# 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.

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LetsSolveMathProblems6 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.

adandap6 months ago^{+1}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 Jelo6 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 singh6 months agoK = -2015

IlTrojo6 months agoWolframalpha unexpectedly manages it: bit.ly/2Qtb3Fd.

Peter6 months agoI 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)!

LetsSolveMathProblems6 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 Gupta6 months agoOn 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

LetsSolveMathProblems6 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.

LetsSolveMathProblems6 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.

attyfarbuckle6 months agoIt's a quadratic in k. You're going to need 2 values matey.!

Evyatar Baranga6 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

LetsSolveMathProblems6 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 Zox6 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 Zox6 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

LetsSolveMathProblems6 months agoJia Ming6 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 Ming6 months ago^{+1}@LetsSolveMathProblems Uwuwu don't worry hahah it's all good and fun

LetsSolveMathProblems6 months ago^{+1}adandap6 months ago^{+1}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.

LetsSolveMathProblems6 months agoINFINITE MEMORY6 months agoWell explaind

Siddharth Sambamoorthy6 months agoI 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.

LetsSolveMathProblems6 months agoadandap6 months agoCurse 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... :)

adandap6 months ago@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'.)

LetsSolveMathProblems6 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?

adandap6 months ago^{+1}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

Aryan6 months ago^{+1}answer is K=2034 .

Parth Pawar6 months agok=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 p6 months agoBy 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 p6 months agoLetsSolveMathProblems 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

LetsSolveMathProblems6 months agoWith 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 p6 months agoLetsSolveMathProblems that’s amazing I would like to meet you in person someday

aby p6 months ago^{+1}LetsSolveMathProblems do you make these questions on your own or do you get it from somewhere???

aby p6 months agoBrian 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 Salinas6 months agoWtf dude lol

Nitro Zox6 months agoThis is the correct way to do it. 😶

Jonathan Liu6 months ago^{+1}brilliant logic and reasoning

Parth Pawar6 months agothis is crazy

LetsSolveMathProblems6 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 Joshi6 months ago^{+5}Why calculus :'( IDK Calculus yet...

Vincent William Rodriguez6 months ago^{+1}Learn it

el tapa6 months agoMaybe trig sub? Or maybe by parts? Hmmm

Nicholas Patel6 months agoI 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

LetsSolveMathProblems6 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.

LetsSolveMathProblems6 months agoParlyne6 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)).

LetsSolveMathProblems6 months agoNo problem! I'm impressed you found k purely by intuitive reasoning.

Nicholas Patel6 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 Wan6 months ago^{+3}Second

Joshua Cohen6 months ago^{+3}First