# Introduction to Integrals

## Contents

#### Problems for "The Area as a Sum"

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.

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

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

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

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

 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 .

Problem : Find

 f (x k)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 (x k)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 .