Calculus: Analytic Geometry and Calculus, with Vectors

4.4 Areas and integrals 239

two regions R1 and R2. Use partitions and Riemann sums to obtain the

formulas

IR1I = f

0

-2[fi(x) - 12(x)] dx, IR21 fog

[j2(x)

• fi(x)] dx,

1811 + IR21 = f221f1(x) -f2(x)1 dx.

Remark: The widths and heights of rectangles are always positive, and mistakes
in sign are undesirable. When hasty calculations indicate that an area or a

R ,

f,( x ) y

o'

R

1

1 fi(x)

_2

1 -3

'- f2(x) 5

Figure 4.491

population of a city is negative, the calculations

should be examined.

2 The graphs in Figure 4.491 are graphs of

y = 3x and y = x3 - x. Find IR1I + 1R21, this

being the sum of the areas of the two regions

bounded by the graphs. 14ns.: 8.

3 With Figure 4.492 to provide assistance,

make a partition of the interval 0 <- x <- 2 to ob-

tain the area IS11 of the set S1 bounded by the

graphs of y = 0, x = 2, and y = x2. Try to repair

Y

Figure 4.492

the work if the result does not have reasonable agreement with an estimate made
by counting squares and partial squares included by S1. Then interchange the
roles of x and y to find the area IS21 of the set S2 bounded by the graphs of x = 0,
y = x2, and y = 4. Make a partition of the interval 0 < y < 4 and be sure that
the correct integrand and limits of integration( appear in the calculation

IS21 = lim I f(y) Ay=

JvVZf(y)

dy

In this case also, try to repair the work if the result clashes with the result of

counting squares. Finally, have another look at Figure 4.492 and see what

IS1I + IS21 should be.

4 Referring again to Figure 4.492, obtain IS21 by starting with a partition

of the interval 0 S x S 2 and using an estimate of the area of the part of S2

that stands above the interval of length Ax (or Axk) containing the point x (or xk).