Solutions - 0070-02

(1) Recall shells formula:

$$V={\int }_a^{b}2\pi Rhdr$$

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(2) Interpret:

$$R=xh=f(x)dr=dx$$

Bounded above by $f(x)$. Bounded below by $x$-axis.

Bounded left by $x=2$. Bounded right by $x=8$.

Obtain:

$$V=2\pi {\int }_2^{8}xf(x)dx$$

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(3) Create table of values to apply Simpson’s Rule:

$x_i$	$f(x_i)$	$x_{i}f(x_i)$
$x_0=2$	$f(x_0)=0$	$x_{0}f(x_0)=0$
$x_1=3$	$f(x_1)=1$	$x_{1}f(x_1)=3$
$x_2=4$	$f(x_2)=3$	$x_{2}f(x_2)=12$
$x_3=5$	$f(x_3)=3$	$x_{3}f(x_3)=15$
$x_4=6$	$f(x_4)=2$	$x_{4}f(x_4)=12$
$x_5=7$	$f(x_5)=3$	$x_{5}f(x_5)=21$
$x_6=8$	$f(x_6)=0$	$x_{6}f(x_6)=0$

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(4) Recall Simpson’s Rule formula:

$$S_n=\frac{1}{3}\mathrm{\Delta }x(y_0+4y_1+2y_2+\cdots +2y_{n-2}+4y_{n-1}+y_n)$$

Here $y_i=x_{i}f(x_i)$ since $y_i$ in this formula represents the integrand values.

Note that $\mathrm{\Delta }x=1$.

Plug in:

$$\begin{array}{cc}{S}_n& =\frac{1}{3}(x_{0}f(x_0)+4x_{1}f(x_1)+2x_{2}f(x_2)+\cdots +4x_{5}f(x_5)+x_{6}f(x_6))\\ & \\ & =\frac{1}{3}(0+4(3)+2(12)+4(15)+2(12)+4(21)+0)=68\end{array}$$

Therefore:

$${\int }_2^{8}xf(x)dx\approx 68$$

Therefore:

$$V=2\pi {\int }_2^{8}xf(x)dx\approx 2\pi \cdot 68\approx 427.26$$