ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: serussell_ on January 20, 2018, 12:51:09 am

Title: integrating with parenthesis
Post by: serussell_ on January 20, 2018, 12:51:09 am: e.g *integration symbol* 6(2x+1)(x^2 + x)^5 ??
Title: Re: integrating with parenthesis
Post by: RuiAce on January 20, 2018, 08:18:09 am: $\text{One possible way of expanding the parentheses (without 3U binomial theorem)}\\ \text{is to gradually build it up.}\\ \begin{align*}6(2x+1)(x^2+x)^2&=(12x+6)(x^4+2x^3+x^2)\\ &= 12x^5+30x^4+24x^3+6x^2\end{align*}$
$\text{.. and so on. Eventually,}\\ 6(2x+1)(x^2+x)^5=12 x^{11} + 66 x^{10} + 150 x^9 + 180 x^8 + 120 x^7 + 42 x^6 + 6 x^5$
$\therefore \int 6(2x+1)(x^2+x)^5 = x^{12}+6x^{11}+15x^{10}+20x^9+15x^8+6x^7+x^6+C$
There are far more elegant approaches to this question as you would expect. However, since this question was not posted in the 3U section, NOR was there any previous part to it, this is the only possible approach under the limitations of 2U.
Title: Re: integrating with parenthesis
Post by: jamonwindeyer on January 20, 2018, 11:58:40 am: You could also be clever and spot that:

$\frac{d}{dx}\left(x^2+x\right)^6=6\left(x^2+x\right)^5\left(2x+1\right)$

That's just using the chain rule of differentiation. Although it is typically more of a 3U thing, definitely appropriate for us to just use this idea to go back to the original LHS when we integrate the RHS:

$\therefore \int6\left(2x+1\right)\left(x^2+1\right)^5dx=\left(x^2+x\right)^6+C$

PS - Welcome to the forums! ;D

In general, if you can see an expression as an integrand that matches a derivative, you can jump immediately back to the anti-derivative when you integrate, just like I've done here :)