- #1
Prove It
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View attachment 7140
View attachment 7141
(a) Since a = 2, that means $\displaystyle \begin{align*} \Delta x = \frac{2}{5} \end{align*}$
(b)
$\displaystyle \begin{align*} f \left( \frac{7\,\Delta x}{2} \right) &= f \left( \frac{7}{5} \right) \\ &= \left( \frac{7}{5} \right) ^2 \\ &= \frac{49}{25} \end{align*}$
(c)
$\displaystyle \begin{align*} A_4 &= f \left( \frac{7}{5} \right) \cdot \Delta x \\ &= \frac{49}{25} \cdot \frac{2}{5} \\ &= \frac{98}{125} \,\textrm{units}^2 \end{align*}$
(d)
$\displaystyle \begin{align*} A_i &= f \left( x_i \right) \cdot \Delta x \end{align*}$
(e)
$\displaystyle \begin{align*} S_5 &= \sum_{i = 1}^5{ \left[ f \left( x_i \right) \cdot \Delta x \right] } \\ &= f \left( x_1 \right) \cdot \Delta x + f \left( x_2 \right) \cdot \Delta x + f \left( x_3 \right) \cdot \Delta x + f \left( x_4 \right) \cdot \Delta x + f \left( x_5 \right) \cdot \Delta x \\ &= \left( \frac{1}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{3}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{5}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{7}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{9}{5} \right) ^2 \cdot \frac{2}{5} \\ &= \frac{1}{25} \cdot \frac{2}{5} + \frac{9}{25} \cdot \frac{2}{5} + \frac{25}{25} \cdot \frac{2}{5} + \frac{49}{25} \cdot \frac{2}{5} + \frac{81}{25} \cdot \frac{2}{5} \\ &= \frac{2}{125} + \frac{18}{125} + \frac{50}{125} + \frac{98}{125} + \frac{162}{125} \\ &= \frac{330}{125} \\ &= \frac{66}{25} \,\textrm{units}^2 \end{align*}$
(f)
$\displaystyle \begin{align*} A &= \int_0^a{f\left( x \right) \,\mathrm{d}x} \\ &= \int_0^2{ x^2\,\mathrm{d}x } \\ &= \left[ \frac{x^3}{3} \right] _0^2 \\ &= \frac{2^3}{3} - \frac{0^3}{3} \\ &= \frac{8}{3} \,\textrm{units}^2 \end{align*}$
(g)
$\displaystyle \begin{align*} \textrm{Error} &= \frac{66}{25} - \frac{8}{3} \\ &= \frac{198}{75} - \frac{200}{75} \\ &= -\frac{2}{75} \end{align*}$
So the percentage error is
$\displaystyle \begin{align*} \frac{-\frac{2}{75}}{\frac{8}{3}} \cdot 100\% &= -\frac{2}{75} \cdot \frac{3}{8} \cdot 100\% \\ &= -\frac{600\%}{600} \\ &= -1\% \end{align*}$
View attachment 7141
(a) Since a = 2, that means $\displaystyle \begin{align*} \Delta x = \frac{2}{5} \end{align*}$
(b)
$\displaystyle \begin{align*} f \left( \frac{7\,\Delta x}{2} \right) &= f \left( \frac{7}{5} \right) \\ &= \left( \frac{7}{5} \right) ^2 \\ &= \frac{49}{25} \end{align*}$
(c)
$\displaystyle \begin{align*} A_4 &= f \left( \frac{7}{5} \right) \cdot \Delta x \\ &= \frac{49}{25} \cdot \frac{2}{5} \\ &= \frac{98}{125} \,\textrm{units}^2 \end{align*}$
(d)
$\displaystyle \begin{align*} A_i &= f \left( x_i \right) \cdot \Delta x \end{align*}$
(e)
$\displaystyle \begin{align*} S_5 &= \sum_{i = 1}^5{ \left[ f \left( x_i \right) \cdot \Delta x \right] } \\ &= f \left( x_1 \right) \cdot \Delta x + f \left( x_2 \right) \cdot \Delta x + f \left( x_3 \right) \cdot \Delta x + f \left( x_4 \right) \cdot \Delta x + f \left( x_5 \right) \cdot \Delta x \\ &= \left( \frac{1}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{3}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{5}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{7}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{9}{5} \right) ^2 \cdot \frac{2}{5} \\ &= \frac{1}{25} \cdot \frac{2}{5} + \frac{9}{25} \cdot \frac{2}{5} + \frac{25}{25} \cdot \frac{2}{5} + \frac{49}{25} \cdot \frac{2}{5} + \frac{81}{25} \cdot \frac{2}{5} \\ &= \frac{2}{125} + \frac{18}{125} + \frac{50}{125} + \frac{98}{125} + \frac{162}{125} \\ &= \frac{330}{125} \\ &= \frac{66}{25} \,\textrm{units}^2 \end{align*}$
(f)
$\displaystyle \begin{align*} A &= \int_0^a{f\left( x \right) \,\mathrm{d}x} \\ &= \int_0^2{ x^2\,\mathrm{d}x } \\ &= \left[ \frac{x^3}{3} \right] _0^2 \\ &= \frac{2^3}{3} - \frac{0^3}{3} \\ &= \frac{8}{3} \,\textrm{units}^2 \end{align*}$
(g)
$\displaystyle \begin{align*} \textrm{Error} &= \frac{66}{25} - \frac{8}{3} \\ &= \frac{198}{75} - \frac{200}{75} \\ &= -\frac{2}{75} \end{align*}$
So the percentage error is
$\displaystyle \begin{align*} \frac{-\frac{2}{75}}{\frac{8}{3}} \cdot 100\% &= -\frac{2}{75} \cdot \frac{3}{8} \cdot 100\% \\ &= -\frac{600\%}{600} \\ &= -1\% \end{align*}$