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 October 2nd, 2011, 05:16 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Trapezoidal Rule I keep getting the wrong answer for this question. $\int_0^13cos(x^2)\,dx$ Estimate the integral, using the Trapezoidal Rule with n = 4. (Round your answer to six decimal places.) $\int_a^bf(x)\,dx=T_n=\dfrac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+ . . . +2f(x_{n-1})+f(x_n)]$ $\Delta x= \dfrac{b-a}{n}=\dfrac{1-0}{4}=\dfrac{1}{4}$ $x_0= 0, \hspace{7mm} x_1= \dfrac{1}{4}, \hspace{7mm} x_2=\dfrac{1}{2}, \hspace{7mm} x_3=\dfrac{3}{4}, \hspace{7mm} x_4=1$ $\int_0^13cos(x^2)\,dx \hspace{7mm}=\hspace{7mm} \dfrac{1}{8}[3cos((0)^2)+3cos((\dfrac{1}{4})^2)+3cos((\dfrac{1} {2})^2)3cos((\dfrac{3}{4})^2)3cos((1)^2)]$ $=\hspace{7mm} \dfrac{1}{8}[3cos(0)+3cos(\dfrac{1}{8})+3cos(\dfrac{1}{4})3cos( \dfrac{9}{16})+3cos(1)$ $=\dfrac{1}{8}(14.999362" /> 1.874920352
 October 2nd, 2011, 05:28 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Trapexoidal Rule I get: $\int_0\,^1 3\cos$$x^2$$\,dx\approx\frac{3}{8}$$\cos\(0^2$$+2\ cos$$0.25^2$$+2\cos$$0.5^2$$+2\cos$$0.75^2$$+\cos\ (1^2\)\)\approx2.687277$
 October 2nd, 2011, 05:32 PM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Trapexoidal Rule yep, you were right.
 October 2nd, 2011, 05:36 PM #4 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Trapexoidal Rule I forgot the 2.
 October 2nd, 2011, 05:37 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Trapexoidal Rule You weren't multiplying all but the first and last iterations by 2, and it appears you may have been multiplying some iterations instead of adding.

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