That's a lot of things to average

Calculus Level 5

On $x \in [a, b]$ , the area under $y = x^2$ is always 5. That is, for any $(a, b)$ you choose, $a$ and $b$ satisfy $\displaystyle \int_{a}^{b} x^2 \, dx = 5$ .

On $x \in [0, 100 ]$ , there is an infinite number of possibilities for $[a, b]$ that satisfy the above conditions. Then, you can get an infinite number of distances between $a$ and $b$ depending on the values of $a$ and $b$ .

If the average distance between $a$ and $b$ given all the above conditions is $\mathfrak{D}$ , find $\left \lceil 1000 \mathfrak{D} \right \rceil$ .

Note: Please when you solve the problem to avoid entering an "incorrect" answer that is actually correct, integrate with respect to $a$ , not $b$ . There seems to be a possible discrepancy with Wolfram Alpha. Also, yes, you may use Wolfram Alpha to compute the integral.

The answer is 54.

1 solution

Hobart Pao
Jul 13, 2016

$\displaystyle \int_{a}^b x^2 \, dx = 5 \to \dfrac{b^3 - a^3}{3} = 5$

Rearrange to get

$b = \sqrt[3]{a^3 + 15}$

The minimum value of $a = 0$ and maximum value of $b$ is 100, so we can solve for max value of a.

$100 = \sqrt[3]{a^3 + 15 }$ $10^6 = a^3 + 15$ $10^6 - 15 = a^3$ $\sqrt[3]{10^6 - 15} = a$

To get the average distance, we want to know the average $b - a$ but we've already figured out $b$ in terms of $a$ .

$\dfrac{1}{\sqrt[3]{10^6 - 15} - 0} \displaystyle \int_{0}^{\sqrt[3]{10^6 - 15}} \sqrt[3]{a^3 + 15} - a \, da = 0.0532...$

Finally, $\left \lceil 1000 \cdot 0.0532 \right \rceil = \boxed{54}$

Nice problem and solution. The discrepancy between integration wrt $a$ and integration wrt $b$ is a curious one; It may be a flaw with WolframAlpha, or perhaps there is something more fundamental going on. Not sure yet .... Before reposting I'd like to figure out what's going on, because at present others who integrate wrt $b$ will be told they have the incorrect answer and will wonder what to do next, (as was the case for me).

Brian Charlesworth - 4 years, 10 months ago

The discrepancy between the two answers comes from the difference between the underlying probability distributions for a and for b. If we select a uniformly at random from a certain interval, then the corresponding values of b such that $b^3 - a^3 = 15$ will not be uniformly distributed; and vice versa. If b is uniformly distributed, then the corresponding values of a will not be. Ultimately, it lies in the unclarity of what probability distribution we are putting on the choices of the pairs (a,b).

Marcus Neal - 2 years, 4 months ago

Ah, ok. Thanks for clearing that up. :)

Brian Charlesworth - 2 years, 4 months ago

How to integrate function cubert(x^3+15) with bounds

D S - 3 years, 5 months ago

You use a computer.

Hobart Pao - 3 years, 4 months ago

That's a lot of things to average

The answer is 54.

1 solution

0 pending reports