_{n}= [(b-a)/2n][f(x

_{0})+2f(x

_{1})+2f(x

_{2})+2f(x

_{3})+...+f(x

_{n})]

_{0}=0, b=x

_{n}=pi, n=4

_{4}= (pi/8)[sin(0)+2sin(pi/4)+2sin(pi/2)+2sin(3pi/2)+sin(pi)]

_{4}

The trapezoidal approximation of ∫ sin x dx from 0 to pi using 4 equal subdivisions of the interval of integration is

a) pi/2

b) pi

c) (pi/4)(1+sqrt(2))

d) (pi/2)(1+sqrt(2))

e) (pi/4)(2+sqrt(2))

Answer: C

Please show all your work.

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The trapezoidal approximation for the definite integral of f(x) between the points a and b is:

T_{n} = [(b-a)/2n][f(x_{0})+2f(x_{1})+2f(x_{2})+2f(x_{3})+...+f(x_{n})]

Where a and b are the limits of the integration and n = the number of trapezoids used in the estimate. In your case:

f(x)= sinx

a=x_{0}=0, b=x_{n}=pi, n=4

T_{4} = (pi/8)[sin(0)+2sin(pi/4)+2sin(pi/2)+2sin(3pi/2)+sin(pi)]

Solve for T_{4}

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