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Key concepts

  • The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value c such that f ( c ) equals the average value of the function. See [link] .
  • The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. See [link] .
  • The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. The total area under a curve can be found using this formula. See [link] .

Key equations

  • Mean Value Theorem for Integrals
    If f ( x ) is continuous over an interval [ a , b ] , then there is at least one point c [ a , b ] such that f ( c ) = 1 b a a b f ( x ) d x .
  • Fundamental Theorem of Calculus Part 1
    If f ( x ) is continuous over an interval [ a , b ] , and the function F ( x ) is defined by F ( x ) = a x f ( t ) d t , then F ( x ) = f ( x ) .
  • Fundamental Theorem of Calculus Part 2
    If f is continuous over the interval [ a , b ] and F ( x ) is any antiderivative of f ( x ) , then a b f ( x ) d x = F ( b ) F ( a ) .

Consider two athletes running at variable speeds v 1 ( t ) and v 2 ( t ) . The runners start and finish a race at exactly the same time. Explain why the two runners must be going the same speed at some point.

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Two mountain climbers start their climb at base camp, taking two different routes, one steeper than the other, and arrive at the peak at exactly the same time. Is it necessarily true that, at some point, both climbers increased in altitude at the same rate?

Yes. It is implied by the Mean Value Theorem for Integrals.

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To get on a certain toll road a driver has to take a card that lists the mile entrance point. The card also has a timestamp. When going to pay the toll at the exit, the driver is surprised to receive a speeding ticket along with the toll. Explain how this can happen.

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Set F ( x ) = 1 x ( 1 t ) d t . Find F ( 2 ) and the average value of F over [ 1 , 2 ] .

F ( 2 ) = −1 ; average value of F over [ 1 , 2 ] is −1 / 2 .

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In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

d d x 1 x e t 2 d t

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d d x 1 x e cos t d t

e cos t

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d d x 3 x 9 y 2 d y

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d d x 4 x d s 16 s 2

1 16 x 2

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d d x 0 x t d t

x d d x x = 1 2

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d d x 0 sin x 1 t 2 d t

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d d x cos x 1 1 t 2 d t

1 cos 2 x d d x cos x = | sin x | sin x

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d d x 1 x t 2 1 + t 4 d t

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d d x 1 x 2 t 1 + t d t

2 x | x | 1 + x 2

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d d x 0 ln x e t d t

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d d x 1 e 2 ln u 2 d u

ln ( e 2 x ) d d x e x = 2 x e x

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The graph of y = 0 x f ( t ) d t , where f is a piecewise constant function, is shown here.

A function with linear segments which goes through the points (0, 0), (1, 3), (2, 2), (3, 0), (4, 3), (5, 3), and (6, 2). The area under the function and above the x axis is shaded.
  1. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, is it equal to zero?
  2. What are the maximum and minimum values of f ?
  3. What is the average value of f ?
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The graph of y = 0 x f ( t ) d t , where f is a piecewise constant function, is shown here.

A graph of a function with linear segments that goes through the points (0, 0), (1, -1), (2, 1), (3, 1), (4, -2), (5, -2), and (6, 0). The area over the function but under the x axis over the interval [0, 1.5] and [3.25, 6] is shaded. The area under the function but over the x axis over the interval [1.5, 3.25] is shaded.
  1. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, is it equal to zero?
  2. What are the maximum and minimum values of f ?
  3. What is the average value of f ?

a. f is positive over [ 1 , 2 ] and [ 5 , 6 ] , negative over [ 0 , 1 ] and [ 3 , 4 ] , and zero over [ 2 , 3 ] and [ 4 , 5 ] . b. The maximum value is 2 and the minimum is −3. c. The average value is 0.

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The graph of y = 0 x ( t ) d t , where is a piecewise linear function, is shown here.

A graph of a function which goes through the points (0, 0), (1, -1), (2, 1), (3, 3), (4, 3.5), (5, 4), and (6, 2). The area over the function and under the x axis over [0, 1.8] is shaded, and the area under the function and over the x axis is shaded.
  1. Over which intervals is positive? Over which intervals is it negative? Over which, if any, is it zero?
  2. Over which intervals is increasing? Over which is it decreasing? Over which, if any, is it constant?
  3. What is the average value of ?
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Questions & Answers

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Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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