Values of a and b

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  • #1
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[tex]\int_a^b (x-x^2)dx[/tex]

What values of a and b make this integral's value maximum? I have tried to do it but cannot get it. I know that the maximum value of [itex]x-x^2[/itex] is 0,25. but I'm stuck from there.
I try to write the integral in the form of a riemann sum but (b-a)s cancel each other.what am I to do?
Please help about this
 

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  • #2
dextercioby
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U can integrate and obtain a function

[tex]F(a,b) [/tex]

whose critical values are found setting the partial derivatives wrt to "a" & "b" to zero. Computing the hessian on the solutions gives the nature of the critical points: extremum or saddle.

Daniel.
 
  • #3
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Well thanks but I just introduced myself to the integral calculus so I do not understand anything from what you say.

edit: btw graphing doesn't count it is too easy using the graph
 
  • #4
dextercioby
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By the looks of the graph, it should be the intercepts a=0 b=1.

Daniel.
 
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Yes but isn't there an algeabric way to calculate those values?
 
  • #6
lurflurf
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wisredz said:
[tex]\int_a^b (x-x^2)dx[/tex]

What values of a and b make this integral's value maximum? I have tried to do it but cannot get it. I know that the maximum value of [itex]x-x^2[/itex] is 0,25. but I'm stuck from there.
I try to write the integral in the form of a riemann sum but (b-a)s cancel each other.what am I to do?
Please help about this
Well that is easy you can make the integral as big as you want take b<a and far apart. For example if a=10 b=-10 the integral is 2000/3. If you also want to require a<b then take a the smaller zero (0) and b the larger zero (1) this gives 1/6 for the integral.
 
  • #7
AKG
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The solution is much easier. The integral is the area under the curve. If the curve is below the x axis, then the area is negative. This is the integral of a simple quadratic polynomial, so it has at most two roots. Also, you can tell that since the coefficient of the quadratic term is negative, that the graph of the polynomial opens downwards, so if you can find two roots, then the graph between those roots lies above the axis, and so you take that entire area to be your maximum area.

You can use the above ideas to produce a more formal argument that makes no reference to the graph, but looking at the graph will suggest to you what facts you should be trying to prove in order to determine a and b.

If you wanted to do it dextercioby's way, you would have to know something about functions of several variables. If you know what a Jacobian is (it's a generalization of the derivative of a single-variable function), then that matrix is zero at the critical points. Then, to determine the nature of the critical points, you determine the eigenvalues of the Hessian, which is analogous to finding the second derivative, i.e. you use the Second Derivative Test, which I'm sure you've heard of for the single variable case.
 
  • #8
lurflurf
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wisredz said:
Yes but isn't there an algeabric way to calculate those values?
yes
x-x^2=0
x(x-1)=0
x=0,1
 
  • #9
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oh yes thank you everybody. I must be getting dumber by each second to miss this...
 

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