<p>Yeah, I have a AP Calculus Question. I missed today, so I really don't know what's going on with the Trapezoidal Rule. Here's a question, if you could, go through the process so i will have an idea what to do.</p>
<p>Integral 0 to 2 , x^(2) dx , n=4 using trapezoidal Rule.</p>
<p>Would be much appreciated</p>
<p>Thanks, </p>
<p>Jerod</p>
<p>The area of a trapezoid is (b1+b2)/2*h. In the context of your problem, the h will lie on the x-axis. You are given the interval of integration and the number of subintervals desired. To find the height length, find the length of the main interval and divide by the number of subintervals. The bases are perpendicular to the x-axis and extend to the curve. Hence, they go from 0 to f(xi), where xi is a bound of a subinterval; so, the length of the a base is f(xi).</p>
<p>You are approximating the area under the curve f(x)=x^2 from 0 to 2 and are using 4 subintervals. The total interval length is 2. Thus each subinterval will be 2/4=1/2 units long.</p>
<p>Here are your subintervals (I will reference the interval numbers later):
1) [0,0+1/2]
2) [1/2,1/2+1/2]
3) [1,1+1/2]
4) [3/2,3/2+1/2]</p>
<p>You should take note that the upper bound of a subinterval is equal to the lower bound of the next subinterval.</p>
<p>Let find the area for (1). The height is 1/2. The first base is f(0), the second is f(1/2). (don't evaluate these values yet). So, the area is 1/2<em>[f(0)+f(1/2)]</em>1/2</p>
<p>Area for (2): Height=1/2. First base is f(1/2), second is f(1).
Area=1/2<em>[f(1/2)+f(1)]</em>1/2</p>
<p>Area for (3)=1/2<em>[f(1)+f(3/2)]</em>1/2</p>
<p>Area for (4)=1/2<em>[f(3/2)+f(2)]</em>1/2</p>
<p>If you add these together, it will be the approxiamation you seek:
First factor you the 1/2<em>1/2,
total area=1/2</em>1/2<em>[f(0)+f(1/2)+f(1/2)+f(1)+f(1)+f(3/2)+f(3/2)+f(2)]
=1/4</em>[f(0)+2<em>f(1/2)+2</em>f(1)+2*f(3/2)+f(2)]</p>
<p>Hope this helps.</p>