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## Calculus 2

### Unit 2: Lesson 7

Improper integrals

# Improper integrals review

AP.CALC:
LIM‑6 (EU)
,
LIM‑6.A (LO)
,
LIM‑6.A.1 (EK)
,
LIM‑6.A.2 (EK)
Review your knowledge of improper integrals.

## What are improper integrals?

Improper integrals are definite integrals that cover an unbounded area.
One type of improper integrals are integrals where at least one of the endpoints is extended to infinity. For example, integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x is an improper integral. It can be viewed as the limit limit, start subscript, b, \to, infinity, end subscript, integral, start subscript, 1, end subscript, start superscript, b, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x.
Another type of improper integrals are integrals whose endpoints are finite, but the integrated function is unbounded at one (or two) of the endpoints. For example, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x is an improper integral. It can be viewed as the limit limit, start subscript, a, \to, 0, start superscript, plus, end superscript, end subscript, integral, start subscript, a, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x.
An unbounded area that isn't infinite?! Is that for real?! Well, yeah! Not all improper integrals have a finite value, but some of them definitely do. When the limit exists we say the integral is convergent, and when it doesn't we say it's divergent.

## Practice set 1: Evaluating improper integrals with unbounded endpoints

Let's evaluate, for example, the improper integral integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x. As mentioned above, it's useful to view this integral as the limit limit, start subscript, b, \to, infinity, end subscript, integral, start subscript, 1, end subscript, start superscript, b, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x. We can use the fundamental theorem of calculus to find an expression for the integral:
\begin{aligned} \displaystyle\int_1^b\dfrac{1}{x^2}\,dx&=\displaystyle\int_1^b x^{-2}\,dx \\\\ &=\left[\dfrac{x^{-1}}{-1}\right]_1^b \\\\ &=\left[-\dfrac{1}{x}\right]_1^b \\\\ &=-\dfrac{1}{b}-\left(-\dfrac{1}{1}\right) \\\\ &=1-\dfrac{1}{b} \end{aligned}
Now we got rid of the integral and we have a limit to find:
\begin{aligned} \displaystyle\lim_{b\to\infty}\int_1^b\dfrac{1}{x^2}\,dx&=\displaystyle\lim_{b\to\infty}\left(1-\dfrac{1}{b}\right) \\\\ &=1-0 \\\\ &=1 \end{aligned}
Problem 1.1
integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, cubed, end fraction, d, x, equals, question mark

Want to try more problems like this? Check out this exercise.

## Practice set 2: Evaluating improper integrals with unbounded function

Let's evaluate, for example, the improper integral integral, start subscript, 0, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x. As mentioned above, it's useful to view this integral as the limit limit, start subscript, a, \to, 0, end subscript, integral, start subscript, a, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x. Again, we use the fundamental theorem of calculus to find an expression for the integral:
\begin{aligned} \displaystyle\int_a^1\dfrac{1}{\sqrt x}\,dx&=\displaystyle\int_a^1 x^{^{\large -\frac{1}{2}}}\,dx \\\\ &=\left[\dfrac{x^{^{\large\frac{1}{2}}}}{\frac{1}{2}}\right]_a^1 \\\\ &=\Bigl[2\sqrt x\Bigr]_a^1 \\\\ &=2\sqrt 1-2\sqrt a \\\\ &=2-2\sqrt a \end{aligned}
Now we got rid of the integral and we have a limit to find:
\begin{aligned} \displaystyle\lim_{a\to 0}\int_a^1\dfrac{1}{\sqrt x}\,dx&=\displaystyle\lim_{a\to 0}(2-2\sqrt a) \\\\ &=2-2\cdot 0 \\\\ &=2 \end{aligned}
Problem 2.1
integral, start subscript, 0, end subscript, start superscript, 8, end superscript, start fraction, 1, divided by, cube root of, x, end cube root, end fraction, d, x, equals, question mark

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• How can i know if the improper integral is divergent?
• If the limit doesn't exist! Say, you evaluate the limit and get infinity (+ or -) then the integral will be divergent. Otherwise the limit should exist and it will be convergent.
• how to evaluate an improper integral whose limits have been given
• thanks. I am going to work through steps. When I graphed it, I see a positive area above the x axis. So, the graph should be under the x axis?
• My calculations for the review problem 2.2 give an answer of 3/2. Sal's answer is -3/2. The question is what is the area over the interval [0,1] for the function y=1/((x-1)^1/3). Can you confirm which is the correct answer?
• This integral has a problem when x=1, so substitute a for 1 and find the limit of the integral as a-->1
If you used u=(x-1) you should have gotten 3/2(x-1)^(2/3) + C
Now evaluate the integral from 0 to a to get:
3/2(a-1)^(2/3) - 3/2(0-1)^(2/3) = 3/2(a-1)^(2/3) - 3/2(-1)^(2/3)                                = 3/2(a-1)^(2/3) - 3/2(-1)^(2/3)                                 = 3/2(a-1)^(2/3) - 3/2(³√(-1)²)                                 = 3/2(a-1)^(2/3) - 3/2(³√1)                                 = 3/2(a-1)^(2/3) - 3/2(1)                                 = 3/2(a-1)^(2/3) - 3/2

Now take the limit as a-->1 to get 3/2(1-1)^(2/3) - 3/2 = 0 - 3/2 = -3/2
• The solution of indefinite integral logsinx is ?
• That function can't be expressed in terms of elementary functions.
• Is the integral from -1 to 1 of of 1/x equal to 0? Is this (cancelling out equal but opposite sized infinite regions) allowed for integrating all unbounded odd functions?
• You are correct. All Odd and Even Function rules apply
• This is a side question. If f(x)=sin (pi*x), then my graph is a wave with a range of -1 to 1. If instead f(x) = sin x, does the graph then have a range of -2pi to 2pi?
(1 vote)
• No, the range of f(x) = sin(k*x) is ALWAYS from y = -1 to y = 1, for any value of k (except the silly value of k = 0). What the k does is change the period of the wave. f(x) = sin(x) has a period (or wavelength) of 2pi, while f(x) = sin(k*x) has a period of 2pi/k. Thus, f(x)=sin (pi*x) has a period of 2.

In order to change the range, you would put a coefficient in front. f(x) = A*sin(k*x) has a range from y = -A to y = A.
• I do not understand why the integral of 1/sqtx from 0 to 1 is an unbounded function. Is it because it was a restriction in its domain (x cannot be equal to 0 even though 0 is in the domain of sqtx)?
• The problem is that 0 is not in the domain of the function we're integrating 1/√x, and consequently the fundamental theorem of calculus does not apply. This is why we have to be a bit more ingenious and use the improper integral:
Lim a->0 ∫ a, 1 (1/√x) dx to convince ourselves that the integral converges.
• I don't understand why he integral of x^(-1/2) is improper with finite endpoints. I can't see how this holds: "the integrated function is unbounded at one (or two) of the endpoints" while when the function is evaluated from 0 to 1, it has a finite value. What am I missing? May I ask your help?
(1 vote)
• I assume you're wondering why an integral like ∫x^(-1/2) from x=0 to x=1 is improper, when you can evaluate all points between 0 and 1 of x^(-1/2). Let me know if I misunderstood your question.

Well, I want to ask you: what do you get when you use x=0 in x^(-1/2). You get 1/(0^(1/2)), or 1/0. So at what y-value should we start finding the area under to get the value of this integral? That's why it's improper.

However, we can find the y-value of x=0.00001 in x^(-1/2). It will be very large, but it will exist. Similarly, we can find almost any value along the curve x^(-1/2), except 0. Thus, let's try to take the limit of the integral.

This gives us lim_{a → 0} of ∫x^(-1/2) from x=a to x=1. That solves to: lim_{a → 0} of (2√x) from x=a to x=1, or lim_{a → 0} of (2 - 2√a). By direct substitution, that evaluates the integral to 2.

How does the integral evaluate to a real number, like 2? Let's take a simpler example.
Take a rectangle with sides of length 2 and 1 (which has an area of 2). Now, divide this rectangle into 2 squares. We can represent the areas of the squares and the rectangles as:
2 = 1 + 1

Let's divide the second square into two rectangles, representing the areas as:
2 = 1 + 2*(1/2) = 1 + 1/2 + 1/2

If we keep dividing, we find:
2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...

That should make sense, if you think about it visually. The same applies to integrals, and allows improper integrals with a seemingly infinite area to have a finite area.
• You substitute for the value that is undefined for the function. Typically, it's usually ∞ or 0. Also, it doesn't really matter which variable you use for the substitution. Hope this helps!