Consider the integral 8 (x2+1) dx 0 (a) Estimate the area under the curve using a left-hand sum with n = 4. Is this sum an overe
stimate or an underestimate of the true value? overestimate underestimate (b) Estimate the area under the curve using a right-hand sum with n = 4. Is this sum an overestimate or an underestimate of the true value? overestimate underestimate
1 answer:
Answer:
(a) 120 square units, underestimate
(b) 248 square units, overestimate
Step-by-step explanation:
(a) <u>left sum</u>
The left sum is the sum of the areas of the rectangles whose width is the total interval width (8-0) divided by the number of divisions (n=4). The height of each rectangle is the function value at its left edge.
We can compute the sum by adding the function values and multiplying that total by the width of the rectangles:
left sum = (1 + 5 + 17 + 37)×2 = 60×2 = 120 . . . square units
The curve is increasing throughout the interval of interest, so the left sum underestimates the area under the curve.
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(b) <u>right sum</u>
The rectangles whose area is the right sum are shown in the attachment, along with the table of function values. The right sum is computed the same way as the left sum, but using the function value on the right side of each subinterval.
right sum = (5 + 17 + 37 + 65)×2 = 124×2 = 248 . . . square units
The curve is increasing throughout the interval of interest, so the right sum overestimates the area under the curve.
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The actual area under the curve on the interval [0, 8] is 178 2/3, just slightly less than the average of the left- and right- sums.
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