6.4 q3.) use simpson's rule to approximate and use the fundamental theorem of calculus to find the exact value of the definite integral.

When we divide the interval from 1 to 5 into 4 subintervals, each has a width of 1. Simpson's rule has us evaluate the integral as

... integral = (1/3)(f(1) +4f(2) +2f(3) +4f(4) +f(5)) = (1/3)(10 +4·25 +2·46 +4·73 +106)

... integral = (1/3)(600) = 200 . . . . . by Simpson's rule

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The integral evaluates to

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Simpson's rule uses a quadratic interpolation, so evaluates quadratics exactly.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)