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f ( x ) = sin x , a = 0 , b = 2 π

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f ( x ) = cos x , a = 0 , b = 2 π

f ave = 0 ; c = π 2 , 3 π 2

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In the following exercises, approximate the average value using Riemann sums L 100 and R 100 . How does your answer compare with the exact given answer?

[T] y = ln ( x ) over the interval [ 1 , 4 ] ; the exact solution is ln ( 256 ) 3 1 .

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[T] y = e x / 2 over the interval [ 0 , 1 ] ; the exact solution is 2 ( e 1 ) .

L 100 = 1.294 , R 100 = 1.301 ; the exact average is between these values.

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[T] y = tan x over the interval [ 0 , π 4 ] ; the exact solution is 2 ln ( 2 ) π .

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[T] y = x + 1 4 x 2 over the interval [ −1 , 1 ] ; the exact solution is π 6 .

L 100 × ( 1 2 ) = 0.5178 , R 100 × ( 1 2 ) = 0.5294

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In the following exercises, compute the average value using the left Riemann sums L N for N = 1 , 10 , 100 . How does the accuracy compare with the given exact value?

[T] y = x 2 4 over the interval [ 0 , 2 ] ; the exact solution is 8 3 .

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[T] y = x e x 2 over the interval [ 0 , 2 ] ; the exact solution is 1 4 ( e 4 1 ) .

L 1 = 0 , L 10 × ( 1 2 ) = 8.743493 , L 100 × ( 1 2 ) = 12.861728 . The exact answer 26.799 , so L 100 is not accurate.

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[T] y = ( 1 2 ) x over the interval [ 0 , 4 ] ; the exact solution is 15 64 ln ( 2 ) .

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[T] y = x sin ( x 2 ) over the interval [ π , 0 ] ; the exact solution is cos ( π 2 ) 1 2 π .

L 1 × ( 1 π ) = 1.352 , L 10 × ( 1 π ) = −0.1837 , L 100 × ( 1 π ) = −0.2956 . The exact answer 0.303 , so L 100 is not accurate to first decimal.

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Suppose that A = 0 2 π sin 2 t d t and B = 0 2 π cos 2 t d t . Show that A + B = 2 π and A = B .

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Suppose that A = π / 4 π / 4 sec 2 t d t = π and B = π / 4 π / 4 tan 2 t d t . Show that A B = π 2 .

Use tan 2 θ + 1 = sec 2 θ . Then, B A = π / 4 π / 4 1 d x = π 2 .

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Show that the average value of sin 2 t over [ 0 , 2 π ] is equal to 1/2 Without further calculation, determine whether the average value of sin 2 t over [ 0 , π ] is also equal to 1/2.

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Show that the average value of cos 2 t over [ 0 , 2 π ] is equal to 1 / 2 . Without further calculation, determine whether the average value of cos 2 ( t ) over [ 0 , π ] is also equal to 1 / 2 .

0 2 π cos 2 t d t = π , so divide by the length 2 π of the interval. cos 2 t has period π , so yes, it is true.

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Explain why the graphs of a quadratic function (parabola) p ( x ) and a linear function ( x ) can intersect in at most two points. Suppose that p ( a ) = ( a ) and p ( b ) = ( b ) , and that a b p ( t ) d t > a b ( t ) d t . Explain why c d p ( t ) > c d ( t ) d t whenever a c < d b .

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Suppose that parabola p ( x ) = a x 2 + b x + c opens downward ( a < 0 ) and has a vertex of y = b 2 a > 0 . For which interval [ A , B ] is A B ( a x 2 + b x + c ) d x as large as possible?

The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A = b b 2 4 a c 2 a and B = b + b 2 4 a c 2 a .

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Suppose [ a , b ] can be subdivided into subintervals a = a 0 < a 1 < a 2 < < a N = b such that either f 0 over [ a i 1 , a i ] or f 0 over [ a i 1 , a i ] . Set A i = a i 1 a i f ( t ) d t .

  1. Explain why a b f ( t ) d t = A 1 + A 2 + + A N .
  2. Then, explain why | a b f ( t ) d t | a b | f ( t ) | d t .
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Suppose f and g are continuous functions such that c d f ( t ) d t c d g ( t ) d t for every subinterval [ c , d ] of [ a , b ] . Explain why f ( x ) g ( x ) for all values of x .

If f ( t 0 ) > g ( t 0 ) for some t 0 [ a , b ] , then since f g is continuous, there is an interval containing t 0 such that f ( t ) > g ( t ) over the interval [ c , d ] , and then d d f ( t ) d t > c d g ( t ) d t over this interval.

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Suppose the average value of f over [ a , b ] is 1 and the average value of f over [ b , c ] is 1 where a < c < b . Show that the average value of f over [ a , c ] is also 1.

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Suppose that [ a , b ] can be partitioned. taking a = a 0 < a 1 < < a N = b such that the average value of f over each subinterval [ a i 1 , a i ] = 1 is equal to 1 for each i = 1 ,…, N . Explain why the average value of f over [ a , b ] is also equal to 1.

The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, a b f ( t ) d t = a 0 a 1 f ( t ) d t + a 1 a 2 f ( t ) d t + + a N + 1 a N f ( t ) d t = a 0 a 1 1 d t + a 1 a 2 1 d t + + a N + 1 a N 1 d t = ( a 1 a 0 ) + ( a 2 a 1 ) + + ( a N a N 1 ) = a N a 0 = b a . Dividing through by b a gives the desired identity.

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Suppose that for each i such that 1 i N one has i 1 i f ( t ) d t = i . Show that 0 N f ( t ) d t = N ( N + 1 ) 2 .

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Suppose that for each i such that 1 i N one has i 1 i f ( t ) d t = i 2 . Show that 0 N f ( t ) d t = N ( N + 1 ) ( 2 N + 1 ) 6 .

0 N f ( t ) d t = i = 1 N i 1 i f ( t ) d t = i = 1 N i 2 = N ( N + 1 ) ( 2 N + 1 ) 6

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[T] Compute the left and right Riemann sums L 10 and R 10 and their average L 10 + R 10 2 for f ( t ) = t 2 over [ 0 , 1 ] . Given that 0 1 t 2 d t = 0. 33 , to how many decimal places is L 10 + R 10 2 accurate?

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[T] Compute the left and right Riemann sums, L 10 and R 10 , and their average L 10 + R 10 2 for f ( t ) = ( 4 t 2 ) over [ 1 , 2 ] . Given that 1 2 ( 4 t 2 ) d t = 1. 66 , to how many decimal places is L 10 + R 10 2 accurate?

L 10 = 1.815 , R 10 = 1.515 , L 10 + R 10 2 = 1.665 , so the estimate is accurate to two decimal places.

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If 1 5 1 + t 4 d t = 41.7133.. . , what is 1 5 1 + u 4 d u ?

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Estimate 0 1 t d t using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value 0 1 t d t ?

The average is 1 / 2 , which is equal to the integral in this case.

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Estimate 0 1 t d t by comparison with the area of a single rectangle with height equal to the value of t at the midpoint t = 1 2 . How does this midpoint estimate compare with the actual value 0 1 t d t ?

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From the graph of sin ( 2 π x ) shown:

  1. Explain why 0 1 sin ( 2 π t ) d t = 0 .
  2. Explain why, in general, a a + 1 sin ( 2 π t ) d t = 0 for any value of a .
    A graph of the function f(x) = sin(2pi*x) over [0, 2]. The function is shaded over [.7, 1] above the curve and below to x axis, over [1,1.5] under the curve and above the x axis, and over [1.5, 1.7] above the curve and under the x axis. The graph is antisymmetric with respect o t = ½ over [0,1].

a. The graph is antisymmetric with respect to t = 1 2 over [ 0 , 1 ] , so the average value is zero. b. For any value of a , the graph between [ a , a + 1 ] is a shift of the graph over [ 0 , 1 ] , so the net areas above and below the axis do not change and the average remains zero.

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If f is 1-periodic ( f ( t + 1 ) = f ( t ) ) , odd, and integrable over [ 0 , 1 ] , is it always true that 0 1 f ( t ) d t = 0 ?

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If f is 1-periodic and 0 1 f ( t ) d t = A , is it necessarily true that a 1 + a f ( t ) d t = A for all A ?

Yes, the integral over any interval of length 1 is the same.

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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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