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17Calculus - Improper Integrals - Endpoint Discontinuity

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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.

Final Answer

\(4\)

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

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

\(\displaystyle{ \int{\frac{dx}{\sqrt{2x}}} }\)

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

\(\displaystyle{ \frac{1}{\sqrt{2}} \frac{x^{1/2}}{1/2} }\)

\(\displaystyle{ \sqrt{2x} }\)

\(\displaystyle{ \lim_{a\to 0^+}{ \left[ \sqrt{2x} \right]_a^8 } }\)

\(\displaystyle{ \lim_{a\to 0^+}{ [ \sqrt{16} - \sqrt{2a} ] } = 4 }\)

Final Answer

\(4\)

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

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

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

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.

Final Answer

\(\pi/2\)

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 \(x=3\sin\theta \to dx = 3\cos\theta~d\theta\).
\(\displaystyle{ \int{ \frac{3\cos\theta~d\theta}{\sqrt{9-9\sin^2\theta}} } = }\) \(\displaystyle{\int{ \frac{\cos\theta}{\cos\theta}~d\theta } = }\) \(\displaystyle{ \int{1~d\theta} = \theta = \arcsin(x/3) }\)
\(\displaystyle{ \lim_{b\to3^-}{ [ \arcsin(x/3)|_0^b ] } = }\) \(\displaystyle{ \lim_{b\to3^-}{ [ \arcsin(b/3) - \arcsin(0/3) ] } = }\) \(\arcsin 1 = \pi/2 \)

Final Answer

\(\pi/2\)

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

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.

Final Answer

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.

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

\(\displaystyle{ \lim_{a \to 3^+}{ [-2(4-3)^{-1/2} + 2(a-3)^{-1/2}] } }\)

\( -2 + \infty = \infty \)

Final Answer

The integral diverges.

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

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.

Final Answer

\(3\)

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.

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

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

\(\displaystyle{ \frac{3}{2} [(8)^{1/3} - (2(-3)+6)^{1/3}] = 3 }\)

Final Answer

\(3\)

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

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.

Final Answer

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.

\(\displaystyle{ \lim_{b\to\pi/2^-}{ \int_{0}^{b}{ \tan\theta ~ d\theta } } }\)

\(\displaystyle{ \lim_{b\to\pi/2^-}{ \left[ - \ln|\cos\theta| \right]_0^b } }\)

\(\displaystyle{ \lim_{b\to\pi/2^-}{ \left[ -\ln|\cos b| + \ln|\cos 0| \right] } }\)

\( -\ln|\cos \pi/2| + \ln|1| \)

\( -\ln|0| + 0 = +\infty \)

Final Answer

The integral diverges.

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

Final Answer

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.

Final Answer

The integral diverges.

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

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.

Final Answer

\( 1/9 \)

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 \(1/9-0 = 1/9\)

Final Answer

\( 1/9 \)

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