Answer :
It just means to approximate the area with rectangles.
First step is to find out what your x values are. The length of the interval from 2 to 14 is 12. Divide it into six equal parts, which is 2. Then the x points for the left endpoints are: 2, 4, 6, 8, 10, 12 (and the last rectangle goes from 12 to 14), but you don't use 14 unless you are doing right endpoints or trapezoids or something.)
Next step is to find the areas of the rectangles. It is just the base times the height, and the height is the function value of the x value, since the rectangle sits on the x axis and extends vertically up or down to the graph of the function. In integrals an area below the axis is considered to contribute a negative quantity to the integral, so we will do the same for rectangles that extend below the x axis.
f(2) = 3 - (1/2) 2 = 3 - 1 = 2
f(4) = 3 - 4/2 = 1
f(6) = 3 - 6/2 = 0
f(8) = 3 - 8/2 = -1
f(10) = 3 - 10/2 = -2
f(12) = 3 - 12/2 = -3
That should be in a table. Next to it you write the area, multiplied by -1 if f(x) < 0 (I'm saying that because areas are always positive.)
That gives 4, 2, 0, -2, -4, -6 for the six rectangles. Add all that up and it mostly cancels but you get -6 for the final answer.
I have explained as I went along. You do not have to explain so much when you give the answer on a test or homework. Just the computations and not so many words.
First step is to find out what your x values are. The length of the interval from 2 to 14 is 12. Divide it into six equal parts, which is 2. Then the x points for the left endpoints are: 2, 4, 6, 8, 10, 12 (and the last rectangle goes from 12 to 14), but you don't use 14 unless you are doing right endpoints or trapezoids or something.)
Next step is to find the areas of the rectangles. It is just the base times the height, and the height is the function value of the x value, since the rectangle sits on the x axis and extends vertically up or down to the graph of the function. In integrals an area below the axis is considered to contribute a negative quantity to the integral, so we will do the same for rectangles that extend below the x axis.
f(2) = 3 - (1/2) 2 = 3 - 1 = 2
f(4) = 3 - 4/2 = 1
f(6) = 3 - 6/2 = 0
f(8) = 3 - 8/2 = -1
f(10) = 3 - 10/2 = -2
f(12) = 3 - 12/2 = -3
That should be in a table. Next to it you write the area, multiplied by -1 if f(x) < 0 (I'm saying that because areas are always positive.)
That gives 4, 2, 0, -2, -4, -6 for the six rectangles. Add all that up and it mostly cancels but you get -6 for the final answer.
I have explained as I went along. You do not have to explain so much when you give the answer on a test or homework. Just the computations and not so many words.
