17Calculus Infinite Series - Integral Test
17Calculus
The Integral Test is easy to use and is good to use when the ratio test and the comparison tests won't work and you are pretty sure that you can evaluate the integral. The idea of this test is to evaluate the improper integral \(\displaystyle{ \int_{k}^{\infty}{f(x)~dx} }\).
If you want a complete lecture on this test, we recommend this video.
Prof Leonard - Calculus 2 Lecture 9.3: Using the Integral Test for Convergence/Divergence of Series, P-Series
video by Prof Leonard |
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Recommended Books on Amazon (affiliate links) | ||
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Integral Test |
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For a series \( \displaystyle{\sum_{n=1}^{\infty}{a_n}} \) where we can find a positive, continuous and decreasing function \(f\) for \(n > k\) and \( a_n = f(n) \), then we know that if \[ \int_{k}^{\infty}{f(x) ~ dx}\] converges, the series also converges. Similarly when the integral diverges, the series also diverges. |
Two Things To Watch For
1. The value of k
First, you need to find a constant k such that the function satisfies all of these conditions for all \( n > k \):
continuous |
positive |
decreasing |
One of the favorite tricks that teachers like to put on exams (which I fell for when I first took the class) is to tell you to use the Integral Test but then not give you k. Many books just show this integral with \( k=1 \), which is not always valid. So be careful.
How To Find k:
The best way is to calculate the critical values of the function and then check that the derivative is negative to the right of the largest critical value. Then, if you have access to a graphing calculator, do a quick plot to check your answer. If everything looks good, choose k to be greater than the largest critical value. Any value will do, so choose one that will be easy to use in the integration.
There is no one value that will always work. It depends on the function.
2. The final value of the integration
Secondly, if you get a finite value for the integral and determine that the series converges, the finite value you got from the integral is NOT what the series converges to. The number itself has no meaning in this context (ie. we don't use the value of the number to tell us anything about the series). The significance of it lies in whether it is finite or not. That's it. That's all the information you can get from that number. So do NOT assume that the series converges to that number.
Okay, let's watch some videos to see how this test works.
In this first video clip, he does a great job explaining the integral test. He uses the integral test to show the divergence of the p-series \( \sum{ 1/n } \).
Dr Chris Tisdell - Intro to series + the integral test [11min-23secs]
video by Dr Chris Tisdell |
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In this next video, the instructor explains the integral test in more detail by using it on the two series \( \sum{ 1/n } \) and \( \sum{ 1/n^2 } \) to show that one diverges and the other converges.
Dr Chris Tisdell - Integral test for Series [13min-43secs]
video by Dr Chris Tisdell |
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Here is another good explanation of the integral test. He looks at the sum \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{n^p}} }\).
PatrickJMT - Integral Test - Basic Idea [3min-26secs]
video by PatrickJMT |
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Here is a great video giving an intuitive understanding on why this works.
PatrickJMT - Integral Test for Series: Why It Works [14min-46secs]
video by PatrickJMT |
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This last video discusses the remainder estimate for the integral test. Although not required to understand how to use the integral test, this video will help you understand more intuitively what is going on.
PatrickJMT - Remainder Estimate for the Integral Test [7min-45secs]
video by PatrickJMT |
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Integral Test Proof
Here are 4 consecutive videos showing a proof of the Integral Test. You do not need to watch these in order to understand and use the Integral Test but we include them here for those of you who are interested.
Chau Tu - Integral Test Proof - Part 1 of 4
video by Chau Tu |
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Chau Tu - Integral Test Proof - Part 2 of 4
video by Chau Tu |
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Chau Tu - Integral Test Proof - Part 3 of 4
video by Chau Tu |
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Chau Tu - Integral Test Proof - Part 4 of 4
video by Chau Tu |
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Practice
Unless otherwise instructed, determine the convergence or divergence of the following series using the Integral Test, if possible.
Basic |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) using the integral test, if possible.
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) using the integral test, if possible.
Solution
This is a p-series with \( p=2 > 1\), so the series converges by the p-series test.
We could also have used the Integral Test, as follows.
\(\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2}dx} }\) |
\(\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{\frac{1}{x^2}dx}} }\) |
\(\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{x^{-2}dx}} }\) |
\(\displaystyle{ \lim_{b \to \infty}{ \left[ -x^{-1} \right]_{1}^{b}} }\) |
\(\displaystyle{ \lim_{b \to \infty}{-b^{-1} + 1^{-1}} }\) |
\( 0 + 1 = 1 \) |
Since the improper integral is finite, the series converges by the Integral Test.
Note: The value \(1\), from the integral is NOT necessarily what the series converges to. The significance of this number is only that it is finite.
Final Answer
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3} } }\) using the integral test, if possible.
Solution
blackpenredpen - 3232 video solution
video by blackpenredpen |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{(n+2)^2} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{(n+2)^2} } }\) using the integral test, if possible.
Solution
The Organic Chemistry Tutor - 3226 video solution
video by The Organic Chemistry Tutor |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{n}{\ln(n)} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{n}{\ln(n)} } }\) using the integral test, if possible.
Solution
UConnQCenter - 3230 video solution
video by UConnQCenter |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(\ln n)^2} } }\) using the integral test, if possible.
Solution
This problem is solved by two different instructors.
Chau Tu - 240 video solution
video by Chau Tu |
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Michel vanBiezen - 240 video solution
video by Michel vanBiezen |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{(n^2+1)^2} } }\) using the integral test, if possible.
Solution
This problem is solved by two different instructors.
turksvids - 241 video solution
video by turksvids |
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Dr Chris Tisdell - 241 video solution
video by Dr Chris Tisdell |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-k} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-k} } }\) using the integral test, if possible.
Solution
The Calculus Professor - 3239 video solution
video by The Calculus Professor |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ ne^{-n^2} } }\) using the integral test, if possible.
Solution
The first video solves the given problem. The second video solves a very similar problem, the same as given in the problem statement but multiplied by a constant. Since multiplication by a constant does not change the result, the solution to the second one is very similar.
PatrickJMT - 242 video solution
video by PatrickJMT |
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Trefor Bazett - 242 video solution
video by Trefor Bazett |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-2k^2} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ k e^{-2k^2} } }\) using the integral test, if possible.
Solution
The Calculus Professor - 3238 video solution
video by The Calculus Professor |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ n^2 e^{-n^3} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ n^2 e^{-n^3} } }\) using the integral test, if possible.
Solution
blackpenredpen - 3233 video solution
video by blackpenredpen |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\) using the integral test, if possible.
Final Answer |
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The series diverges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\) using the integral test, if possible.
Solution
PatrickJMT - 243 video solution
video by PatrickJMT |
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Final Answer
The series diverges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+1} } }\) using the integral test, if possible.
Solution
Krista King Math - 248 video solution
video by Krista King Math |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }\) using the integral test, if possible.
Final Answer |
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The series diverges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{n}} } }\) using the integral test, if possible.
Solution
This problem is solved by two different instructors.
blackpenredpen - 249 video solution
video by blackpenredpen |
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Educator.com - 249 video solution
video by Educator.com |
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Final Answer
The series diverges by the integral test.
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\(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n-2}} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n-2}} } }\) using the integral test, if possible.
Solution
The Organic Chemistry Tutor - 3229 video solution
video by The Organic Chemistry Tutor |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\) using the integral test, if possible.
Final Answer |
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The series diverges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\) using the integral test, if possible.
Solution
This problem is solved by two different instructors.
The Organic Chemistry Tutor - 250 video solution
video by The Organic Chemistry Tutor |
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MIP4U - 250 video solution
video by MIP4U |
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Final Answer
The series diverges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.1}} } }\) using the integral test, if possible.
Solution
MIP4U - 251 video solution
video by MIP4U |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n-2} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{5n-2} } }\) using the integral test, if possible.
Solution
MIP4U - 1213 video solution
video by MIP4U |
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\(\displaystyle{ 1 + 1/3 + 1/5 + 1/7 + 1/9 + . . . }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ 1 + 1/3 + 1/5 + 1/7 + 1/9 + . . . }\) using the integral test, if possible.
Solution
blackpenredpen - 3236 video solution
video by blackpenredpen |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n}{3n^2+4} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n}{3n^2+4} } }\) using the integral test, if possible.
Solution
The Organic Chemistry Tutor - 3227 video solution
video by The Organic Chemistry Tutor |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(\sqrt{n})}{n} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(\sqrt{n})}{n} } }\) using the integral test, if possible.
Solution
William Lindsey - 3231 video solution
video by William Lindsey |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\sqrt{\ln n}} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\sqrt{\ln n}} } }\) using the integral test, if possible.
Solution
blackpenredpen - 3235 video solution
video by blackpenredpen |
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Intermediate |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2-4n+5}}}\) using the integral test, if possible.
Solution
Although this problem would be more easily solved using one of the comparison tests, this is a great example using the integral test.
PatrickJMT - 244 video solution
video by PatrickJMT |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2+6n+10} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2+6n+10} } }\) using the integral test, if possible.
Solution
The Organic Chemistry Tutor - 3228 video solution
video by The Organic Chemistry Tutor |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n} } }\) using the integral test, if possible.
Solution
PatrickJMT - 245 video solution
video by PatrickJMT |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n[(\ln n)^2+4]} } }\) using the integral test, if possible.
Solution
PatrickJMT - 246 video solution
video by PatrickJMT |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^4+1} } }\)
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^4+1} } }\) using the integral test, if possible.
Solution
blackpenredpen - 3234 video solution
video by blackpenredpen |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\) using the integral test, if possible.
Final Answer |
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The series diverges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4}{n\sqrt[3]{n}}+\frac{5}{n}\right] } }\) using the integral test, if possible.
Solution
PatrickJMT - 247 video solution
video by PatrickJMT |
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Final Answer
The series diverges by the integral test.
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\) using the integral test, if possible.
Final Answer |
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The series converges by the integral test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\) using the integral test, if possible.
Solution
This problem is solved by two different instructors.
turksvids - 252 video solution
video by turksvids |
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MIP4U - 252 video solution
video by MIP4U |
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Final Answer
The series converges by the integral test.
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\(\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }\)
Problem Statement |
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Determine the convergence or divergence of the series \(\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }\) using the integral test, if possible.
Final Answer |
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The series diverges by the Integral Test.
Problem Statement
Determine the convergence or divergence of the series \(\displaystyle{ \sum_{k=2}^{\infty}{ \frac{2}{k\ln k} } }\) using the integral test, if possible.
Solution
Math Cabin - 2443 video solution
video by Math Cabin |
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Final Answer
The series diverges by the Integral Test.
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Advanced |
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Find \(p\) so that \(\displaystyle{ \sum_{n=1}^{\infty}{ n(1+n^2)^p } }\) will converge.
Problem Statement
Find \(p\) so that \(\displaystyle{ \sum_{n=1}^{\infty}{ n(1+n^2)^p } }\) will converge.
Solution
blackpenredpen - 3237 video solution
video by blackpenredpen |
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integral test 17calculus youtube playlist
Really UNDERSTAND Calculus
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Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Integral Test Quick Notes
used to prove convergence |
yes |
used to prove divergence |
yes |
can be inconclusive |
yes |
\(a_n\) must be positive | |
\(a_n\) must be decreasing | |
requires that the integrand must be integrable (not always possible) | |
requires the evaluation of infinite limits (after integration) | |
if the result of the limit (after integration) does not exist (different than diverges), this test is inconclusive | |
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