Calculus Question, help please!

Question

Estimate the area under the graph of  over the interval  using four approximating rectangles and right endpoints.

 

Repeat the approximation using left endpoints.

 

Answer 1

Ln: (Left Reimann Sum)

Area of 1st rectangle = (3/4)(1/3) = 1/4

Area of 2nd rectangle = (3/4)(4/15) = 1/5

Area of 3rd rectangle = (3/4)(2/9) = 1/6

Area of 4th rectangle: (3/4)(4/21) = 1/7

Total = 1/4 + 1/5 + 1/6 + 1/7 = 319/420 or approx 0.7595

For Rn, (Right Reimann Sum), move the tops of each rectangle down so the upper right corner touches the point on the function and redo the sum:

Area of 1st rectangle = (3/4)(4/15) = 1/5

Area of 2nd rectangle = (3/4)(2/9) = 1/6

Area of 3rd rectangle = (3/4)(4/21) = 1/7

Area of 4th rectangle: (3/4)(1/6) = 1/8

Total = 1/5 + 1/6 + 1/7 + 1/8 = 533/840 or approx 0.6345

Answer 2

First, you divide the interval into 4 equal intervals, so since the interval has size 3, each would be 3/4 wide, so they would be [-1,-1/4],[-1/4,1/2],[1/2,5/4],[5/4,2]. Next to calculate the left endpoint approximation, you approximate the average value of the function on each interval by the value of the function at the left endpoint and you use that as the height of a rectangle where the width is the width of the interval, 3/4. Therefore, for example, the first rectangle would have height f(-1) for the left endpoint approximation, the second would have height f(-1/4), and so on. You then multiply the width of 3/4 by the height and sum up the areas of each of these rectangles.For the right endpoint approximation, you instead utilize the value of the function at the right endpoint to approximate the average value. Thus, for the first rectangle, it would instead have a height of f(-1/4), while the second would have a height of f(1/2), the third f(5/4) and the fourth f(2). You again multiply the width of 3/4 by the heights for each rectangle and sum them up to find the approximation.