# Homework Help: Find the minimum value of area under y = 4x - x^3

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1. Jul 9, 2018

### issacnewton

1. The problem statement, all variables and given/known data
Find the minimum value of the area of the region under the curve $y=4x - x^3$ from $x=a$ to $x=a+1$, for all $a>0$. This problem is from Stewart's Calculus

2. Relevant equations
Finding the area under the curve....

3. The attempt at a solution
I can set up the equation for the area as
$$A = \int_a^{a+1} (4x-x^3) \; dx$$ Solving this, we get,
$$A = (a+1)^2 \left[ 2 - \frac{(a+1)^2}{4} \right ]$$ And I need to maximize this function. I can plug $\alpha = (a+1)^2$ and the area becomes $A = \alpha (2 - \frac{\alpha}{4})$. But this looks like an inverted parabola and it would have a maxima and a minima of zero area. Does that make sense ?

2. Jul 9, 2018

### Delta²

I think this needs some extra caution. $f(x)=4x-x^3$ becomes negative for $x>2$. You should treat integral in a special way for the $a$ such that $1<a<2$ because for such $a$, $f(x)>0$ for $a<x<2$ but $f(x)<0$ for $2<x<a+1$ .

The problem asks for the area "under" the curve so I don't know when f becomes negative whether you should ignore that part of integral or just take $\int|f(x)|dx$

Last edited: Jul 9, 2018
3. Jul 9, 2018

### Staff: Mentor

Is this the exact wording of the problem, in particular the part about "under the curve"?

For positive x, the graph of $y = 4x = x^3$ is nonnegative only on the interval [0, 2]. If x > 2, the y values are negative, so all of the integrals of the form $\int_a^{a + 1} f(x) dx$ for a > 2 are unbounded.

4. Jul 9, 2018

### vela

Staff Emeritus
It looks like you plugged in just the upper limit. What about the lower limit?

5. Jul 9, 2018

### Ray Vickson

If you weaken the inequality $a > 0$ to $a \geq 0$, you have a problem with a possible end-point solution. If that happens, the derivative will not necessarily vanish at the optimum. For example, the solution of the problem $\min f(a) = a$, subject to the constraint $a \geq 0$ is at $a = 0$, where the derivative is $f'(a) = 1,$ not 0.

If we re-interpret your problem as involving the integral $I(a) = \int_a^{a+1} |4x - x^3| \, dx$ there is a local minimum at a positive value of $a$.

6. Jul 13, 2018

### issacnewton

Hello
Sorry for late reply. I was busy. But vela is right. I don't have the integral evaluated correctly. I think the problem meant to take the absolute value of the integrand while calculating the area as people have pointed out. Otherwise the minimum area would be just zero.The problem statement is exact from the James Stewart's Calculus , 8th edition page no 353. So after giving some thought to this problem, I think we need to consider three cases here. Case 1) $0 < a \leqslant 1$. In this case, the area would be
$$A = \int_a^{a+1} f(x)\; dx = \int_a^{a+1} (4x - x^3) \; dx$$
$$A = -a^3-\frac{3}{2}a^2 + 3a + \frac{7}{4} \cdots\cdots (1)$$

Case 2) $1 < a< 2$
Here part of the area would be positive and part would be negative. So we will need to take the absolute value of the negative area.
$$A = \int_a^{a+1} |f(x)|\; dx = \int_a^{2} (4x - x^3) + \int_2^{a+1}(-1)(4x - x^3)\; dx$$
$$A = \frac{a^4}{2} + a^3 - \frac{5a^2}{2} -3a + \frac{25}{4}\cdots\cdots (2)$$
And finally the third case would be
Case 3) $a \geqslant 2$
Here the function is negative and we need to take the absolute value of the integrand. So the area would be
$$A = \int_a^{a+1} |f(x)|\; dx = - \int_a^{a+1} f(x)\; dx$$
$$A = \int_a^{a+1} (x^3 - 4x) \; dx$$
$$A = a^3 + \frac{3a^2}{2} - 3a - \frac{7}{4}\cdots\cdots (3)$$
So to get the minimum, we will need to find the minimum of the area over three cases and then get the overall minimum. For this I used the plotting function and minimize function in the WolframAlpha. For case 1, it turns out that the area is minimum when $a=0$. So the minimum area is $A = 7/4$. For case 2, it turns out that the minimum area is when $a = \frac{\sqrt{13}}{2} - \frac{1}{2}=1.303$ and the minimum area is $A = 7/4$. For the case 3, the minimum area is when $a = 2$ and this minimum area is $A = 25/4$.
So the overall minimum area for all $a > 0$ is $A = 7/4$.
Does this look Ok ?
Thanks

7. Jul 14, 2018

### Ray Vickson

Yes, it is correct, and the method is correct as well.

If you stick to the original question, exactly as you wrote it, the point $a=0$ is not "allowed", so the minimum on $a > 0$ is at $a = \sqrt{13}/2 - 1/2.$

8. Jul 15, 2018

### issacnewton

Thanks Ray. I missed that.