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Key equations

  • Integrals of Exponential Functions
    e x d x = e x + C
    a x d x = a x ln a + C
  • Integration Formulas 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

In the following exercises, compute each indefinite integral.

e −3 x d x

−1 3 e −3 x + C

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3 x d x

3 x ln 3 + C

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2 x d x

ln ( x 2 ) + C

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In the following exercises, find each indefinite integral by using appropriate substitutions.

d x x ( ln x ) 2

1 ln x + C

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d x x ln x ( x > 1 )

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d x x ln x ln ( ln x )

ln ( ln ( ln x ) ) + C

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cos x x sin x x cos x d x

ln ( x cos x ) + C

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ln ( sin x ) tan x d x

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ln ( cos x ) tan x d x

1 2 ( ln ( cos ( x ) ) ) 2 + C

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x 2 e x 3 d x

e x 3 3 + C

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e tan x sec 2 x d x

e tan x + C

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e ln ( 1 t ) 1 t d t

t + C

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In the following exercises, verify by differentiation that ln x d x = x ( ln x 1 ) + C , then use appropriate changes of variables to compute the integral.

ln x d x ( H i n t : ln x d x = 1 2 x ln ( x 2 ) d x )

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x 2 ln 2 x d x

1 9 x 3 ( ln ( x 3 ) 1 ) + C

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ln x x 2 d x ( H i n t : Set u = 1 x . )

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ln x x d x ( H i n t : Set u = x . )

2 x ( ln x 2 ) + C

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Write an integral to express the area under the graph of y = 1 t from t = 1 to e x and evaluate the integral.

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Write an integral to express the area under the graph of y = e t between t = 0 and t = ln x , and evaluate the integral.

0 ln x e t d t = e t | 0 ln x = e ln x e 0 = x 1

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In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

sin ( 3 x ) cos ( 3 x ) sin ( 3 x ) + cos ( 3 x ) d x

1 3 ln ( sin ( 3 x ) + cos ( 3 x ) )

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x sin ( x 2 ) cos ( x 2 ) d x

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x csc ( x 2 ) d x

1 2 ln | csc ( x 2 ) + cot ( x 2 ) | + C

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ln ( cos x ) tan x d x

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ln ( csc x ) cot x d x

1 2 ( ln ( csc x ) ) 2 + C

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e x e x e x + e x d x

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In the following exercises, evaluate the definite integral.

1 2 1 + 2 x + x 2 3 x + 3 x 2 + x 3 d x

1 3 ln ( 26 7 )

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0 π / 3 sin x cos x sin x + cos x d x

ln ( 3 1 )

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π / 6 π / 2 csc x d x

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π / 4 π / 3 cot x d x

1 2 ln 3 2

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In the following exercises, integrate using the indicated substitution.

x x 100 d x ; u = x 100

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y 1 y + 1 d y ; u = y + 1

y 2 ln | y + 1 | + C

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1 x 2 3 x x 3 d x ; u = 3 x x 3

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sin x + cos x sin x cos x d x ; u = sin x cos x

ln | sin x cos x | + C

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e 2 x 1 e 2 x d x ; u = e 2 x

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ln ( x ) 1 ( ln x ) 2 x d x ; u = ln x

1 3 ( 1 ( ln x 2 ) ) 3 / 2 + C

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In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R 50 and solve for the exact area.

[T] y = e x over [ 0 , 1 ]

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[T] y = e x over [ 0 , 1 ]

Exact solution: e 1 e , R 50 = 0.6258 . Since f is decreasing, the right endpoint estimate underestimates the area.

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[T] y = ln ( x ) over [ 1 , 2 ]

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[T] y = x + 1 x 2 + 2 x + 6 over [ 0 , 1 ]

Exact solution: 2 ln ( 3 ) ln ( 6 ) 2 , R 50 = 0.2033 . Since f is increasing, the right endpoint estimate overestimates the area.

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[T] y = 2 x over [ −1 , 0 ]

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[T] y = 2 x over [ 0 , 1 ]

Exact solution: 1 ln ( 4 ) , R 50 = −0.7164 . Since f is increasing, the right endpoint estimate overestimates the area (the actual area is a larger negative number).

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In the following exercises, f ( x ) 0 for a x b . Find the area under the graph of f ( x ) between the given values a and b by integrating.

f ( x ) = log 10 ( x ) x ; a = 10 , b = 100

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f ( x ) = log 2 ( x ) x ; a = 32 , b = 64

11 2 ln 2

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f ( x ) = 2 x ; a = 1 , b = 2

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f ( x ) = 2 x ; a = 3 , b = 4

1 ln ( 65 , 536 )

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Find the area under the graph of the function f ( x ) = x e x 2 between x = 0 and x = 5 .

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Compute the integral of f ( x ) = x e x 2 and find the smallest value of N such that the area under the graph f ( x ) = x e x 2 between x = N and x = N + 10 is, at most, 0.01.

N N + 1 x e x 2 d x = 1 2 ( e N 2 e ( N + 1 ) 2 ) . The quantity is less than 0.01 when N = 2 .

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Find the limit, as N tends to infinity, of the area under the graph of f ( x ) = x e x 2 between x = 0 and x = 5 .

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Show that a b d t t = 1 / b 1 / a d t t when 0 < a b .

a b d x x = ln ( b ) ln ( a ) = ln ( 1 a ) ln ( 1 b ) = 1 / b 1 / a d x x

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Suppose that f ( x ) > 0 for all x and that f and g are differentiable. Use the identity f g = e g ln f and the chain rule to find the derivative of f g .

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Use the previous exercise to find the antiderivative of h ( x ) = x x ( 1 + ln x ) and evaluate 2 3 x x ( 1 + ln x ) d x .

23

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Show that if c > 0 , then the integral of 1 / x from ac to bc ( 0 < a < b ) is the same as the integral of 1 / x from a to b .

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The following exercises are intended to derive the fundamental properties of the natural log starting from the definition ln ( x ) = 1 x d t t , using properties of the definite integral and making no further assumptions.

Use the identity ln ( x ) = 1 x d t t to derive the identity ln ( 1 x ) = ln x .

We may assume that x > 1 , so 1 x < 1 . Then, 1 1 / x d t t . Now make the substitution u = 1 t , so d u = d t t 2 and d u u = d t t , and change endpoints: 1 1 / x d t t = 1 x d u u = ln x .

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Use a change of variable in the integral 1 x y 1 t d t to show that ln x y = ln x + ln y for x , y > 0 .

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Use the identity ln x = 1 x d t x to show that ln ( x ) is an increasing function of x on [ 0 , ) , and use the previous exercises to show that the range of ln ( x ) is ( , ) . Without any further assumptions, conclude that ln ( x ) has an inverse function defined on ( , ) .

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Pretend, for the moment, that we do not know that e x is the inverse function of ln ( x ) , but keep in mind that ln ( x ) has an inverse function defined on ( , ) . Call it E . Use the identity ln x y = ln x + ln y to deduce that E ( a + b ) = E ( a ) E ( b ) for any real numbers a , b .

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Pretend, for the moment, that we do not know that e x is the inverse function of ln x , but keep in mind that ln x has an inverse function defined on ( , ) . Call it E . Show that E ' ( t ) = E ( t ) .

x = E ( ln ( x ) ) . Then, 1 = E ' ( ln x ) x or x = E ' ( ln x ) . Since any number t can be written t = ln x for some x , and for such t we have x = E ( t ) , it follows that for any t , E ' ( t ) = E ( t ) .

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The sine integral, defined as S ( x ) = 0 x sin t t d t is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x . Show that for k 1 , | S ( 2 π k ) S ( 2 π ( k + 1 ) ) | 1 k ( 2 k + 1 ) π . ( H i n t : sin ( t + π ) = sin t )

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[T] The normal distribution in probability is given by p ( x ) = 1 σ 2 π e ( x μ ) 2 / 2 σ 2 , where σ is the standard deviation and μ is the average. The standard normal distribution in probability, p s , corresponds to μ = 0 and σ = 1 . Compute the left endpoint estimates R 10 and R 100 of −1 1 1 2 π e x 2 / 2 d x .

R 10 = 0.6811 , R 100 = 0.6827
A graph of the function f(x) = .5 * ( sqrt(2)*e^(-.5x^2)) / sqrt(pi). It is a downward opening curve that is symmetric across the y axis, crossing at about (0, .4). It approaches 0 as x goes to positive and negative infinity. Between 1 and -1, ten rectangles are drawn for a right endpoint estimate of the area under the curve.

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[T] Compute the right endpoint estimates R 50 and R 100 of −3 5 1 2 2 π e ( x 1 ) 2 / 8 .

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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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