Thomas Calc 11th ed., Problem 5.63

[noclueatol]

I'm just working through this book for self-study, so I hope this isn't a stupid question.

Homework Statement

What values of a and b maximize the value of \int_{a}^{b}(x-x^{2}){dx}?

Homework Equations

(Hint: Where is the integrand positive?)

The Attempt at a Solution

Well, the integrand is positive for 0 < x < 1, and the answer in the back of the book is a=0, b=1. That gives a value of (1/2 - 1/3) = 1/6.

But the problem asks about the expression as a whole, not the integrand. And it doesn't specify that a<b. For large values of x, the integrand will take on a large negative value. Then if you reverse the order of the limits, you get a large positive value. So why wouldn't, say, a=10,000 and b=1 yield a much larger value for the integral than the given answer? Or is it always assumed that a<b?

 

[Mark44]

I'm just working through this book for self-study, so I hope this isn't a stupid question.

Homework Statement

What values of a and b maximize the value of \int_{a}^{b}(x-x^{2}){dx}?

Homework Equations

(Hint: Where is the integrand positive?)

The Attempt at a Solution

Well, the integrand is positive for 0 < x < 1, and the answer in the back of the book is a=0, b=1. That gives a value of (1/2 - 1/3) = 1/6.

But the problem asks about the expression as a whole, not the integrand. And it doesn't specify that a<b. For large values of x, the integrand will take on a large negative value. Then if you reverse the order of the limits, you get a large positive value. So why wouldn't, say, a=10,000 and b=1 yield a much larger value for the integral than the given answer? Or is it always assumed that a<b?

I believe that they're tacitly assuming that a and b in the limits of integration are in the order a < b.

 

[Bacle2]

I'm just working through this book for self-study, so I hope this isn't a stupid question...
.

But the problem asks about the expression as a whole, not the integrand

Right, but a Riemann sum will have negative terms when the integrand is negative.

 

[noclueatol]

I believe that they're tacitly assuming that a and b in the limits of integration are in the order a < b.

So are you saying that I'm correct that I could get a larger value for the integral with b>a, but that would be violating the tacit assumption?

 

[noclueatol]

Right, but a Riemann sum will have negative terms when the integrand is negative.

Yes, but won't reversing the limits make the integral positive?

 

[Mark44]

Definite integrals are usually written with the lower integration limit being less than the upper limit, and I think it's reasonable to assume that that's what Thomas had in mind for this problem.

The value of the integral will be largest on the largest interval for which the integrand is greater than or equal to zero: namely, the interval [0, 1].

Don't overthink this.

 

[noclueatol]

Don't overthink this.

I wouldn't if it was a drill problem, but as the high number indicates, it's in the "theory" part of the exercises, which sort of encourages second thoughts. Also, the concept of reversing the limits to change the sign of the integral was introduced in this same section, so I would have thought they expected me to use it in solving the problems.

Oh well, if all I'm missing is a tacit assumption, then I guess I'm satisfied that I understood the material. Thanks for the responses.

 

[Bacle2]

If you reverse the limits of integration,you will basically be seeking to

maximize ∫ (x²-x)dx, instead of ∫ x-x²dx .

This seems a different problem, tho maybe I misunderstood

your goal. Still, the idea would be the same, in that the intervals that maximize

your integral are those where x²-x >0 .