User:Egm6341.s10.team4.anandankala/HW2-9


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 * $$E_n:=I_b-I_n=\displaystyle\int^b_a \underbrace{[f(x)-f_n(x)]}_{e_n(x)} \,dx$$
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 * $$|E_n|\leq\displaystyle\int^b_a |e_n(x)| \,dx$$
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 * $$e_n(x)=\frac{q_{n+1}(x)}{(n+1)!}f^{n+1}(\zeta)$$       $$\zeta\in[a,b]$$
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 * $$\Rightarrow |E_n|\leq \frac{M_{n+1}}{(n+1)!}\displaystyle\int^b_a |q_{n+1}(x)| \,dx $$
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 * $$Where M_(n+1):= max f^{n+1}(\zeta)$$
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We are considering the case of simple trapezoidal rule where n=1 and $$q_2(x)=(x-x_0)(x-x_1)=(x-a)(x-b).$$


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 * $$ |E_1|\leq \frac{M_2}{2!}\displaystyle\int^b_a \underbrace{|q_2(x)|}_{\geq0} \,dx $$
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 * $$ |E_1|\leq \frac{M_2}{2!}\displaystyle\int^b_a \underbrace{(x-a)}_{\geq0}\underbrace{(b-x)}_{\geq0} \,dx $$
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 * $$ |E_1|\leq \frac{M_2}{2!}\displaystyle\int^b_a (bx+ax-x^2-ab) \,dx $$
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 * $$ |E_1|\leq \frac{M_2}{2!} [\frac{bx^2}{2}+\frac{ax^2}{2}-\frac{x^3}{3}-abx]^b{_a} $$
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 * $$ |E_1|\leq \frac{M_2}{2!} [(\frac{b^3}{2}+\frac{ab^2}{2}-\frac{b^3}{3}-ab^2)-(\frac{ba^2}{2}+\frac{a^3}{2}-\frac{a^3}{3}-ba^2)] $$
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 * $$ |E_1|\leq \frac{M_2}{2!} [\frac{b^3}{6}+\frac{3ab^2}{6}-\frac{a^3}{6}-\frac{3a^2b}{6}] $$
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 * $$ |E_1|\leq \frac{M_2}{12} [b^3+3ab^2-a^3-3a^2b] $$
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 * $$ |E_1|\leq \frac{M_2}{12} (b-a)^3 $$
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 * $$ |E_1|\leq \frac{M_2}{12} h^3$$     $$ where h:= b-a $$
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