724
Chapter 6 | Applications of Integration
∫ 1
u du = ln | u | + C .
Example 6.36 Calculating Integrals Involving Natural Logarithms Calculate the integral ∫ x x 2 +4 dx . Solution Using u -substitution, let u = x 2 +4. Then du =2 xdx and we have ∫ x x 2 +4 dx = 1 2 ∫ 1 u du = 1 2 ln | u | + C = 1 2 ln |
x 2 +4 | + C = 1
2 ln ⎛
⎞ ⎠ + C .
⎝ x 2 +4
Calculate the integral ∫ x 2 x 3 +6
6.36
dx .
Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.
Theorem 6.18: Properties of the Natural Logarithm If a , b >0 and r is a rational number, then i. ln1=0 ii. ln( ab ) = ln a +ln b iii. ln ⎛ ⎝ a b ⎞ ⎠ = ln a −ln b iv. ln( a r ) = r ln a
Proof i. By definition, ln1= ∫ 1 1 1
t dt =0.
ii. We have () ln( ab ) = ∫ 1
ab
ab
a
1 t dt = ∫
1 t dt + ∫
1 t dt .
a
1
Use u -substitution on the last integral in this expression. Let u = t / a . Then du = (1/ a ) dt . Furthermore, when t = a , u =1, and when t = ab , u = b . So we get () ln( ab ) = ∫ 1 a 1 t dt + ∫ a ab 1 t dt = ∫ 1 a 1 t dt + ∫ a ab a t · 1 a dt = ∫ 1 a 1 t dt + ∫ 1 b 1 u du = ln a +ln b .
This OpenStax book is available for free at http://cnx.org/content/col11964/1.12
Made with FlippingBook - professional solution for displaying marketing and sales documents online