Challenge 71: Integral Squared is Eight

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• Published on Dec 13, 2018
• * The problem should state, "Find the largest possible value of k," not "Find all values of k." I sincerely apologize for the error.
Congratulations to PRAKHAR AGARWAL, Nicholas Patel, Hiren Bavaskar, santosh tripathy, fmakofmako, Nachiket Bhagade, Mr L, and Allaizn for successfully solving the last week's math challenge question! PRAKHAR AGARWAL was the first person to solve the question.
Your support is truly a huge encouragement.
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Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.
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• LetsSolveMathProblems  6 months ago +4

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Although finding all values of k is possible, doing so is much more difficult than the former and requires a calculator. You have my sincere apologies; I was not thinking properly when I was transferring this problem from my notebook to computer. Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted.

• Of course there's a second solution for a large -ve value of k. D'oh! Very obvious now. Mathematica says it's at -6091.98

• Kostas Jelo 6 months ago +1

ANS: K=2034.
We break the integral and get (kA+2019B)^2=8.
Int by parts in A gives 2034A+2019B=√8 => 2034A +(kA +2019B)-kA=√8 => (2034-k)A=√8 ± √8. We have obious answer k=2034. For the second value of k we will need the first integral to be negative. That happens IFF k is negative since all the other terms are positive and the powers of x even. We care for the biggest value of k and since we have a positive one we found it.

• sachin singh 6 months ago

K = -2015

• IlTrojo 6 months ago

Wolframalpha unexpectedly manages it: bit.ly/2Qtb3Fd.

• Peter 6 months ago

I found two solutions for k = 2034 and -6091.975388 to 6 decimal places.
By ignoring the first half of the integral we can use integration by parts on the integrand k*x^2028*Sqrt(1+x^10). We can integrate x^9*Sqrt(1+x^10) and differentiate the remaining k*x^2019. By evaluating this and rewriting the new integral found we find that the integral of the aforementioned k*x^2028*Sqrt(1+x^10) between 0 and 1 is equal to k*Sqrt(2)/1017 - (k/2034)*A where A is the same integral as the first half of the original integrand (2019*x^2018*sqrt(1+x^10) between 0 and 1).
Some rearranging gives [(1-k/2034)*A + k*sqrt(2)/1017]^2 = 8. We can take the square root of both sides such that the left expression is equal to 2*Sqrt(2) or -2*sqrt(2). Solving each equation separately we see that they are linear equations in terms of k where A (the previously mentioned integral) is unknown. We can deduce that A > 0 as the integrand involves only coefficients of x^2n, but I know of no methods to solve this integral for a closed form.
We see that in the first equation where we want the expression to equal 2*sqrt(2), that when k=2034, the expression becomes 0*A + 2034*sqrt(2)/1017 which is exactly 2*sqrt(2) as required. So k = 2034 is a solution.
For the negative value of k we can use an approximation for the integral A to find another value of k. This gets a value of k ≈ -6092 as previously mentioned.
As a side note evaluating A gives ≈ 1.412468917 which is astoundingly close to sqrt(2)!

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• Rishav Gupta 6 months ago

On resolving we get the integral inside=
2019×(X^2018(y^3/2-(n-1)(x^10)(y^1/2)) dx
Where n=k/2019
And y=(x^10+1)^1/2
We see if we differentiate k(x)=x^2019×y^3/2
We get similar form as above
So if we consider above integral with
Differnentiation of k(x)
We get that
2019(n-1)=15
=k-2019=15
=k=2034

• LetsSolveMathProblems  6 months ago

@attyfarbuckle * I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• attyfarbuckle 6 months ago

It's a quadratic in k. You're going to need 2 values matey.!

• Evyatar Baranga 6 months ago +1

K = 2034
Trying to u sub, i found the anti derivtive of the function: x^2019*(1+x^10)^1.5
Its derivative is:
x^2018(1+x^10)^0.5*(2019+ 2034 *x^10)
And if you check its values at 1 and 0 you realy get that it is sqrt(8)
And thus k = 2034

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• Nitro Zox 6 months ago +2

Ok, I think this is a correct solution. Let L = x^2018(2019+kx^10)*sqrt*(1+x^10)
Now, as we cannot remove a root from this equation, we need a way so that we get rid of x^2018. So we split x^2018 into 2 parts, x^p and x^2018-p. First divide both sides by 2019 to get
L/2019 = x^2018(1+ nx^10)sqrt(1+x^10) where n = k/2019. Multiply x^p to first bracket and take x^2018-p inside the sqrt. Inside , it will become x^(4036-2p) . 1st bracket becomes :
x^p + nx^p+10. Inside the sqrt we get : x^4036-2p + x^4046-2p. Now we need the 1st bracket to be the derivative of the term in the sqrt otherwise the integral will be not solvable. Hence the derivative of term inside sqrt is (4036-2p)x^4035-2p + (4046-2p)x^4045-2p . Now compare the coefficient of x^4035-2p and x^p. They must be equal for outside term to be the derivative. Hence 4035-2p = p=> p = 1345. Hence inside sqrt we get x^1346 + x^1356. Let u = x^1346 + x^1356. ; du = 1346x^1345 + 1356x^1355. Divide both sides by 1346 to get du/1346 = x^1345 + 1356/1346 (x^1355) . Now, see the 1st bracket which was x^p + nx^p+10 which is x^1345 + nx^1355. But remember as this is the derivative, coefficient of x^1355 is equal. So , n = 1356/1346 => k/2019 = 1356/1346 and hence k = 2034. Also elegantly on putting values , L = 2sqrt2 which implies L^2 = 8

• Nitro Zox 6 months ago

@LetsSolveMathProblems What I found intriguing was that even if the value of the integral wasn't given, I think we can reach k = 2034

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• Jia Ming 6 months ago +1

Let *I = int_0^1 x^2018 sqrt(1+x^10) dx*
and *J = int_0^1 x^2028 sqrt(1+x^10) dx*
Using by parts on J, where u = x^2019 and v' = x^9 sqrt(1+x^10),
and so, u' = 2019x^2018 and v = 1/15 (1+x^10)^(3/2),

We get J = sqrt(8)/15 - 2019/15 * int_0^1 x^2018(1+x^10)^(3/2) dx.
We can split (1+x^10)^(3/2) into (1+x^10)sqrt(1+x^10) and simplify the above into
J = sqrt8 /15 - 2019I/15 -2019J/15,
2034J/15 = sqrt8 / 15 -2019I/15,

and thus, *2019I + 2034J = sqrt8*
Which gives us an immediate solution of *k = 2034
*
But we're still missing one more solution,
Note that J = (2sqrt2 -2019I) /2034,
And let 2019I + kJ = -sqrt8.

Solving for k gives us an exact value in terms of I: *k = (2019I + sqrt8) / (2019I - sqrt8) * 2034*
And since I is approximately 0.000699588369,
The other solution is approximately *k = -6091.9754*

• Jia Ming 6 months ago +1

@LetsSolveMathProblems Uwuwu don't worry hahah it's all good and fun

• LetsSolveMathProblems  6 months ago +1

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• In one of my failed attempts, I noted that integrating by parts reduced the power of (1+x^10) by unity, which would allow me to write Integral(x^a (1+x^10)^3/2 in terms of x^(a+10) (1+x)^1/2, which corresponds to the 'k' term in the integral. Then use 2019 + k x^10 = 2019 (1+x^10) + (k-2019)x^10 and use parts on the first two terms to get I = 2^(3/2) +(k-2034) * (a postitive definite quantity). If I^2 = 8, then k=2034.

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• INFINITE MEMORY 6 months ago

Well explaind

• Siddharth Sambamoorthy 6 months ago

I got k=2034
This is obtained by simplifying the given integral to the form of a trigonometric integral in the form of sines and cosines which could be evaluated to find k.However,the integral is not a simple one, so to save time i solved it numerically.

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• Curse you let's solve math problems guy. I'm knee deep in hypergeometric functions and - not for the first time - have a sneaking suspicion I'm not doing this the easiest way... :)

• @LetsSolveMathProblems Done - apologies. By now I have a hunch you'd almost expect "the old physics guy" to be the one who blunders head first into hypergeometrics rather than high school calculus! You got me to dig out my copy of Abramowitz and Stegun from under the dust and spiders. (And BTW, my hypergeometric fn relationship is actually between 2F1(-1/2, a, a+1, -1) and 2F1(-1/2, a+1, a+2, -1) - sort of 'two step contiguous functions'.)

• LetsSolveMathProblems  6 months ago +1

I knew someone would try utilizing hypergeometric functions. I'm glad you got the answer in the end. For this particular problem, I do remark that there is actually a very elementary solution that only requires the u-substitution, but it is not easy to see.

Having stated that, I remind you that reply comments will not be accepted. Could you copy-and-paste your reply to an entirely new comment?

• OK, got it now, but by going back to basics. In one of my failed attempts, I noted that integrating by parts reduced the power of (1+x^10) by unity, which would allow me to write Integral(x^a (1+x^10)^3/2 in terms of x^(a+10) (1+x)^1/2, which corresponds to the 'k' term in the integral. Then use 2019 + k x^10 = 2019 (1+x^10) + (k-2019)x^10 and use parts on the first two terms to get I = 2^(3/2) +(k-2034) * (a postitive definite quantity). If I^2 = 8, then k=2034.
Incidentally, my failed attempt lets me deduce a relationship between 2F1(-1/2,a,b,-1) and 2F1(-1/2,a+1,b+1,-1) which doesn't seem to be in my standard references. Of course, it also doesn't seem to be very useful. :-D

• Aryan 6 months ago +1

• Parth Pawar 6 months ago

k=2034
Substitute u=x^2019(1+x^10)^3/2 and then we get du same as the integrand if k=2034, which on after putting limits does give us 8

• aby p 6 months ago

By sheer brute force I have got the ans in the first step in added 2019x^10 and subtracted inside the bracket then I took 2019 common with the plus sign and separated the integral the first integral becomes x^2018 2019(1+x^10) we integrate this by parts and we get the ans as 2root2 the other integral converts into (K-2034) multiplied by other terms so for the ans to come as 2root2 we must get k-2034=0 therefore k=2034 is the ans.

• aby p 6 months ago

LetsSolveMathProblems I really want to know how do you think to make such good problems that’s the reason I want to meet you someday in person

• LetsSolveMathProblems  6 months ago

With the exception of approximately 12 problems (out of 69 so far), all Weekly Math Challenge problems--including this one--were made by myself. The exceptions owe their existence to my younger brother or both of us working in conjunction.

• aby p 6 months ago

LetsSolveMathProblems that’s amazing I would like to meet you in person someday

• aby p 6 months ago +1

LetsSolveMathProblems do you make these questions on your own or do you get it from somewhere???

• aby p 6 months ago

Brian talks about I don’t do mathematics for money I do it because of my love for this subject just never ends. Mathematics keeps me in awe because after every problem i solve I get to think of new ideas and mathematics is one the most unpredictable subjects.

• Smokie Bear 🔴🔵 6 months ago +14

8 is the same thing as 2 to the power of 3, which means 2 of the digits for the value of k are ‘2’ and ‘3’. Any integral gives areas in two dimensions. So since we are squaring it, then we get four dimensions, so another digit is 4. Then, since the acceleration is 0 for a mass moving by a spring with spring constant ‘k’ when the displacement is 0, another digit is ‘0’. so our digits are 2,3,4,0. and rearranging this we get k=2034. The mitochondria is the powerhouse of the cell

• Fabian Salinas 6 months ago

Wtf dude lol

• Nitro Zox 6 months ago

This is the correct way to do it. 😶

• Jonathan Liu 6 months ago +1

brilliant logic and reasoning

• Parth Pawar 6 months ago

this is crazy

• LetsSolveMathProblems  6 months ago +9

I cannot accept your answer because you have not proven that the displacement of the spring is 0 in this particular configuration.

• Prathmesh Joshi 6 months ago +5

Why calculus :'( IDK Calculus yet...

• el tapa 6 months ago

Maybe trig sub? Or maybe by parts? Hmmm

• Nicholas Patel 6 months ago

I got k=2034
First notice that when we differentiate,wrt x, x^2019(1+x^10)^3/2, we get a very similar expression to the given integral where the coefficient of x^10 in the middle bracket is 2034, so maybe k=2034. Testing this value of k gives that the integral evaluates to 2^1.5, so it squared gives 8 as desired and this k works. I don’t think any other k will work for the integral to evaluate to +-2sqrt2,, since I think what I found is the only antiderivative that keeps everything else the same/matches everything, so this is the only value

• LetsSolveMathProblems  6 months ago

@Parlyne * I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• LetsSolveMathProblems  6 months ago

* I just realized that I meant to state "Find the largest possible value of k," NOT "Find all values of k." Any explanations with the answer k = 2034, even if the smaller value of k is not found, will be accepted, and do not need to be edited. I sincerely apologize for the error.

• Parlyne 6 months ago +1

But, it's not unique. The equation can be broken up into the form (I_1 + k I_2)^2 = 8 (where I_1 and I_2 are integrals to be evaluated), which is clearly quadratic in k and, by subtracting 8 from both sides and using difference of squares, this can be recast as (I_1 + k I_2 + 2\sqrt{2})*(I_1 + k I_2 - 2\sqrt{2}) = 0. This clearly means there must be two solutions for k (given that the integrands of both integrals are positive and finite on (0,1)).

• LetsSolveMathProblems  6 months ago

No problem! I'm impressed you found k purely by intuitive reasoning.

• Nicholas Patel 6 months ago +1

LetsSolveMathProblems I completely agree that the explanation is not backed up or proved, so thank you for accepting the solution still :)

• Obi Wan 6 months ago +3

Second

• Joshua Cohen 6 months ago +3

First