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โ™พ๏ธap calculus ab/bc review

6.13 Evaluating Improper Integrals

Verified for the 2025 AP Calculus AB/BC examโ€ขCitation:

Remember earlier in this unit when you had to evaluate definite integrals? Now, imagine trying to solve a definite integral with one of the boundaries, instead of being an integer, being to infinity. How would you go about doing that? Today, weโ€™ll learn about improper integrals and how to deal with these cases. ๐Ÿš€

๐Ÿšง This is an AP Calculus BC topic only! If you are taking Calculus AB, you can skip this material. If youโ€™re taking AP Calculus BC, here you go! โฌ‡๏ธ


๐ŸŸฅ Recap: Definite Integrals

When coming across integration problems, you will encounter one of the following: an indefinite integral or a definite integral. Unlike indefinite integrals, which do not have boundaries, definite integrals have an upper and lower bound that can be plugged into your antiderivative to get a numerical value. Here are some examples of typical definite integrals you may have encountered in AP Calculus AB:

โˆซ0ฯ€sin(x)dx=โˆ’cos(ฯ€)+cos(0)=2โˆซ^ฯ€_0sin(x)dx=-cos(ฯ€)+cos(0)=2 โˆซ02eโˆ’xdx=โˆ’eโˆ’2+e0=โˆ’1e2+1โˆซ^2_0e^{-x}dx=-e^{-2}+e^0=-\frac{1}{e^2}+1

โ™พ๏ธ Evaluating Improper Integrals

An improper integral occurs when the limits of integration involve infinity or when the function being integrated becomes unbounded within the integration interval. This basically means that in the interval being evaluated, the function becomes unbounded to infinity.

In order to properly evaluate them, we must go back to Unit 1 and use limits. Limits allow for the evaluation of integral expressions where one or both of the boundaries are unbounded. When taking the limit of an improper integral, set the unbounded boundary to approach a letter and then take the limit from the lower bound to the upper bound:

โˆซ0โˆžf(x)dx=limโกbโ†’โˆžโˆซ0bf(x)dxโˆซ^โˆž_0f(x)dx=\lim_{b\toโˆž}โˆซ^b_0f(x)dx

Here are some steps to follow when evaluating improper integrals:

  1. Identify the Issue ๐Ÿค”: Determine whether the improper integral has one or both limits extending to infinity or if the integrand is unbounded within the integration interval.

  2. Express as a Limit โœ๏ธ : If one or both limits are infinite, express the improper integral as a limit. For example, if integrating from a to infinity, rewrite it as the limit as b approaches infinity of the integral from a to b.

  3. Evaluate the Integral ๐Ÿ“: Integrate the function as you normally would and apply the First Fundamental Theorem of Calculus:

The First Fundamental Theorem of Calculus. Image Courtesy of Calcworkshop
  1. Evaluate the Limit ๐Ÿ“: Now, evaluate the limit that you've introduced in step 2. This involves considering what happens to the integral as the bound approaches its infinite or unbounded value.
  2. Check for Convergence ๐Ÿ: Determine whether the limit converges or diverges. If the limit approaches a finite value, the improper integral converges; if it approaches infinity or doesn't exist, it diverges.

๐Ÿง  Improper Integrals Example 1

โˆซ011xdxโˆซ^1_0\frac{1}{\sqrt{x}}dx

Step 1) Identify the Issue ๐Ÿค”:

In the integral above, the graph is unbounded at x=0x=0.

Step 2) Express as a Limit โœ๏ธ:

limโกaโ†’0โˆซa11xdx\lim_{a\to0}โˆซ^1_a\frac{1}{\sqrt{x}}dx

Step 3) Evaluate the Integral ๐Ÿ“:

limโกaโ†’0โˆซa1xโˆ’12dx\lim_{a\to0}โˆซ^1_ax^{-\frac{1}{2}}dx limโกaโ†’0[2xโˆฃa1]\lim_{a\to0}[2\sqrt{x}|^1_a] limโกaโ†’0[2โˆ’2a]\lim_{a\to0}[2-2\sqrt{a}]

Step 4) Evaluate the Limit ๐Ÿ“:

limโกaโ†’0[2โˆ’2a]\lim_{a\to0}[2-2\sqrt{a}] 22

Step 5) Check for Convergence ๐Ÿ:

The limit approaches 2, a finite number, which means the improper integral converges at 2.

Screenshot 2024-01-20 121928.png
Graph Displaying f(x)=1/sqrt(x)f(x)=1/sqrt(x). Image Courtesy of Desmos

๐Ÿง  Improper Integrals Example 2

Find the area of the function y=ex4+e2xy=\frac{e^x}{4+e^{2x}} from โˆ’โˆž-โˆž to โˆžโˆž or state that it converges.

Screenshot 2024-01-20 122104.png
Graph displaying the function yy is equal to e2/4+e2xe^2/4+e^2x. Image Courtesy of Desmos A=โˆซโˆ’โˆžโˆžex4+e2xdxA=โˆซ^โˆž_{-โˆž}\frac{e^x}{4+e^{2x}}dx

Since both the upper and lower boundaries are unbounded, we must split the integral into two by choosing an arbitrary point in between the bounds. By choosing lnโก2\ln2 (the global maximum of the function), we make it so both to the left and the right of lnโก2\ln2 are equal in area, so we only have to manipulate one side of the integral:

=2limโกbโ†’โˆžโˆซln2bex4+e2xdx=2\lim_{b\toโˆž}โˆซ^b_{ln2}\frac{e^x}{4+e^{2x}}dx u=exu=e^x u2=e2xu^2=e^{2x} 2udu=2e2xdx2udu=2e^{2x}dx dx=duexdx=\frac{du}{e^x} =2limโกbโ†’โˆžโˆซln2b14+u2du=2\lim_{b\toโˆž}โˆซ^b_{ln2}\frac{1}{4+u^2}du =2limโกbโ†’โˆž[12arctanโก(ex2)โˆฃlnโก2b]=2\lim_{b\toโˆž}[\frac{1}{2}\arctan(\frac{e^x}{2})|^b_{\ln2}] =limโกbโ†’โˆž[arctanโก(eb2)โˆ’arctanโก(elnโก22)]=\lim_{b\toโˆž}[\arctan(\frac{e^b}{2})-\arctan(\frac{e^{\ln2}}{2})] =ฯ€2โˆ’ฯ€4=\frac{ฯ€}{2}-\frac{ฯ€}{4} =ฯ€4=\frac{ฯ€}{4}

So, the area of this unbounded shaded region is ฯ€4\frac{ฯ€}{4}. Great job! ๐Ÿค—


๐Ÿ–Š๏ธ 2017 AP Calc Exam #5c Walkthrough

The following free-response question (FRQ) is Question 5 (c) from the 2017 AP Calculus BC Examination administered by College Board. All credit to College Board.

Let ff be the function defined byโ€ฆ

f(x)=32x2โˆ’7x+5f(x)=\frac{3}{2x^2-7x+5}

Using the following identity, evaluate โˆซ5โˆžf(x)dxโˆซ^โˆž_5f(x)dx or show that the integral diverges.

32x2โˆ’7x+5=22xโˆ’5โˆ’1xโˆ’1\frac{3}{2x^2-7x+5}=\frac{2}{2x-5}-\frac{1}{x-1}

Letโ€™s go through it! โฌ‡๏ธ

To solve this question, all we must do is express the integral as a limit and determine whether or not it diverges.

limโกbโ†’โˆžโˆซ5bf(x)dx\lim_{b\toโˆž}โˆซ^b_5f(x)dx

Because the partial fraction is already given to us, all we must do is substitute it into our integrand and solve normally:

limโกbโ†’โˆžโˆซ5b[22xโˆ’5โˆ’1xโˆ’1]dx\lim_{b\toโˆž}โˆซ^b_5[\frac{2}{2x-5}-\frac{1}{x-1}]dx limโกbโ†’โˆž[lnโกโˆฃ2xโˆ’5โˆฃโˆ’lnโกโˆฃxโˆ’1โˆฃ]5b\lim_{b\to\infin}[\ln|2x-5|-\ln|x-1|]^b_5 limโกbโ†’โˆž[lnโกโˆฃ2bโˆ’5โˆฃโˆ’lnโกโˆฃbโˆ’1โˆฃโˆ’lnโก5+lnโก4]\lim_{b\toโˆž}[\ln|2b-5|-\ln|b-1|-\ln5+\ln4] lnโก2โˆ’lnโก(54)=lnโก(85)\ln2-\ln(\frac{5}{4})=\ln(\frac{8}{5})

Awesome work! ๐Ÿ‘


โญ๏ธ Closing

Well done! You have now learned how to successfully evaluate improper integrals and distinguish whether or not the area of these integrals converge or diverge. Make sure to keep practicing and improving your skills. ๐Ÿ˜Š

Key Terms to Review (15)

Antiderivative: An antiderivative is essentially the reverse process of differentiation. It refers to finding a function whose derivative matches a given function.
Comparison Test: The comparison test is used to determine whether an infinite series converges or diverges by comparing it with another known series whose convergence behavior is already established.
Constant of Integration: The constant of integration is an arbitrary constant added when finding indefinite integrals or antiderivatives. It accounts for the fact that there are infinitely many functions with the same derivative.
Constraining the Interval of Integration: Constraining the interval of integration refers to limiting the range over which a definite integral is evaluated. It involves specifying the lower and upper bounds within which the function is integrated.
Convergent Function: A convergent function is a function that approaches a specific value as the input approaches a certain value or infinity.
Improper Integral Tables: Improper integral tables are reference tables that provide formulas and solutions for evaluating improper integrals. They are useful tools for quickly finding solutions without having to go through lengthy calculations each time.
Improper Integrals: Improper integrals are definite integrals where one or both of the limits of integration are infinite, or the function being integrated is unbounded within the interval. They require special techniques to evaluate.
Improper Indefinite Integrals: Improper indefinite integrals involve integrating functions that have singularities within their domain or functions whose antiderivatives are not defined over their entire domain.
Improper Definite Integrals: Improper definite integrals are integrals with infinite limits of integration or integrands that become unbounded within the interval of integration.
Integrand: The term "integrand" refers to the function being integrated within an integral. It represents what we want to find the antiderivative of.
Integrating by Parts: Integrating by parts is a technique used to evaluate integrals involving products of functions. It involves breaking down the integral into two parts and applying a specific formula to simplify the integration process.
L'Hรดpital's Rule: L'Hรดpital's Rule is a method used to evaluate limits of indeterminate forms, such as 0/0 or โˆž/โˆž. It involves taking the derivative of the numerator and denominator separately and then evaluating the limit again.
Limits of Integration: The limits of integration are the values that determine the range over which an integral is evaluated. They specify the starting and ending points on the x-axis for finding the area or calculating other quantities using integration.
Singularity: A singularity is a point in a function where it becomes undefined or approaches infinity. It is a point of discontinuity.
Trigonometric Substitution: Trigonometric substitution is a technique used in calculus to simplify integrals involving radical expressions or quadratic forms. It involves substituting trigonometric functions for variables in order to transform the integral into a more manageable form.