- #1

issacnewton

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## 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 ?