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Problem :

Prove that the integral $\int^{10}_0 \frac{x}{x^3+16}dx$ is less than $\frac{5}{6}$

Getting no clue how to proceed as I am not getting any factor so that by manipulating I will be able to cancel the numerator or denominator :

I the denominator could have been $x^3+32 = (x+4)(x^2-4x+16)$ could have been the factor and we will write numerator as x +4-4 but .... not getting the right clue here..Please suggest thanks...

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  • $\begingroup$ Substitute 10 with a w... Compute the Taylor series for your new F(w) centered at 0. And then evaluate at 10 $\endgroup$ – frogeyedpeas Jun 19 '13 at 3:39
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HINT: Recall $\int_a^bf(x)\mathrm{d}x\le(b-a)M$ where $f(x)\le M $ for $a\lt x \lt b$

(You can use calculus techniques to find $M$)

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  • $\begingroup$ how can I find the value of M here, please guide little bit more... thanks.. $\endgroup$ – sultan Jun 19 '13 at 6:39
  • $\begingroup$ well here $f(x)=\frac{x}{x^3+16}$ and we essentially want to find the maximum value of $f(x)$ for $0\le x \le 10$ which must occur at $x=0, x=10$ or where $f'(x)=0$ (since $f(x)$ is defined and continuous over this domain) $\endgroup$ – john Jun 19 '13 at 6:45
  • $\begingroup$ That means, $f'(x) =\frac{-2x^3+16}{(x^3+16)^2}$ Now if I put this equal to zero $\Rightarrow -2x^3=-16 \Rightarrow x = \pm 2$ Ignoring the -ve value we take +2, so function will attain maximum value at 2 within the given limits, therefore putting this value in given function I got : $\frac{1}{12} < \frac{5}{6}$ Is that correct.. thanks.. $\endgroup$ – sultan Jun 19 '13 at 15:40
  • $\begingroup$ @sultan: You forgot to multiply with $(b-a)$. Side note: It's true that you don't use $x=-2$ regardless, but note that $-2$ is not a solution to the equation $f'(x)=0$, because $(-2)^3=-8$. The only real solution is $x=2$. $\endgroup$ – Jonas Meyer Jun 19 '13 at 18:20
  • $\begingroup$ thanks a lot that was great..I got it..thanks once again.. $\endgroup$ – sultan Jun 20 '13 at 2:48
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Some bounding of the function by easier to integrate functions gives stronger bounds than $\frac56$. When $x$ is small, the upper bound $\dfrac{x}{16}$ is useful. When $x$ is large, more useful is the upper bound $\dfrac{1}{x^2}$. For each $a$ between $0$ and $10$, we have $$\int^{10}_0 \frac{x}{x^3+16}\,dx\leq\int_0^a\frac{x}{16}\,dx+\int_a^{10}\frac{1}{x^2}\,dx=\frac{a^2}{32}+\frac{1}{a}-\frac{1}{10}.$$ Because this is true for all $a$, it will hold when $a$ is chosen to make the last quantity as small as possible. It has its minimum when $a=\sqrt[3]{16}$, giving a bound just under $\frac12$. However, it would also be enough here to use the upper bound when $a=2$, which gives $$\int^{10}_0 \frac{x}{x^3+16}\,dx\leq\frac{4}{32}+\frac{1}{2}-\frac1{10}<\frac58<\frac56.$$

This would also work to give an upper bound of $\frac58$ (or a little smaller if we optimize) for $\int^{\infty}_0 \frac{x}{x^3+16}\,dx$.

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  • $\begingroup$ $$\int^{10}_0 \frac{x}{x^3+16}\,dx\leq\int_0^a\frac{x}{16}\,dx+\int_a^{10}\frac{1}{x^2}\,dx= \frac{a^2}{32}+ \frac{1}{a}-\frac{1}{10}.$$ How did you get this inequality , please share some detail I will be thankful to you.,, $\endgroup$ – sultan Jun 19 '13 at 15:42
  • $\begingroup$ sultan: Both of the inequalities $\dfrac{x}{x^3+16}\leq\dfrac{x}{16}$ and $\dfrac{x}{x^3+16}\leq\dfrac{1}{x^2}$ hold for each $x>0$. If $f(x)\leq g(x)$ for all $x$ with $a<x<b$, then $\int_a^bf(x)\,dx\leq\int_a^bg(x)\,dx$. First breaking up the integral, using $\int_0^{10}f(x)\,dx=\int_0^af(x)\,dx+\int_a^{10}f(x)\,dx$, you can then apply the different bounds to the different intervals, where they are most useful. $\endgroup$ – Jonas Meyer Jun 19 '13 at 18:18
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Factor $ x^3 + 16 $

Note that if $x^3 + 16 = 0 $

Then:

$x^3 = -16$

And so therefore:

$x = 2*2^{1/3}$ times the three cube roots of unity

Lets call this number involving the 2's w for short

The roots of unity are 1, (1 + i $3^{1/2}$)/2, and (1 - i $3^{1/2}$)/2 lets call these numbers 1, $o_1$ and $o_2$ respectively to keep it simple

Thus you now factor the x^3 + 16 as....

(x - 1w)(x - $o_1w$)(x - $o_2w$)

At this point a standard partial fraction decomposition approach will work.

Comment below if you want me to elaborate on that... Or you can do it yourself.

This of course is a brute force approach of actually solving the integral itself. I'm sure there are more elegant ways out there

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  • $\begingroup$ Just edited for a mistake $\endgroup$ – frogeyedpeas Jun 19 '13 at 3:55
  • $\begingroup$ Yeah,,please provide some detail on this...its something new for me....but I want to learn this as well.. thanks a lot... $\endgroup$ – sultan Jun 19 '13 at 15:48
  • $\begingroup$ This is a bit late but I'll still mention it. So after factoring the bottom half into the product abc (each of the letter corresponds to (x - 1w)(x- o1w) and (x - o2w) respectively. your goal is to to rewrite x/(abc) into s/a + d/b + f/c. So basically: sbc + dac + f*ab = x. But we note that for a product of 2 letters, ac, ab, bc we end up with a quadratic (ax^2 + bx + c). So we can now set up a system of 3 linear equations in 3 variables and solve for s, d, and f independently. Once done we can now break up the 3 integrals, evaluate them (This is really easy to do) and add up our answers $\endgroup$ – frogeyedpeas Jun 26 '13 at 16:12

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