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In the following exercises, express the limits as integrals.

lim n i = 1 n ( x i * ) Δ x over [ 1 , 3 ]

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lim n i = 1 n ( 5 ( x i * ) 2 3 ( x i * ) 3 ) Δ x over [ 0 , 2 ]

0 2 ( 5 x 2 3 x 3 ) d x

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lim n i = 1 n sin 2 ( 2 π x i * ) Δ x over [ 0 , 1 ]

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lim n i = 1 n cos 2 ( 2 π x i * ) Δ x over [ 0 , 1 ]

0 1 cos 2 ( 2 π x ) d x

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In the following exercises, given L n or R n as indicated, express their limits as n as definite integrals, identifying the correct intervals.

L n = 1 n i = 1 n i 1 n

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R n = 1 n i = 1 n i n

0 1 x d x

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L n = 2 n i = 1 n ( 1 + 2 i 1 n )

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R n = 3 n i = 1 n ( 3 + 3 i n )

3 6 x d x

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L n = 2 π n i = 1 n 2 π i 1 n cos ( 2 π i 1 n )

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R n = 1 n i = 1 n ( 1 + i n ) log ( ( 1 + i n ) 2 )

1 2 x log ( x 2 ) d x

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In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x -axis.

A graph of three isosceles triangles corresponding to the functions 1 - |x-1| over [0,2], 2 - |x-4| over [2,4], and 3 - |x-9| over [6,12]. The first triangle has endpoints at (0,0), (2,0), and (1,1). The second triangle has endpoints at (2,0), (6,0), and (4,2). The last has endpoints at (6,0), (12,0), and (9,3). All three are shaded.

1 + 2 · 2 + 3 · 3 = 14

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A graph with three shaded parts. The first is a triangle with endpoints at (0, 0), (2, 0), and (1, 1), which corresponds to the function 1 - |x-1| over [0, 2] in quadrant 1. The second is the lower half of a circle with center at (4, 0) and radius two, which corresponds to the function –sqrt(-12 + 8x – x^2) over [2, 6]. The last is a triangle with endpoints at (6, 0), (12, 0), and (9, 3), which corresponds to the function 3 - |x-9| over [6, 12].

1 2 π + 9 = 10 2 π

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In the following exercises, evaluate the integral using area formulas.

2 3 ( 3 x ) d x

The integral is the area of the triangle, 1 2

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−3 3 ( 3 | x | ) d x

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0 6 ( 3 | x 3 | ) d x

The integral is the area of the triangle, 9.

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−2 2 4 x 2 d x

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1 5 4 ( x 3 ) 2 d x

The integral is the area 1 2 π r 2 = 2 π .

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0 12 36 ( x 6 ) 2 d x

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−2 3 ( 3 | x | ) d x

The integral is the area of the “big” triangle less the “missing” triangle, 9 1 2 .

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In the following exercises, use averages of values at the left ( L ) and right ( R ) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.

{ ( 0 , 0 ) , ( 2 , 1 ) , ( 4 , 3 ) , ( 5 , 0 ) , ( 6 , 0 ) , ( 8 , 3 ) } over [ 0 , 8 ]

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{ ( 0 , 2 ) , ( 1 , 0 ) , ( 3 , 5 ) , ( 5 , 5 ) , ( 6 , 2 ) , ( 8 , 0 ) } over [ 0 , 8 ]

L = 2 + 0 + 10 + 5 + 4 = 21 , R = 0 + 10 + 10 + 2 + 0 = 22 , L + R 2 = 21.5

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{ ( −4 , −4 ) , ( −2 , 0 ) , ( 0 , −2 ) , ( 3 , 3 ) , ( 4 , 3 ) } over [ −4 , 4 ]

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{ ( −4 , 0 ) , ( −2 , 2 ) , ( 0 , 0 ) , ( 1 , 2 ) , ( 3 , 2 ) , ( 4 , 0 ) } over [ −4 , 4 ]

L = 0 + 4 + 0 + 4 + 2 = 10 , R = 4 + 0 + 2 + 4 + 0 = 10 , L + R 2 = 10

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Suppose that 0 4 f ( x ) d x = 5 and 0 2 f ( x ) d x = −3 , and 0 4 g ( x ) d x = −1 and 0 2 g ( x ) d x = 2 . In the following exercises, compute the integrals.

0 4 ( f ( x ) + g ( x ) ) d x

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2 4 ( f ( x ) + g ( x ) ) d x

2 4 f ( x ) d x + 2 4 g ( x ) d x = 8 3 = 5

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0 2 ( f ( x ) g ( x ) ) d x

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2 4 ( f ( x ) g ( x ) ) d x

2 4 f ( x ) d x 2 4 g ( x ) d x = 8 + 3 = 11

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0 2 ( 3 f ( x ) 4 g ( x ) ) d x

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2 4 ( 4 f ( x ) 3 g ( x ) ) d x

4 2 4 f ( x ) d x 3 2 4 g ( x ) d x = 32 + 9 = 41

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In the following exercises, use the identity A A f ( x ) d x = A 0 f ( x ) d x + 0 A f ( x ) d x to compute the integrals.

π π sin t 1 + t 2 d t ( H i n t : sin ( t ) = sin ( t ) )

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π π t 1 + cos t d t

The integrand is odd; the integral is zero.

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1 3 ( 2 x ) d x ( Hint: Look at the graph of f .)

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2 4 ( x 3 ) 3 d x ( Hint: Look at the graph of f .)

The integrand is antisymmetric with respect to x = 3 . The integral is zero.

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In the following exercises, given that 0 1 x d x = 1 2 , 0 1 x 2 d x = 1 3 , and 0 1 x 3 d x = 1 4 , compute the integrals.

0 1 ( 1 + x + x 2 + x 3 ) d x

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0 1 ( 1 x + x 2 x 3 ) d x

1 1 2 + 1 3 1 4 = 7 12

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0 1 ( 1 x ) 2 d x

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0 1 ( 1 2 x ) 3 d x

0 1 ( 1 2 x + 4 x 2 8 x 3 ) d x = 1 1 + 4 3 2 = 2 3

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0 1 ( 6 x 4 3 x 2 ) d x

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0 1 ( 7 5 x 3 ) d x

7 5 4 = 23 4

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In the following exercises, use the comparison theorem .

Show that 0 3 ( x 2 6 x + 9 ) d x 0 .

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Show that −2 3 ( x 3 ) ( x + 2 ) d x 0 .

The integrand is negative over [ −2 , 3 ] .

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Show that 0 1 1 + x 3 d x 0 1 1 + x 2 d x .

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Show that 1 2 1 + x d x 1 2 1 + x 2 d x .

x x 2 over [ 1 , 2 ] , so 1 + x 1 + x 2 over [ 1 , 2 ] .

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Show that 0 π / 2 sin t d t π 4 . ( H i n t : sin t 2 t π over [ 0 , π 2 ] )

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Show that π / 4 π / 4 cos t d t π 2 / 4 .

cos ( t ) 2 2 . Multiply by the length of the interval to get the inequality.

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In the following exercises, find the average value f ave of f between a and b , and find a point c , where f ( c ) = f ave .

f ( x ) = x 2 , a = −1 , b = 1

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f ( x ) = x 5 , a = −1 , b = 1

f ave = 0 ; c = 0

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f ( x ) = 4 x 2 , a = 0 , b = 2

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f ( x ) = ( 3 | x | ) , a = −3 , b = 3

3 2 when c = ± 3 2

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Practice Key Terms 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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