Consider the function f (x) = x2 + 1 on the interval [0, 2].
Using four subdivisions, find the left-hand approximation, L4, of the
area under the curve of f on the interval indicated.
|Δx||= = =|
|L4||= f (0) + f () + f (1) + f ()|
|= 1 + +2 +|
For the same function, using four subdivisions, find the right-hand sum, R4.
|R4||= f () + f (1) + f () + f (2)|
|= +2 + + 5|
For the same function, using four subdivisions, find the midpoint sum, M4.
|M4||= 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 Ln < Mn < Rn. If f is decreasing on an interval, then Rn < Mn < Ln.
Problem : Find
|f (xk)Δx for f (x) = 2x on [0, 2]|
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|
Problem : Find
|f (xk)Δx for f (x) = on [0, 3]|
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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