;dx-\displaystyle \in... ...style \int^{4}_{2.372}(x+1)\;dx-\displaystyle \int^{4}_{2.372}(9-x^2)\;dx\Bigg]$$

$$= \displaystyle \int^{2.372}_{0}(8-x^2-x)\;dx+\displaystyle \int^{4}_{2.372}(x^2+x-8)\;dx$$

$$= \left(8x-\displaystyle \frac{x^3}{3}-\displaystyle \frac{x^2}{2}\... ...{x^3}{3}+\displaystyle \frac{x^2}{2}\displaystyle \Bigg\vert^{4}_{2.372}\right)$$

$$= 11.7142+9.04753$$

$$= 20.7671$$

One application of this technique is the following problem: Suppose the following graphs are velocity functions for two automobiles, car A and car B. How can we determine which car has tavelled further in a given time interval?

The area between the two functions tells us the difference in their distances travelled. So after 1 minute car B is ahead since it has greater area under its curve and it is ahead the same distance as the area between the curves. If the area between the curves on $[0,1]$ is the same as the area between the curves on $[1,2]$ then the cars are side by side after two minutes. If this area on the interval $[1,2]$ is less than the area on the interval $[0,1]$, car B is still ahead after two minutes, even though car A has been travelling faster for the second minute. Without thinking of area between the velocity functions there is no way to decide which car is ahead after a certain time period.

This idea can be extended to many applications. Suppose we had two runoff functions (with units $ft.^3/sec.$), measuring water flow for an entire season, for two streams in different drainages. We want to know which drainage produced more runoff. Since the units for the area under the curve are $ft.^3/sec.\;\times\;sec.=ft.^3$, we can compare the areas under the two runoff curves. If the two curves are placed on the same graph then the area between the curves would indicate how much more water the dominant stream delivered.