Beat the Machines!

An advertisement for a symbolic algebra program claims that an engineer worked for three weeks on the integral ∫(kln⁡x−2x3+3x2+b)4 dx\int {(k \ln x-2x^3+3x^2+b)^4 \text{ d}x} which deals with turbulence in an aerospace application. The advertisement said that the engineer never got the same answer twice in three weeks.

  1. Give two reduction formulas that will be of use in finding the above integral. Establish each formula.

  2. Explain (only) how you could use those two reduction formulas in finding the integral, i.e. what is your strategy?

  3. Challenge! With your strategy in #2, compute the engineer's integral by hand. You may comment only the final answer here, as the solution is very very lengthy.

  4. Compare your answer with those produced by the following computer algebra systems. Use k=b=1k=b=1. (Links to them are provided for easy access. More computer algebra systems will be added here later):

  5. Is there any reason for the engineer to take three weeks?

[The problem information was taken from Edwards and Penney, Calculus 6th EditionCalculus\text{ }6th\text{ }Edition.]

#Calculus #Integration #ComputerAlgebraSystems #ReductionFormulas

Note by Jaydee Lucero
6 years, 10 months ago

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1 vote

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Comments

That's interesting.

I'm working on it.

Kishlaya Jaiswal - 6 years, 2 months ago

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Yup! It is indeed interesting . But I am more interested in guessing the train of thoughts in that engineer's mind .

It's always fascinating to find(deduce) what someone else thinks :D

A Former Brilliant Member - 6 years, 2 months ago

b4−4b3k+12b2k2−24bk3+24k4)x+49(9b3−9b2k+6bk2−2k3)x3+116(−32b3+24b2k−12bk2+3k3)x4+54125(25b2−10bk+2k2)x5−23(18b2−6bk+k2)x6+12343(98b2−7b(−63+4k)+k(−63+4k))x7−278(8b−k)x8+19(81+144b−16k)x9−425(135+20b−2k)x10+(216x11)11−8x12+(16x13)13−(114700)kx(−58800b3+352800k3+1225k2x2(−32+9x)+48kx4(1323−1225x+300x2)+88200b2(2k+(−2+x)x2)+420x6(−540+945x−560x2+112x3)−420b(840k2+35kx2(−8+3x)+12x4(63−70x+20x2)))Log[x]+170k2x(420b2+840k2+35kx2(−8+3x)+12x4(63−70x+20x2)−420b(2k+(−2+x)x2))Log[x]2−2k3x(−2b+2k+(−2+x)x2)Log[x]3+k4xLog[x]4b^4 - 4 b^3 k + 12 b^2 k^2 - 24 b k^3 + 24 k^4) x + \frac{4}{9} (9 b^3 - 9 b^2 k + 6 b k^2 - 2 k^3) x^3 + \frac{1}{16} (-32 b^3 + 24 b^2 k - 12 b k^2 + 3 k^3) x^4 + \frac{54}{125} (25 b^2 - 10 b k + 2 k^2) x^5 - \frac{2}{3} (18 b^2 - 6 b k + k^2) x^6 + \frac{12}{343} (98 b^2 - 7 b (-63 + 4 k) + k (-63 + 4 k)) x^7 - \frac{27}{8} (8 b - k) x^8 + \frac{1}{9} (81 + 144 b - 16 k) x^9 - \frac {4}{25} (135 + 20 b - 2 k) x^10 + \frac{(216 x^11)}{11} - 8 x^12 +\frac{ ( 16 x^13)}{13} - (\frac{1}{14700}) k x (-58800 b^3 + 352800 k^3 + 1225 k^2 x^2 (-32 + 9 x) + 48 k x^4 (1323 - 1225 x + 300 x^2) + 88200 b^2 (2 k + (-2 + x) x^2) + 420 x^6 (-540 + 945 x - 560 x^2 + 112 x^3) - 420 b (840 k^2 + 35 k x^2 (-8 + 3 x) + 12 x^4 (63 - 70 x + 20 x^2))) Log[x] + \frac{1}{70} k^2 x (420 b^2 + 840 k^2 + 35 k x^2 (-8 + 3 x) + 12 x^4 (63 - 70 x + 20 x^2) - 420 b (2 k + (-2 + x) x^2)) Log[ x]^2 - 2 k^3 x (-2 b + 2 k + (-2 + x) x^2) Log[x]^3 + k^4 x Log[x]^4

I think this is the answer as long as I didn't do any mistake in arithmetic calculations.

Akshay Bodhare - 6 years, 2 months ago
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