and to the left of the line x = 3.
30.23
30.24
and
250
CHAPTER 30
30.21
Let M = 1 + e", du = e" dx. Then
Let
Then 1 =
and
30.22
Find the area of the region in the first quadrant under the curve
Note that x = -3 is a root of x
3 + 21, and, dividing the latter by x — 3. we
Then 1=
which is irreducible. Let
obtain x
2 - 3x + 9,
Let x = -3. Then 1=27 A, A=&. Equate the coefficients of x
2 :
Thus,
Equate the constant coefficients: 1 = 9/4 + 3C, C = ^.
Hence,
But
In
In (x
2 - 3x + 9) +
In
In
Thus, the complete integral is
In
In
In
In
Note that
In
Use the substitution of Problem 29.45, followed by the method of partial fractions. Thus, let z =
Evaluate
Then
In
In
In
Find
Using the same approach as in Problem 30.23, we have:
Let
Then 2 = 2,4,
Then z + I = A(\ + z
2 ) + (z - l)(Bz + C). Let z = l.
Equate coefficients of z
2 : 0 = A + B, B = -A = -l. Equate constant coefficients: 1 = A + C,
Thus,
A(u-l) + Bu. Let u = 0. Then l = -A, A = -l. Let u = 1. Then 1 = B. So
= -ln|w| + ln|M-l| + C = -In (1 + e*) + In e' + C= -ln(l + e*) + x + C.
A(x
2 - 3* + 9) + (x + 3)(Bx + C).
0=A + B, B=-A = -&.
A = \.
C=1-A = Q.
z
2 )] + C, = In |z - 1|
2 - In (1 + z
2 ) + C, = In
+ C l =ln(l -sinx)+ C t .
dz = 2[ln|z-l|- |ln(l +
In
In