3. Consider the function f(x) = 3x + 1 on the interval [0, 2]. Let P be the part

Question

3.

Consider the function f(x) = 3x + 1 on the interval [0, 2]. Let P be the partition of (0, 2] consisting of 7 equally spaced points. These points divide [0, 2] into six subintervals of equal width. The first graph below depicts the rectangles obtained by evaluating f at the left endpoints of the subintervals, and the second graph depicts the rectangles obtained by evaluating f at the fight endpoints of the subintervals. Fill in the blanks: The width of each of the six subintervals is _____. (Keep in fraction form). Written as integers or fractions in lowest terms, the 7 points in the partition of [0,2] are Find the value of (f, p), the lower sum of f with respect to P. Show some work. Note that the lower sum is the sum of the areas of the rectangles depicted in the first graph. Find the value of U(f, P), the upper sum of f with respect to P. Show some work. Note that the upper sum is the sum of the areas of the rectangles depicted in the second graph.

Explanation / Answer

a.) width of each partition = (2-0)/(7-1) = 1/3

The 7 points are 0,1/3,2/3,1,4/3,5/3,2

Area of a rectangle from x to x+1/3 is f(x)*1/3

b.) L(f,P)=(f(0)*1/3+f(1/3)*1/3+f(2/3)*1/3+f(1)*1/3+f(4/3)*1/3+f(5/3)*1/3)

=(f(0)+f(1/3)+f(2/3)+f(1)+f(4/3)+f(5/3))*1/3=(1+2+3+4+5+6)*1/3 = 7

b.) U(f,P)=(f(1/3)*1/3+f(2/3)*1/3+f(1)*1/3+f(4/3)*1/3+f(5/3)*1/3+f(2)*1/3)

=(f(1/3)+f(2/3)+f(1)+f(4/3)+f(5/3)+f(2))*1/3=(2+3+4+5+6+7)*1/3 = 9

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