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Suppose the rate of growth of the fly population is given by g ( t ) = e 0.01 t , and the initial fly population is 100 flies. How many flies are in the population after 15 days?

There are 116 flies.

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Evaluating a definite integral using substitution

Evaluate the definite integral using substitution: 1 2 e 1 / x x 2 d x .

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x , then bring the x 2 in the denominator up to the numerator using a negative exponent. We have

1 2 e 1 / x x 2 d x = 1 2 e x −1 x −2 d x .

Let u = x −1 , the exponent on e . Then

d u = x −2 d x d u = x −2 d x .

Bringing the negative sign outside the integral sign, the problem now reads

e u d u .

Next, change the limits of integration:

u = ( 1 ) −1 = 1 u = ( 2 ) −1 = 1 2 .

Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,

1 1 / 2 e u d u = 1 / 2 1 e u d u = e u | 1 / 2 1 = e e 1 / 2 = e e .
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Evaluate the definite integral using substitution: 1 2 1 x 3 e 4 x −2 d x .

1 2 1 x 3 e 4 x −2 d x = 1 8 [ e 4 e ]

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Integrals involving logarithmic functions

Integrating functions of the form f ( x ) = x −1 result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as f ( x ) = ln x and f ( x ) = log a x , are also included in the rule.

Rule: integration formulas involving logarithmic functions

The following formulas can be used to evaluate integrals involving logarithmic functions.

x −1 d x = ln | x | + C ln x d x = x ln x x + C = x ( ln x 1 ) + C log a x d x = x ln a ( ln x 1 ) + C

Finding an antiderivative involving ln x

Find the antiderivative of the function 3 x 10 .

First factor the 3 outside the integral symbol. Then use the u −1 rule. Thus,

3 x 10 d x = 3 1 x 10 d x = 3 d u u = 3 ln | u | + C = 3 ln | x 10 | + C , x 10 .

See [link] .

A graph of the function f(x) = 3 / (x – 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.
The domain of this function is x 10 .
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Find the antiderivative of 1 x + 2 .

ln | x + 2 | + C

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Finding an antiderivative of a rational function

Find the antiderivative of 2 x 3 + 3 x x 4 + 3 x 2 .

This can be rewritten as ( 2 x 3 + 3 x ) ( x 4 + 3 x 2 ) −1 d x . Use substitution. Let u = x 4 + 3 x 2 , then d u = 4 x 3 + 6 x . Alter du by factoring out the 2. Thus,

d u = ( 4 x 3 + 6 x ) d x = 2 ( 2 x 3 + 3 x ) d x 1 2 d u = ( 2 x 3 + 3 x ) d x .

Rewrite the integrand in u :

( 2 x 3 + 3 x ) ( x 4 + 3 x 2 ) −1 d x = 1 2 u −1 d u .

Then we have

1 2 u −1 d u = 1 2 ln | u | + C = 1 2 ln | x 4 + 3 x 2 | + C .
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Finding an antiderivative of a logarithmic function

Find the antiderivative of the log function log 2 x .

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

log 2 x d x = x ln 2 ( ln x 1 ) + C .
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Find the antiderivative of log 3 x .

x ln 3 ( ln x 1 ) + C

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[link] is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.

Evaluating a definite integral

Find the definite integral of 0 π / 2 sin x 1 + cos x d x .

We need substitution to evaluate this problem. Let u = 1 + cos x , , so d u = sin x d x . Rewrite the integral in terms of u , changing the limits of integration as well. Thus,

u = 1 + cos ( 0 ) = 2 u = 1 + cos ( π 2 ) = 1 .

Then

0 π / 2 sin x 1 + cos x = 2 1 u −1 d u = 1 2 u −1 d u = ln | u | | 1 2 = [ ln 2 ln 1 ] = ln 2.
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Key concepts

  • Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and decay.
  • Substitution is often used to evaluate integrals involving exponential functions or logarithms.

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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