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Proof

Applying the definition of the derivative, we have

F ( x ) = lim h 0 F ( x + h ) F ( x ) h = lim h 0 1 h [ a x + h f ( t ) d t a x f ( t ) d t ] = lim h 0 1 h [ a x + h f ( t ) d t + x a f ( t ) d t ] = lim h 0 1 h x x + h f ( t ) d t .

Looking carefully at this last expression, we see 1 h x x + h f ( t ) d t is just the average value of the function f ( x ) over the interval [ x , x + h ] . Therefore, by [link] , there is some number c in [ x , x + h ] such that

1 h x x + h f ( x ) d x = f ( c ) .

In addition, since c is between x and h , c approaches x as h approaches zero. Also, since f ( x ) is continuous, we have lim h 0 f ( c ) = lim c x f ( c ) = f ( x ) . Putting all these pieces together, we have

F ( x ) = lim h 0 1 h x x + h f ( x ) d x = lim h 0 f ( c ) = f ( x ) ,

and the proof is complete.

Finding a derivative with the fundamental theorem of calculus

Use the [link] to find the derivative of

g ( x ) = 1 x 1 t 3 + 1 d t .

According to the Fundamental Theorem of Calculus, the derivative is given by

g ( x ) = 1 x 3 + 1 .
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Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of g ( r ) = 0 r x 2 + 4 d x .

g ( r ) = r 2 + 4

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Using the fundamental theorem and the chain rule to calculate derivatives

Let F ( x ) = 1 x sin t d t . Find F ( x ) .

Letting u ( x ) = x , we have F ( x ) = 1 u ( x ) sin t d t . Thus, by the Fundamental Theorem of Calculus and the chain rule,

F ( x ) = sin ( u ( x ) ) d u d x = sin ( u ( x ) ) · ( 1 2 x −1 / 2 ) = sin x 2 x .
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Let F ( x ) = 1 x 3 cos t d t . Find F ( x ) .

F ( x ) = 3 x 2 cos x 3

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Using the fundamental theorem of calculus with two variable limits of integration

Let F ( x ) = x 2 x t 3 d t . Find F ( x ) .

We have F ( x ) = x 2 x t 3 d t . Both limits of integration are variable, so we need to split this into two integrals. We get

F ( x ) = x 2 x t 3 d t = x 0 t 3 d t + 0 2 x t 3 d t = 0 x t 3 d t + 0 2 x t 3 d t .

Differentiating the first term, we obtain

d d x [ 0 x t 3 d t ] = x 3 .

Differentiating the second term, we first let u ( x ) = 2 x . Then,

d d x [ 0 2 x t 3 d t ] = d d x [ 0 u ( x ) t 3 d t ] = ( u ( x ) ) 3 d u d x = ( 2 x ) 3 · 2 = 16 x 3 .

Thus,

F ( x ) = d d x [ 0 x t 3 d t ] + d d x [ 0 2 x t 3 d t ] = x 3 + 16 x 3 = 15 x 3 .
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Let F ( x ) = x x 2 cos t d t . Find F ( x ) .

F ( x ) = 2 x cos x 2 cos x

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Fundamental theorem of calculus, part 2: the evaluation theorem

The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.

After finding approximate areas by adding the areas of n rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.

The 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 ) .

We often see the notation F ( x ) | a b to denote the expression F ( b ) F ( a ) . We use this vertical bar and associated limits a and b to indicate that we should evaluate the function F ( x ) at the upper limit (in this case, b ), and subtract the value of the function F ( x ) evaluated at the lower limit (in this case, a ).

Practice Key Terms 4

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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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