# Homework Help: Integration by Substitution

1. Sep 11, 2011

### Awesomesauce

1. The problem statement, all variables and given/known data
(i) find $\int$$^{X}_{0}$ xe$^{-x^{2}}$ dx in terms of X.
(ii) Find $\int$$^{X}_{0}$ xe$^{-x^{2}}$ dx for X= 1, 2, 3 and 4.

2. Relevant equations
-

3. The attempt at a solution
(i) $\int$$^{X}_{0}$ xe$^{-x^{2}}$dx

-x$^{2}$ = X

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

so, -$\frac{1}{2}$ $\int$$^{X}_{0}$ e$^{x}$ dx

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

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

I assume substituting -x$^{2}$ = 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 aswell with negative powers of x.
Using the answer the textbook got, I input, $\frac{1}{2}$ [1 - e$^{(-1)^{2}}$ ] and get -0.859. The answer should be 0.3161, and to get this I need to remove the brackets from (-1) hence$\frac{1}{2}$ [1 - e$^{-1^{2}}$ ]

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!

2. Sep 11, 2011

### SteamKing

Staff Emeritus
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).

3. Sep 11, 2011

### Awesomesauce

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?

Last edited: Sep 11, 2011
4. Sep 11, 2011

### SteamKing

Staff Emeritus
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.

5. Sep 11, 2011

### Awesomesauce

Ah sorry, didn't make my workings clear. :p

(ii) Find $\int$$^{X}_{0}$ xe$^{-x^{2}}$ 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.

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

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

Thanks again :)

6. Sep 11, 2011

### SteamKing

Staff Emeritus
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