# Thread: Find the minimum of the integral

1. ## Find the minimum of the integral

Hi all, I got set this question as part of my calculus of variations homework, but the lecturer has told me that it's not actually a calculus of variations problem (just a calculus problem).

Find the minimum of the integral:

$\displaystyle I = \int{(2(y+2)^3\frac{dy}{dx}}+(x-7))dx$

I'm not sure whether I should do something involving Leibniz's rule, or whether I should try integrating, then differentiating twice.

So far I got:

$\displaystyle I = {\frac{1}{2}(y+2)^4+\frac{1}{2}x^2-7x$,

since

$\displaystyle \frac{d}{dx}(\frac{1}{2}(y+2)^4) = 2(y+2)^3\frac{dy}{dx}$

Any hints?

2. Set dI/dx = 0. What do you get?

3. $\displaystyle 2(y+2)^3\frac{dy}{dx}}+(x-7) = 0$

Take the (x-7) to the other side and integrate both sides, gives the same as in my first post; $\displaystyle I = {\frac{1}{2}(y+2)^4+\frac{1}{2}x^2-7x$, which confuses me, does dI/dx = I here?

4. Correct. For finding out about min or max, try taking the second derivative and see what you get. If you can show that the second derivative is positive for the solution of the first-order DE above that you need to solve, then you've got a min. In CoV problems, I'm afraid they often don't do the second-order checking necessary to discover if you're dealing with a max or a min.