Find the minimum value of area under y = 4x - x^3

In summary, the attempt at a solution is to find the area under the curve from x=a to x=a+1 for all a>0. The equation for the area is found and it is a maximum when plugged in just the upper limit. The lower limit is then plugged in and the absolute value of the integral is taken to get the minimum area.
  • #1
issacnewton
1,010
31

Homework Statement


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

Homework Equations


Finding the area under the curve...

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 ?
 
Physics news on Phys.org
  • #2
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:
  • #3
IssacNewton said:
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
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
IssacNewton said:

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,
$$ A = (a+1)^2 \left[ 2 - \frac{(a+1)^2}{4} \right ] $$
It looks like you plugged in just the upper limit. What about the lower limit?
 
  • Like
Likes Chestermiller
  • #5
IssacNewton said:

Homework Statement


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

Homework Equations


Finding the area under the curve...

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 ?

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
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
 
  • Like
Likes Delta2
  • #7
IssacNewton said:
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

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.##
 
  • Like
Likes issacnewton
  • #8
Thanks Ray. I missed that.
 

Related to Find the minimum value of area under y = 4x - x^3

What is the equation for the given graph?

The equation is y = 4x - x^3.

What does the minimum value of area represent?

The minimum value of area represents the smallest possible area that can be enclosed by the graph.

How do you find the minimum value of area?

To find the minimum value of area, you can use calculus and take the derivative of the given equation. Set the derivative equal to 0 and solve for x. Substitute the obtained value of x into the original equation to find the corresponding y-value, which is the minimum value of area.

What is the significance of finding the minimum value of area?

Finding the minimum value of area helps in understanding the behavior of the given function and identifying its important features, such as the turning points and the intervals of increase and decrease.

Can the minimum value of area be negative?

No, the minimum value of area cannot be negative. It represents the smallest enclosed area, which cannot be smaller than 0. If the calculated minimum value of area is negative, it indicates that there is no minimum value of area for the given function.

Similar threads

Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
527
  • Calculus and Beyond Homework Help
Replies
3
Views
651
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Replies
2
Views
517
Back
Top