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Q: If $M$ be the maximum value of $$72\int_0^y \sqrt{x^4+(y-y^2)^2}dx$$ for $y\in [0,1]$, then find $\frac{M}{6}$ $$$$ A: $4$

My first thoughts involved differentiating it and equating it to zero, but since the integrand is dependent on the upper bound, I'm unable to do so.

Then I thought of actually integrating it, and the trying to find its maximum value, but the $x^4$ is throwing me off track; had it been $x^2$ I'd be able to carry out my thoughts. But since it isn't I'm stuck.

Thanks in advance.

Edit: My question is a teeny bit different from this question; the solutions there use multivariable calculus, while I'm looking for a solution involving single variable calculus, as I encountered this one while studying for an exam whose syllabus doesn't involve multivariable calculus.

Edit 2: Since a user pointed out that this is a possibile duplicate, I edited the question body to show how my question is different from that one, yet the question was marked duplicate.

AryanSonwatikar
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1 Answers1

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You can perform a suitable scaling transformation; e.g., suppose $u = x/y$, $du = 1/y \, dx$. Then $$f(y) = \int_{u=0}^1 \sqrt{(yu)^4 + (y - y^2)^2} \, y \, du = \int_{u=0}^1 y^2 \sqrt{(y+1)^2 + y^2 u^4} \, du.$$ Now how would you proceed?

Consider the function $$g(y, u) = y^2 \sqrt{(y+1)^2 + y^2 u^4}.$$ Then $$f(y) = \int_{u=0}^1 g(y, u) \, du.$$ We aim to show that for each fixed $u \in [0,1]$, $$0 \le y_1 < y_2 \le 1 \implies g(y_1, u) \le g(y_2, u). \tag{1}$$ This is equivalent to showing $\frac{\partial g}{\partial y} \ge 0$ for all $y \in [0,1]$. To this end, $$\frac{\partial g}{\partial y} = \frac{3(u^4 + 1)y^3 + 5y^2 + 2y}{\sqrt{(y+1)^2 + u^4 y^2}},$$ and since the numerator is clearly nonnegative for $u \in [0,1]$ and $y \in [0,1]$, and the denominator is always strictly positive, $(1)$ follows. This immediately implies $f(y_1) \le f(y_2)$ whenever $y_1 < y_2$. So the maximum value of $f$ is attained when $y = 1$.

heropup
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  • I appreciate your help, but I'm unable to see how to proceed :( – AryanSonwatikar Sep 11 '20 at 03:53
  • @AryanSonwatikar Since the integral is now over a fixed interval, what happens to the value of the integral as $y$ ranges from $0$ to $1$? Can you show that $f(y)$ is monotonically increasing? – heropup Sep 11 '20 at 05:17
  • I'm sorry, but what is $f(y)$? Is it the integrand as a whole? Or just the part inside the root? Also, I am unable to see how we can comment on the maximum value of an integral based on the monotonicity of the integrand. Sorry if my questions seem too childish, but please keep in mind I'm seventeen. – AryanSonwatikar Sep 11 '20 at 05:59
  • The problem is you cannot apply Leibniz's rule when the integrand is a function of two variables. At least to the level of my knowledge of single variable calculus – DatBoi Sep 11 '20 at 11:23
  • @Heropup The question was posed under single variable calculus. As I said in a reply to Alan Abraham, we have been taught the differentiation of $\int_{a(x)}^{b(x)} f(t) dt$ but not of $\int_{a(x)}^{b(x)} f(x,t) dt$. {Insert heavy crying} – AryanSonwatikar Sep 11 '20 at 12:59
  • @AryanSonwatikar Learning mathematics is not simply about using what you already know. It is about using reasoning and logic to extend what you already know. I have not used any concepts in my solution that a first-year calculus student could not follow. Use your brain, rather than objecting to the appearance of new notation or concepts. Think about the meaning and motivation behind the solution. – heropup Sep 11 '20 at 19:04
  • @AryanSonwatikar The idea here is that we have a continuous family of functions $g(y, u)$ indexed by a variable $y \in [0,1]$, for which the area under the curve plotted on $u \in [0,1]$ increases monotonically for increasing $y$. This is because when we look at the curve pointwise, increasing $y$ increases $g(y,u)$. – heropup Sep 11 '20 at 19:07
  • @heropup I appreciate your thoughts and efforts, but please understand that I'm limited to single variable calculus. – AryanSonwatikar Sep 12 '20 at 02:33