So, for your problem, the correct value of f(-1) is 1/e, not e.

[Awesomesauce]

Homework Statement

(i) find [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex] dx in terms of X.
(ii) Find [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex] dx for X= 1, 2, 3 and 4.

Homework Equations

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The Attempt at a Solution

(i) [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex]dx

-x[itex]^{2}[/itex] = X

dX/dx=-2x hence -1/2 dX = xdx

so, -[itex]\frac{1}{2}[/itex] [itex]\int[/itex][itex]^{X}_{0}[/itex] e[itex]^{x}[/itex] dx

then [itex]\frac{1}{2}[/itex] [1 - e[itex]^{x^{2}}[/itex] ]

Ok, here is the first problem I have encountered. For my answer, [itex]\frac{1}{2}[/itex] [1 - e[itex]^{x^{2}}[/itex] ], this is wrong according to wolfram and my textbook. the answer should be
[itex]\frac{1}{2}[/itex] [1 - e[itex]^{-x^{2}}[/itex] ], where there is a minus before the x[itex]^{2}[/itex]. I can not think of how to come to this! :(

I assume substituting -x[itex]^{2}[/itex] = X is correct, as that is what i normally do with standard u-substitution problems.

Ok, second part (ii) So this problem also applies to other problems as well with negative powers of x.
Using the answer the textbook got, I input, [itex]\frac{1}{2}[/itex] [1 - e[itex]^{(-1)^{2}}[/itex] ] and get -0.859. The answer should be 0.3161, and to get this I need to remove the brackets from (-1) hence[itex]\frac{1}{2}[/itex] [1 - e[itex]^{-1^{2}}[/itex] ]

I'm not sure if my brain is playing up and this is a stupid question, or my calculator; but I always thought you place brackets around the negative number when squaring hence -1 x -1 = 1... rather than -1. This problem also happens with other integration problems I have gone through. Someone help me!
Thanks for your time!

 

[SteamKing]

You are jumping ahead of yourself in the choice of substitution. Rather than using the upper limit X of the definite integral, instead choose u = x^2 (where x is the variable of integration).

 

[Awesomesauce]

You are jumping ahead of yourself in the choice of substitution. Rather than using the upper limit X of the definite integral, instead choose u = x^2 (where x is the variable of integration).

Hurray I got it! Thank you x1000. But why do I not substitute the minus aswell, so u= -x^2? rather than just u=x^2
Edit: Checked over it, seems like it only works if I substitute u=-x^2 not u=x^2, ...or not? Confusion! :(
Do you know why my second part is wrong?

 

[SteamKing]

The substitution u = -x^2 is probably better.

I don't know why your second part is wrong because nothing of you work is shown.

 

[Awesomesauce]

Ah sorry, didn't make my workings clear. :p(ii) Find [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex] dx for X= 1, 2, 3 and 4
I will just do X=1, so I understand how to do the others on my own later.

[itex]\frac{1}{2}[/itex] [1 - e[itex]^{(-1)^{2}}[/itex] ] and get -0.859.
The answer should be 0.3161, and to get this I need to remove the brackets from (-1) hence[itex]\frac{1}{2}[/itex] [1 - e[itex]^{-1^{2}}[/itex] ]

Why do I not include brackets around -1? That would be -1 x 1, hence not (-1)^2 then in my calculator.

Thanks again :)

 

[SteamKing]

In evaluating e^-x^2, for x = 1,2,3,4, the convention is the value of x is squared before application of the minus sign. Similarly, if you were given the polynomial f(x) = -x^2+2x-4 to evaluate at x = -1, you would calculate as follows:

f(-1) = -[(-1)^2)] + 2*(-1) - 4 = -1-2-4 = -7