trapezoidal approximation of int [2 - cos(x)] dx

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tsh44

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Using the subintervals [1,5], [5,8], and [8,10] what is the trapezoidal approximation to the integral of (2-cos x) dx from 1 to 10.

Since the intervals are of different length I split the graph into 3 trapezids and did the following.

A= 0.5 h (b1 +b2)
= .5 (4) ((2-cos 5) + (2-cos 1))
= 6.353

I did the same for the other two chnaging h from 4 to 3 and to 2 for the last one. I also calcualted the new bases usign the function given and plugging in the appropraite numbers. I ended up with 18. something. The answer should be 17.129. Where did I go wrong?
 
don't know what you've done wrong ... I get 17.129

\(\displaystyle \L f(x) = 2 - \cos{x}\)

\(\displaystyle \L \int_1^{10} f(x) dx \approx 2[f(1)+f(5)] + 1.5[f(5)+f(8)]+1[f(8)+f(10)] = 17.129\)
 
\(\displaystyle \L A = \frac{1}{2}h(b_1 + b_2)\)

your first interval is [1,5] ... h = 4 ... (1/2)h = 2

your second interval is [5,8] ... h = 3 ... (1/2)h = 1.5

your third interval is [8,10] ... h = 2 ... (1/2)h = 1
 
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