Question 1

how to evaluate an improper integral whose limits have been given

Accepted Answer

I assume you're asking how it is an improper integral if it is being evaluated using defined numbers, rather than infinity?

To be a proper integral, the area being calculated must be an enclosed space (bounded on all sides) - you need to be able to draw an outline with no openings around the area. When you are integrating between two x-values, the right and left side are enclosed by vertical lines at those x-values. But if the curve itself has a vertical asymptote, then the top edge of the area is open (instead of an x-value of infinity, you have a y-value of infinity - infinite height instead of infinite width, if you will).

Luckily if you have a vertical boundary set at the vertical asymptote, the vertical asymptote will converge on the vertical line, so it's possible to calculate the area just like with converging horizontal asymptotes. For evaluation, you calculate it just like any other definite integral. If the x-value boundaries are not at the asymptote, split it into two integrals, one evaluated from the lower bound to the asymptote and the other from the asymptote to the upper bound.

Question 2

How can i know if the improper integral is divergent?

Accepted Answer

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.

Question 3

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?

Accepted 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`

Question 4

The solution of indefinite integral logsinx is ?

Accepted Answer

That function can't be expressed in terms of elementary functions.

Question 5

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)?

Accepted Answer

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.

Question 6

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?

Accepted Answer

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.

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.

Question 7

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?

Accepted Answer

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.

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 6

Improper integrals review

What are improper integrals?

Practice set 1: Evaluating improper integrals with unbounded endpoints

Practice set 2: Evaluating improper integrals with unbounded function

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