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## Class 12 math (India)

### Course: Class 12 math (India)>Unit 10

Lesson 10: Improper integrals

# Improper integrals review

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
• Current
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
• Current
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 to evaluate an improper integral whose limits have been given •  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.
• How can i know if the improper integral is divergent? • 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 ? • 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? • 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)?  • For problem 2.2, why is it lim b->1- not 0? • 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.