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Evaluate $$\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $$

The question look's like there is a nice method to do it, but I can't figure out. Can someone provide some hint or answer?.

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  • $\begingroup$ Split into odd function and an even function? $\endgroup$ – Xiaolang Apr 11 '13 at 15:55
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    $\begingroup$ There is indeed a nice method, but not so easy to see! $\endgroup$ – Jean-Claude Arbaut Apr 11 '13 at 16:04
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    $\begingroup$ You can assume the answer is $f(x) (4x^3 + 5x^2 + 10)^{3/2}$ and try to solve for $f(x)$. The answer magically comes out to be simple (I believe $\frac{x^3}{30}$) for these specific values. But I don't expect this to work in general (hence just a comment). Perhaps there is a trick with definite integrals which works. Where did you get this integral from? $\endgroup$ – Aryabhata Apr 11 '13 at 16:32
  • $\begingroup$ @Aryabhata, actually this is a multiple choice question. The actual question is .The value of the integral is $a\cdot 19 \sqrt{19}$ Solve for $a$ , The options given were bad(We can easily solved for the range of answers and only one option suits it). But, can you please tell me what motivated you for such a method. $\endgroup$ – jdoicj Apr 11 '13 at 16:39
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    $\begingroup$ It is just a guess. Since you see $\sqrt{Q(x)}$, you naturally try to use $Q(x)^{3/2}$. Just doing that does not work, so you try a multiplicative factor, which as it turns out, works. $\endgroup$ – Aryabhata Apr 11 '13 at 16:47
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$$\begin{align} \int_0^1(x^5+x^4+x^2)\sqrt{4x^3+5x^2+10}dx &=\int_0^1(x^4+x^3+x)\sqrt{4x^5+5x^4+10x^2}dx\cr &={1\over20}\int_0^{19}\sqrt{u}du={19^{3/2}\over30}\cr \end{align}$$

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  • $\begingroup$ The is the best one !!!. $\endgroup$ – Felix Marin Sep 19 '13 at 20:55
  • $\begingroup$ @Barry, very good. This deserves upvote $\endgroup$ – Ömer Sep 19 '13 at 20:55
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    $\begingroup$ This is definitely the way it was "meant" to be done; the cascading sequence of coefficients under the square root looked like a tipoff, but I didn't see the right magic at first. Nicely done! $\endgroup$ – Steven Stadnicki Sep 19 '13 at 20:59
  • $\begingroup$ Thanks for the comments, everyone. It helped a lot knowing that there was a simple answer. From there I just figured that the stuff outside the square root somehow had to be the derivative of what's inside, but that meant it had to be of lower degree. Obviously, the problem was composed so that everything works out. $\endgroup$ – Barry Cipra Sep 19 '13 at 21:06
  • $\begingroup$ Nice solution....Barry Cipra..... $\endgroup$ – juantheron Dec 15 '13 at 7:21
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Set R = 4 x3 + 5 x2 + 10 for brevity. Maybe based on Liouville's theorem, we can guess $$\int \sqrt{R} \left( x^5 + x^4 + x^2 \right) = PR^{3/2} + c$$ where P is a polynomial of x and c is the constant of integration. Differentiate on both sides, we get $$\sqrt{R} \left( x^5 + x^4 + x^2 \right) = P'R^{3/2} + PR' \sqrt{R}.$$

Rationalize the formula $$\left( x^5 + x^4 + x^2 \right) = \left( 4x^3 + 5x^2 + 10 \right) P' + \left( 12x^2 + 10x \right) P.$$

P is cubic, so set P = a3x3 + a2x2 + a1x + a0 and solve the linear system. The system is overdeterined, but luckily there is still a solution $$P = \frac{x^3}{30}.$$

As a result, $$\int \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{x^3 \left( 4x^3 + 5x^2 + 10 \right)^{3/2}}{30} + c.$$

Therefore $$\int_0^1 \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{19^{3/2}}{30}.$$

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    $\begingroup$ Not bad, but the interesting (and apparently difficult) part is how to find the indefinite integral. $\endgroup$ – Jean-Claude Arbaut Apr 12 '13 at 18:55
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    $\begingroup$ But, how did you evaluate the definite integral?.Aryabhata suggested a method to do it, if your method is different please post it. $\endgroup$ – jdoicj Apr 13 '13 at 14:08
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    $\begingroup$ @boywholived: I got the indefinite integral from Maxima, not by hand. I am looking into its source code to see what happened. I will post the intermediate steps as soon as I understand them. $\endgroup$ – jdh8 Apr 13 '13 at 16:09
  • $\begingroup$ @jdh8 $\large \left(PR^{3/2}\right)' = P'R^{3/2} + {3 \over 2}\,PR^{1/2}R'$ $\endgroup$ – Felix Marin Sep 19 '13 at 20:15
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$\displaystyle{\large% \int_{0}^{1} \left(x^{5} + x^{4} + x^{2}\right)\, \sqrt {4x^{3} + 5x^2 + 10\;}\;{\rm d}x\quad:{\Huge ?}}$

Following @jdh8 ( $\color{#0000ff}{\mbox{There is a missing}\; \color{#ff0000}{\large 3/2}\ \mbox{factor in his formula}}$ ).

$P \equiv a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$

\begin{align} \left(x^{5} + x^{4} + x^{2}\right) &= \left(4x^{3} + 5x^{2} + 10 \right)P' + \left(18x^{2} + 15x\right)P \\[3mm] {x^{4} + x^{3} + x \over 18x + 15} &= {4x^{3} + 5x^{2} + 10 \over 18x^{2} + 15x}\;P' + P \end{align} Take the limit $x \to 0$. In order to save a divergence we'll get $a_{1} = 0$ and it follows that $a_{0} = 0$. Then $$ {x^{4} + x^{3} + x \over 18x + 15} = {4x^{3} + 5x^{2} + 10 \over 18x + 15}\;\left(3a_{3} x + 2a_{2}\right) + \left[a_{3}x^{3} + a_{2}x^{2}\right] $$ Again, take the limit $x \to 0$ and we get $a_{2} = 0$. The last expression is reduced to $$ {x^{3} + x^{2} + 1 \over 18x + 15} = {4x^{3} + 5x^{2} + 10 \over 18x + 15}\;\left(3a_{3}\right) + \left[a_{3}x^{2}\right] $$ One more time $$ x \to 0 \quad\Longrightarrow\quad {1 \over 15} = {10 \over 15}\left(3a_{3}\right) \quad\Longrightarrow\quad a_{3} = {1 \over 30} \quad\Longrightarrow\quad \color{#ff0000}{\large P = {x^{3} \over 30}} $$

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