Author Topic: Extension 1 help (Read 1066 times) Tweet Share

88siege · « **on:** May 01, 2018, 06:53:48 pm »

I have no clue on how to do this question or even how to start off.

RuiAce · « **Reply #1 on:** May 01, 2018, 07:15:34 pm »

$\text{We note that}\\ \begin{align*} \int_{-2h}^{2h} 1\,dx &= 4h\\\int_{-2h}^{2h} x\,dx& = 0\\ \int_{-2h}^{2h} x^2\,dx &= \frac{16h^3}{3}.\end{align*}\\ \text{These computations are left as your exercise.}$
$\text{If }\int_{-2h}^{2h} f(x)\,dx = A \times f(-h) + B \times f(0) + C \times f(h)\\ \text{for the }\textbf{three different functions }f(x) = 1, x\text{ and }x^2,\\ \text{then upon subbing each of them in we get:}$
\begin{align*}4h &= A + B + C\tag{1}\\ 0 &= -hA + hC\tag{2}\\ \frac{16h^3}{3} &= h^2A + h^2C \tag{3}\end{align*}

To make things clearer...

We're not subbing in for $x$. We're subbing in for $f(x)$.

For example, if we sub in $f(x) = x^2$, what we are really saying is that $ \int_{-2h}^{2h} x^2\,dx = A (-h)^2 + B(0)^2 + Ch^2 $. This gives us equation (3).

$\text{To solve these equations, note that }(2)\times h\text{ gives }\boxed{0 = -h^2A + h^2 C}.\\ \text{Adding this to equation (3) gives }\frac{16h^3}{3} = 2h^2 C \implies \boxed{C = \frac{8h}{3}}\\ \text{Then working back up, we have }\boxed{A = -\frac{8h}{3}}\text{ and }\boxed{B = 4h}.$
Parts ii) and iii) should now be easy. In part ii), you just need to show that LHS = RHS when you sub in $f(x) = x^3$ instead, and part iii) is essentially replacing $ = $ with $\approx$, and subbing in $f(x) = 2^x$ and $h=1$.

Aside: This is no coincidence. This is just a variation on Simpson's rule.

88siege · « **Reply #2 on:** May 01, 2018, 07:57:42 pm »

Thanks allottttttt!!!

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Author Topic: Extension 1 help (Read 1066 times) Tweet Share

88siege

Extension 1 help

RuiAce

Re: Extension 1 help

88siege

Re: Extension 1 help

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