Problems 39–66 are mixed—some may require use of the integration -by-parts formula along with techniques we have considered earlier; others may require repeated use of the integration-by-parts formula. Assume that g ( x ) > 0 whenever ln g ( x ) is involved. 61. ∫ ( ln x ) 4 x d x
Problems 39–66 are mixed—some may require use of the integration -by-parts formula along with techniques we have considered earlier; others may require repeated use of the integration-by-parts formula. Assume that g ( x ) > 0 whenever ln g ( x ) is involved. 61. ∫ ( ln x ) 4 x d x
Solution Summary: The author explains how to find the value of integrals, such as displaystyleint(mathrm
Problems 39–66 are mixed—some may require use of the integration-by-parts formula along with techniques we have considered earlier; others may require repeated use of the integration-by-parts formula. Assume that g (x) > 0 whenever ln g(x) is involved.
61.
∫
(
ln
x
)
4
x
d
x
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
A function is defined on the interval (-π/2,π/2) by this multipart rule:
if -π/2 < x < 0
f(x) =
a
if x=0
31-tan x
+31-cot x
if 0 < x < π/2
Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0.
a=
b= 3
Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
f(x) = (x + 4x4) 5,
a = -1
lim f(x)
X--1
=
lim
x+4x
X--1
lim
X-1
4
x+4x
5
))"
5
))
by the power law
by the sum law
lim (x) + lim
X--1
4
4x
X-1
-(0,00+(
Find f(-1).
f(-1)=243
lim (x) +
-1 +4
35
4 ([
)
lim (x4)
5
x-1
Thus, by the definition of continuity, f is continuous at a = -1.
by the multiple constant law
by the direct substitution property
Chapter 6 Solutions
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
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