4.4 Areas and integrals 243
and this is equivalent to the relation
!6)
At
Jim 09 = 1 - t2 = 1/1 - sine B = cos 6
ne-.o
or to the firstof the formulas
(7) dBsin0=coscosdB^0 -sing.
The second follows from the first and the calculation
(8) dxcos 0 = dx sin (2-O)=- cos (2 -6)= - sin B.
21 Show how formulas of the preceding problem can be used to obtain the
integration formulas
f a=- x2dx=-Ix a2-x2+ a2 sin x+c
a
f
1/az1- x2
dx = sins a + c.
Then keep in contact with the external world by finding these formulas in your
book of tables.
22 Sketch a few figures which illustrate applications of the following fact.
If f is integrable (and hence bounded) over a 5 x < b, we can choose a positive
constant B such that f(x) + B > 0 when a 5 x <- b and write the formula
lbf(x) dx=fab [f(x) + B] dx-fab B dx,
which shows that fabf(x) dx is the result of subtracting the area of a rectangle
from the area of the set of points (x,y) for which a 5 x <--_ b and -B 5 y < f(x).
Remark: This problem and the next provide ways of reducing questions involving
integrals to questions involving integrals with nonnegative integrands.
23 Sketch a few figures which illustrate applications of the following fact.
If f is integrable (and hence bounded) over a 5 x S b, so also are the functions
g and Is defined by
g(x) _
1f(x)I
2
f(x) h(x) =If(x)I 2 f(x)
Moreover, g(x)? 0, h(x) Z 0, f(x) = g(x) - h(x), If(x)I = g(x) + h(x), and
f.
b
AX) dx =f ab g(x) dx- f.'h(x) dx
fab If (x) I dx=f ab g(x) dx + f ab h(x) dx.