User:Egm6341.s11.team5.reiss/HW3

=Problem 8 - Verification of $$\left| E_2 \right|$$= From (meeting 17 page 3)

Given

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\left| \right| \le \frac\int_a^b {\left| {(x - a)\left( {x - \frac{2}} \right)(x - b)} \right|dx} $$ $$
 * $$\displaystyle (8-1)
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Find
Show that Eq (8-1) is
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\displaystyle \begin{align} &=\frac{M_3} \\ &= \frac{M_3} \end{align} $$ $$
 * $$\displaystyle (8-2)
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Solution

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\displaystyle \begin{align} \frac\int_a^b {\left| {(x - a)\left( {x - \frac{2}} \right)(x - b)} \right|dx}&=\frac\int_a^{\frac{2}} {(x - a)\left( {-x + \frac{2}} \right)(-x + b)dx} \\ &+ \frac\int_{\frac{2}}^b {(x - a)\left( {x - \frac{2}} \right)(-x + b)dx} \end{align} $$ $$ Solving the first term
 * $$\displaystyle (8-3)
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\frac\int_a^{\frac{2}} {(x - a)\left( {-x + \frac{2}} \right)(-x + b)dx}$$ $$= \int_a^{\frac{2}}- \frac{2} + \frac{2} - \frac{2} + 2abx - \frac{2} + \frac{2} - \frac{2} + {x^3} $$ $$=\left. { - \frac{2} + \frac{4} - \frac{2} + ab{x^2} - \frac{2} + \frac{4} - \frac{2} + \frac{4}} \right|_a^{\frac{2}}$$ $$\displaystyle (8-4) $$ Evaluating at $$x=\frac{b+a}{2}$$ and $$x=a$$ gives respectively
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$$=- \frac{1}{4}{a^2}{b^2}$$ $$\displaystyle (8-5) $$ Expanding these terms and subtracting gives
 * $$=\frac{1}({a^2} + {b^2}){(a + b)^2} + \frac{3}{(a + b)^4}$$
 * $$=\frac{1}({a^2} + {b^2}){(a + b)^2} + \frac{3}{(a + b)^4}$$
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$$\displaystyle (8-6) $$ Solving the second term
 * $$\frac{1}\left( {{a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}} \right) = \frac{1}{(b - a)^4}$$
 * $$\frac{1}\left( {{a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}} \right) = \frac{1}{(b - a)^4}$$
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$$=\frac{2} - \frac{2} + \frac{2} - 2abx + \frac{2} - \frac{2} + \frac{2} - {x^3}$$ $$=\left. {\frac{2} - \frac{4} + \frac{2} - ab{x^2} + \frac{2} - \frac{4} + \frac{2} - \frac{4}} \right|_{\frac{2}}^b$$ $$\displaystyle (8-7) $$ Evaluating at $$x=b$$ and $$x=\frac{b+a}{2}$$ gives respectively
 * $$\frac\int_{\frac{2}}^b {(x - a)\left( {x - \frac{2}} \right)(-x + b)dx} $$
 * $$\frac\int_{\frac{2}}^b {(x - a)\left( {x - \frac{2}} \right)(-x + b)dx} $$
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$$=- \frac{1}({a^2} + {b^2}){(a + b)^2} - \frac{3}{(a + b)^4}$$ $$\displaystyle (8-8) $$ Expanding these terms and subtracting gives
 * $$= \frac{1}{4}{a^2}{b^2}$$
 * $$= \frac{1}{4}{a^2}{b^2}$$
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$$\displaystyle (8-9) $$ Adding Eq (8-6) and (8-9) as Eq (8-3) shows, we get
 * $$\frac{1}\left( {{a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}} \right) = \frac{1}{(b - a)^4}$$
 * $$\frac{1}\left( {{a^4} - 4{a^3}b + 6{a^2}{b^2} - 4a{b^3} + {b^4}} \right) = \frac{1}{(b - a)^4}$$
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$$\displaystyle (8-10) $$ Then
 * $$\frac{1}{(b - a)^4}$$
 * $$\frac{1}{(b - a)^4}$$
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$$\displaystyle (8-11) $$ Eq (8-11) is equivalent to Eq (8-2) therefor Eq (8-1) is equivalent to Eq (8-2)
 * $$\left( {\frac} \right)\left( {\frac{1}{{(b - a)}^4}} \right) = \frac$$
 * $$\left( {\frac} \right)\left( {\frac{1}{{(b - a)}^4}} \right) = \frac$$
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