Problem : Consider the function f (x) = x ² + 1 on the interval [0, 2] . Using four subdivisions, find the left-hand approximation, L ₄ , of the area under the curve of f on the interval indicated.

Solution for Problem 1 >>

Δx		= = =
L 4		= f (0) + f () + f (1) + f ()
		= 1 + +2 +
		= =

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Problem : For the same function, using four subdivisions, find the right-hand sum, R ₄ .

Solution for Problem 2 >>

R 4		= f () + f (1) + f () + f (2)
		= +2 + + 5
		= =

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Problem : For the same function, using four subdivisions, find the midpoint sum, M ₄ .

Solution for Problem 3 >>

M 4		= f () + f () + f () + f ()
		= + + +
		= =

Notice that on the interval in question, f is a strictly increasing function. If f is increasing on an interval, then L _n < M _n < R _n . If f is decreasing on an interval, then R _n < M _n < L _n .

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Problem : Find

f (x k)Δx for f (x) = 2x on [0, 2]

Solution for Problem 4 >>

To solve this problem, notice that the graph of f (x) is a line, and the area in question is in the shape of a right triangle with base 2 and height f (2) = 4 . So, the limit of the right hand sum, which is the area under the curve, is

(2)(4) = 4

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Problem : Find

f (x k)Δx for f (x) = on [0, 3]

Solution for Problem 5 >>

To solve this problem, notice that the graph of f (x) is a semicircle of radius 3 centered at the origin. The interval [0, 3] contains a section of the curve that is equivalent to a quarter-circle. Thus, the area under the curve is equal to one-fourth the area of a circle with radius 3, or Π(3²) = Π .

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