# expansion of integrand and resulting integral

I was watching a math tutorial on youtube and saw the following. I was wondering if I can get some advice on the difference between the two workings.

Why does the one on the left become fractions of (1/3, 1/2 & 1/4), while the one on the right only has a fraction for (1/5)?

Apologies for my course english, I thought I inserted the print screen with the pink arrows to make it clearer. =)

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One formula computes the mean $E(X)$ and the other the $Var(X) = E(X^2)$. That is, the formulae being integrated are different, so you should expect the evaluation of the integrals to be different.

$\displaystyle E(X) = \int_1^3 x(1-x)(x-3) dx = \int_1^3 x(3 + 4x - x^2) dx = \int_1^3 (3x + 4x^2 -x^3) dx$

$\displaystyle Var(X) = \int_1^3 x^2(1-x)(x - 3) dx = \int_1^3 (3x^2 + 4x^3 - x^4) dx.$

Notice that in the first case, we are integrating a third-degree polynomial, and in the second case, we are integrating a fourth-degree polynomial.

Recall: For all nonnegative integers $n$ ,$$\int_a^b x^n \, dx = \left[\frac{1}{n+1}x^{n+1}\right]_a^b= \frac{1}{n+1}b^{n+1} - \frac{1}{n+1}a^{n+1}$$

If you need more help with how to integrate polynomials, the link will take you to PlanetMath, where you'll find a tutorial to refresh your memory.

For a video tutorial: see Khan Academy's video, which discusses integration of polynomials. The Khan Academy's website also provides interactive tutorials so you can practice with actual problems, and brush-up on your math skills (here's a link to help you review Calculus, e.g.).

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Thanks for the speedy reply @amWhy. Really appreciate it. I've not done integration for a long time and have lost touch with it. Appreciate your kind advice please. –  user1079950 Dec 3 '12 at 13:46

It's just simplified. On the right side you have $$\frac{3}{4}=\int_1^3 4x^3-x^4-3x^2 = \frac{3}{4}\left[\frac{4x^4}{4}-\frac{x^5}{5}-\frac{3x^3}{3}\right]_1^3=\frac{3}{4}\left[{x^4}-\frac{x^5}{5}-{x^3}\right]_1^3$$

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