17Calculus - Improper Integrals - Endpoint Discontinuity

17Calculus

On this page we discuss the second type of improper integrals, integrals with a discontinuity at an endpoint.

Case 2 - Discontinuity at an Endpoint

This case occurs when we have a continuous function within a finite interval but the function is discontinuous at one of the endpoints. For example, we might have an integral like $$\displaystyle{ \int_{0}^{1}{ \frac{1}{x} ~ dx} }$$. In this case, $$f(x)=1/x$$ is discontinuous at one of the endpoints, $$x=0$$. So we can't integrate. However, we can use the same technique above that we used for an infinite interval and use limits. It looks something like this.

Case 2: Finite Integral with a Discontinuity at One Endpoint

$$\displaystyle{ \int_{a}^{b}{f(x) ~ dx} }$$

$$a$$ and $$b$$ are finite

$$f(x)$$ is continuous on the interval $$[a,b)$$ but discontinuous at $$x=b$$

$$\displaystyle{\lim_{t \to b}{ \int_{a}^{t}{f(x) ~ dx} } }$$

$$\displaystyle{ \int_{a}^{b}{g(x) ~ dx} }$$

$$a$$ and $$b$$ are finite

$$g(x)$$ is continuous on the interval $$(a,b]$$ but discontinuous at $$x=a$$

$$\displaystyle{\lim_{t \to a}{ \int_{t}^{b}{g(x) ~ dx} } }$$

Notice that we do the same thing as in Case 1, i.e. we replace the 'problem' point with a limit.

So far, so good. Now, there is only one minor twist to these two cases that is easily handled. If you have a 'problem' (a discontinuity or infinity) at both ends, you just break the integral into two integrals at a convenient point. We know from basic integration that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = }$$ $$\displaystyle{ \int_{a}^{k}{f(x)~dx} + \int_{k}^{b}{f(x)~dx} }$$, right? So we just do the same thing with the improper integral and handle each case and limit separately. For example, we might want to evaluate $$\displaystyle{ \int_{0}^{\infty}{\frac{1}{x}~dx} }$$, so we just split the integral into two and handle each end separately. We can break it anywhere in the interval, I will choose $$2$$. We end up with
$$\displaystyle{ \int_{0}^{\infty}{\frac{1}{x}~dx} = }$$ $$\displaystyle{ \int_{0}^{2}{\frac{1}{x}~dx} + \int_{2}^{\infty}{\frac{1}{x}~dx} = }$$ $$\displaystyle{ \lim_{a \to 0}{\int_{a}^{2}{\frac{1}{x}~dx} } + \lim_{b \to \infty}{ \int_{2}^{b}{\frac{1}{x}~dx} } }$$

Okay, try these practice problems before going on.

Practice

Unless otherwise instructed, evaluate these integrals using proper notation. Give your answers in exact, simplified and factored form.

$$\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$4$$

Problem Statement

Evaluate $$\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

$$\displaystyle{ \int_0^8{ \frac{dx}{\sqrt{2x}} } = \lim_{a\to 0^+}{ \int_a^8{ \frac{dx}{\sqrt{2x}} } } }$$

 $$\displaystyle{ \int{\frac{dx}{\sqrt{2x}}} }$$ $$\displaystyle{ \frac{1}{\sqrt{2}} \int{ x^{-1/2}~dx } }$$ $$\displaystyle{ \frac{1}{\sqrt{2}} \frac{x^{1/2}}{1/2} }$$ $$\displaystyle{ \sqrt{2x} }$$ $$\displaystyle{ \lim_{a\to 0^+}{ \left[ \sqrt{2x} \right]_a^8 } }$$ $$\displaystyle{ \lim_{a\to 0^+}{ [ \sqrt{16} - \sqrt{2a} ] } = 4 }$$

$$4$$

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{4}{ \frac{1}{\sqrt{4-x}} dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{4}{ \frac{1}{\sqrt{4-x}} dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

The Organic Chemistry Tutor - 1306 video solution

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$\pi/2$$

Problem Statement

Evaluate $$\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

We start by rewriting the integral using a limit.
$$\displaystyle{ \int_0^3{ \frac{dx}{\sqrt{9-x^2}} } = \lim_{b\to3^-}{ \int_0^b{ \frac{dx}{\sqrt{9-x^2}} } } }$$
Now we will evaluate the indefinite integral
$$\displaystyle{ \int{ \frac{dx}{\sqrt{9-x^2}} } }$$
Use trig substitution with $$x=3\sin\theta \to dx = 3\cos\theta~d\theta$$.
$$\displaystyle{ \int{ \frac{3\cos\theta~d\theta}{\sqrt{9-9\sin^2\theta}} } = }$$ $$\displaystyle{\int{ \frac{\cos\theta}{\cos\theta}~d\theta } = }$$ $$\displaystyle{ \int{1~d\theta} = \theta = \arcsin(x/3) }$$
$$\displaystyle{ \lim_{b\to3^-}{ [ \arcsin(x/3)|_0^b ] } = }$$ $$\displaystyle{ \lim_{b\to3^-}{ [ \arcsin(b/3) - \arcsin(0/3) ] } = }$$ $$\arcsin 1 = \pi/2$$

$$\pi/2$$

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{3}^{4}{ \frac{dx}{(x-3)^{3/2}} } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{3}^{4}{ \frac{dx}{(x-3)^{3/2}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

The integral diverges.

Problem Statement

Evaluate $$\displaystyle{ \int_{3}^{4}{ \frac{dx}{(x-3)^{3/2}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

Comparing the limits of integration to the integrand, we can see that there is a discontinuity at the lower limit. This is the only problem point. So we rewrite the integral as
$$\displaystyle{ \lim_{a \to 3^+}{ \int_{a}^{4}{ \frac{dx}{(x-3)^{3/2}} } } }$$
We now evaluate the indefinite integral.

 $$\displaystyle{ \int{ \frac{dx}{(x-3)^{3/2}} } }$$ $$\int{ (x-3)^{-3/2} ~dx }$$ $$(x-3)^{-1/2} / (-1/2)$$ $$-2(x-3)^{-1/2}$$

Now plug this result back into the limit and evaluate.

 $$\displaystyle{ \lim_{a \to 3^+}{ [-2(x-3)^{-1/2}]_a^4 } }$$ $$\displaystyle{ \lim_{a \to 3^+}{ [-2(4-3)^{-1/2} + 2(a-3)^{-1/2}] } }$$ $$-2 + \infty = \infty$$

The integral diverges.

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{-3}^{1}{ \frac{dx}{(2x+6)^{2/3}} } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-3}^{1}{ \frac{dx}{(2x+6)^{2/3}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$3$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-3}^{1}{ \frac{dx}{(2x+6)^{2/3}} } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

At $$x=-3$$, the denominator of the integrand is zero. This is the only problem point and it occurs at the lower limit of integration. So we rewrite the integral as $$\displaystyle{ \lim_{a \to -3^+}{ \int_{a}^{1}{ \frac{dx}{(2x+6)^{2/3}} } } }$$
Now we evaluate the indefinite integral.

 $$\displaystyle{ \int{ \frac{dx}{(2x+6)^{2/3}} } }$$ Use integration by substitution. $$u=2x+6 \to du=2dx$$ $$\displaystyle{ \int{ \frac{du}{2 u^{2/3}} } }$$ $$\displaystyle{ \frac{1}{2} \int{ u^{-2/3} ~du } }$$ $$\displaystyle{ \frac{1}{2} \frac{u^{1/3}}{1/3} }$$ $$\displaystyle{ \frac{3}{2} u^{1/3} }$$ $$\displaystyle{ \frac{3}{2} (2x+6)^{1/3} }$$

Now plug this result back into the limit and evaluate.

 $$\displaystyle{ \lim_{a \to -3^+}{ \frac{3}{2} [(2x+6)^{1/3}]_a^1 } }$$ $$\displaystyle{ \frac{3}{2} \lim_{a \to -3^+}{ [(2+6)^{1/3} - (2a+6)^{1/3}] } }$$ $$\displaystyle{ \frac{3}{2} [(8)^{1/3} - (2(-3)+6)^{1/3}] = 3 }$$

$$3$$

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{\pi/2}{ \tan\theta ~ d\theta } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\pi/2}{ \tan\theta ~ d\theta } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

The integral diverges.

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{\pi/2}{ \tan\theta ~ d\theta } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

Looking at the integrand, we know that tangent is not defined at $$\pi/2$$. So that is the only problem point. We rewrite the integral with the limit and solve.

 $$\displaystyle{ \lim_{b\to\pi/2^-}{ \int_{0}^{b}{ \tan\theta ~ d\theta } } }$$ $$\displaystyle{ \lim_{b\to\pi/2^-}{ \left[ - \ln|\cos\theta| \right]_0^b } }$$ $$\displaystyle{ \lim_{b\to\pi/2^-}{ \left[ -\ln|\cos b| + \ln|\cos 0| \right] } }$$ $$-\ln|\cos \pi/2| + \ln|1|$$ $$-\ln|0| + 0 = +\infty$$

The integral diverges.

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{1}{ \frac{1}{x^2} dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{1}{ \frac{1}{x^2} dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

The Organic Chemistry Tutor - 1290 video solution

Log in to rate this practice problem and to see it's current rating.

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

Problem Statement

Evaluate $$\displaystyle{ \int_{-1}^{0}{ \frac{1}{x^2} dx } }$$ giving your answer in exact, simplified, factored form. Make sure to use correct notation in every step of your work.

Solution

PatrickJMT - 1461 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{6}{\frac{1}{ x^{0.8}} dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{6}{\frac{1}{ x^{0.8}} dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

MIP4U - 1467 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{3}{ \frac{1}{x^{1.2}} dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{3}{ \frac{1}{x^{1.2}} dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

MIP4U - 1468 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{5}^{8}{ \frac{2}{\sqrt[3]{x-5}} dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{5}^{8}{ \frac{2}{\sqrt[3]{x-5}} dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

MIP4U - 1474 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{1}{ \frac{\ln(x)}{x^{1/2}} dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{1}{ \frac{\ln(x)}{x^{1/2}} dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

MIT OCW - 1479 video solution

video by MIT OCW

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{-1}^1{ \frac{x}{x^2+2x+1} ~dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{-1}^1{ \frac{x}{x^2+2x+1} ~dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

The integral diverges.

Problem Statement

Evaluate $$\displaystyle{ \int_{-1}^1{ \frac{x}{x^2+2x+1} ~dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

$$\displaystyle{ \frac{x}{x^2+2x+1} = \frac{x}{(x+1)^2} }$$
The domain of the integrand is all real numbers except for $$x=-1$$, which is the lower limit of integration. This is the only problem point. So we rewrite the integral as
$$\displaystyle{ \lim_{a\to-1^+}{ \int_{a}^1{ \frac{x}{x^2+2x+1} ~dx } } }$$
Now we integrate the indefinite integral.

 $$\displaystyle{ \int{ \frac{x}{x^2+2x+1} ~dx } }$$ $$u=x+1 \to du=dx$$ and $$x=u-1$$ $$\displaystyle{ \int{ \frac{u-1}{u^2} ~du } }$$ $$\displaystyle{ \int{ \frac{1}{u} - \frac{1}{u^2} ~du } }$$ $$\displaystyle{ \ln|u| - \frac{u^{-1}}{-1} }$$ $$\displaystyle{ \ln|u| + \frac{1}{u} }$$ $$\displaystyle{ \ln|x+1| + \frac{1}{x+1} }$$

Now plug this result back into the limit.

 $$\displaystyle{ \lim_{a\to-1^+}{ \left[ \ln|x+1| + \frac{1}{x+1} \right]_a^1 } }$$ $$\displaystyle{ \lim_{a\to-1^+}{ \left[ \ln 2 + 1/2 - \ln|a+1| - 1/(a+1) \right] } }$$ $$\ln 2 + 1/2 - (-\infty) - \infty = \infty - \infty$$

Since $$\infty - \infty$$ is indeterminate, we need to drop back and see where those terms came from and evaluate them differently. The second to the last step shows us that they come from the last two terms. So we will evaluate them separately using L'Hopitals Rule.
Note: Since $$a \gt -1$$, we can write $$|a+1|$$ without the absolute values signs, i.e. $$|a+1| = (a+1)$$.

 $$\displaystyle{ \lim_{a\to-1^+}{ \left[ -\ln(a+1) - 1/(a+1) \right] } }$$ $$\displaystyle{ -\lim_{a\to-1^+}{ \left[ \ln(a+1) + 1/(a+1) \right] } }$$ $$\displaystyle{ -\lim_{a\to-1^+}{ \left[ \frac{(a+1)\ln(a+1)+1}{a+1} \right] } }$$ Now apply L'Hopitals Rule $$\displaystyle{ -\lim_{a\to-1^+}{ \left[ \frac{(a+1)[1/(a+1)] + \ln(a+1)(1) }{1} \right] } }$$ $$\displaystyle{ -\lim_{a\to-1^+}{ \left[ 1+\ln(a+1) \right] } }$$ $$-[ 1 + (-\infty) ] = \infty$$

Combining results, we get $$\ln 2 + 1/2 + \infty = \infty$$. So the integral diverges.

The integral diverges.

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \int_{0}^{1}{ x^2\ln(1/x) ~dx } }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{1}{ x^2\ln(1/x) ~dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

$$1/9$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{1}{ x^2\ln(1/x) ~dx } }$$ giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

Comparing the limits of integration to the integrand, we can see that the only problem point is the lower limit, $$x=0$$. So we rewrite the limit as follows.
$$\displaystyle{ \lim_{a\to0^+}{ \int_{a}^{1}{ x^2\ln(1/x) ~dx } } }$$
We will now evaluate the indefinite integral.

 $$\int{ x^2\ln(1/x) ~dx }$$ $$\int{ -x^2\ln(x) ~dx }$$ Use integration by parts. $$u=\ln x \to du=(1/x)dx$$ $$dv = x^2~dx \to v=x^3/3$$ $$\displaystyle{ - \left[ \frac{x^3}{3} \ln x - \int{ \frac{x^3}{3} \frac{1}{x} ~dx } \right] }$$ $$\displaystyle{ - \frac{x^3}{3} \ln x + \frac{1}{3} \int{ x^2 ~dx } }$$ $$\displaystyle{ - \frac{x^3}{3} \ln x + \frac{1}{3} \frac{x^3}{3} }$$ $$\displaystyle{ \frac{x^3}{9} [ 1 - 3\ln x ] }$$

Now we take this result and plug it back into the limit.

 $$\displaystyle{ \lim_{a\to0^+}{ \left. \frac{x^3}{9} ( 1 - 3\ln x ) \right|_a^1 } }$$ $$\displaystyle{ \lim_{a\to0^+}{ \left[ \frac{1}{9} ( 1 - 3\ln 1 ) - \frac{a^3}{9} ( 1 - 3\ln a ) \right] } }$$ $$\displaystyle{ \left[ \frac{1}{9} - \frac{0^3}{9} ( 1 - 3\ln 0 )\right] }$$ $$\displaystyle{ \left[ \frac{1}{9} - 0 \cdot \infty \right] }$$

In the last line above, $$0 \cdot \infty$$ is indeterminate. So we have to drop back and see where they came from so that we can evaluate the limit differently. We need to look at this limit.

 $$\displaystyle{ \lim_{a\to0^+}{ a^3\ln a } }$$ To use L'Hopitals Rule, we need a fraction. $$\displaystyle{ \lim_{a\to0^+}{ \frac{\ln a}{a^{-3}} } }$$ Apply L'Hopitals Rule. $$\displaystyle{ \lim_{a\to0^+}{ \frac{1/a}{-3a^{-4}} } }$$ $$\displaystyle{ \lim_{a\to0^+}{ \frac{-1}{3} a^3 } = 0 }$$

Combining our results, we have $$1/9-0 = 1/9$$

$$1/9$$

Log in to rate this practice problem and to see it's current rating.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.