Question 1140897
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I read this as the integral from 1 to 2 of (19-4x).  [(dx..., not dt)]<br>
The interval 1 to 2 with 4 subintervals means the x values of the endpoints of the intervals are 1, 1.25, 1.5, 1.75, and 2.<br>
The function values at those endpoints are 15, 14, 13, 12, and 11.<br>
The integral by the trapezoidal rule is<br>
{{{(1/4)((15+14)/2+(14+13)/2+(13+12)/2+(12+11)/2) = (1/8)(15+28+26+24+11) = 104/8 = 13}}}<br>
The indefinite integral of the function using calculus is<br>
{{{19x-2x^2}}}<br>
The definite integral for the interval 1 to 2 is<br>
{{{(19(2)-2(2^2))-(19(1)-2(1^2)) = (38-8)-(19-2) = 30-17 = 13}}}<br>
The "estimate" using the trapezoidal rule is exactly the same as the result using calculus.  This is as it should be, because the given function is linear.  For any linear function, the trapezoidal rule will give the exact integral.