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INTEGRALS
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[CHAP. 5
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x2 g x f x dx
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By idealizing the plane region, R, as a volume with uniform density one, the expression f x g x dx stands in for mass and r2 has the coordinate representation x2 . (See Problem 5.25(b) for more details.)
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DEFINITION OF A DEFINITE INTEGRAL 5.1. If f x is continuous in a; b prove that   b n b aX k b a lim f a f x dx n!1 n n a k 1
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Since f x is continuous, the limit exists independent of the mode of subdivision (see Problem 5.31). Choose the subdivision of a; b into n equal parts of equal length x b a =n (see Fig. 5-1, Page 90). Let k a k b a =n, k 1; 2; . . . ; n. Then   b n n X b aX k b a lim f k xk lim f a f x dx n!1 n!1 n k 1 n a k 1
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5.2. Express lim
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  n 1X k f as a de nite integral. n k 1 n
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Then lim   1 n 1X k f f x dx n k 1 n 0
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Let a 0, b 1 in Problem 1.
1 5.3. (a) Express
x2 dx as a limit of a sum, and use the result to evaluate the given de nite integral.
(b) Interpret the result geometrically.
(a) If f x x2 , then f k=n k=n 2 k2 =n2 . Thus by Problem 5.2, 1 1 k2 x2 dx lim n!1 n n2 0 k 1
This can be written, using Problem 1.29 of 1, ! 1 1 12 22 n2 12 22 n2 2 x dx lim 2 2 lim n!1 n n2 n!1 n n n3 0 n n 1 2n 1 6n3 1 1=n 2 1=n 1 lim n!1 6 3 lim
which is the required limit. By using the fundamental 1 2Note: 3 1 3 3 0 x dx x =3 j0 1 =3 0 =3 1=3.
theorem
calculus,
observe
that
(b) The area bounded by the curve y x2 , the x-axis and the line x 1 is equal to 1. 3
CHAP. 5]
INTEGRALS
& 5.4. Evaluate lim
' 1 1 1 . n 1 n 2 n n
The required limit can be written & ' n 1 1 1 1 1X 1 lim lim n!1 n 1 1=n n!1 n 1 2=n 1 n=n 1 k=n k 1 1 dx ln 1 x j1 ln 2 0 1 x 0 using Problem 5.2 and the fundamental theorem of the calculus.
& ' 1 t 2t n 1 t 1 cos t sin sin sin . 5.5. Prove that lim n!1 n n n n t
Let a 0; b t; f x sin x in Problem 1.
Then t t kt sin sin x dx 1 cos t lim n!1 n n 0 k 1
and so lim
n 1 1X kt 1 cos t sin n k 1 n t
using the fact that lim
sin t 0. n
MEASURE ZERO 5.6. Prove that a countable point set has measure zero.
Let the point set be denoted by x1 ; x2 ; x3 ; x4 ; . . . and suppose that intervals of lengths less than =2; =4; =8; =16; . . . respectively enclose the points, where  is any positive number. Then the sum of the lengths of the intervals is less than =2 =4 =8  (let a =2 and r 1 in Problem 2.25(a) of 2 2), showing that the set has measure zero.
PROPERTIES OF DEFINITE INTEGRALS   b   b   j f x j dx if 5.7. Prove that  f x dx @   a a
a < b.
By absolute value property 2, Page 3,   n n n X  X X   f k xk  @ j f k xk j j f k j xk   k 1  k 1 k 1 Taking the limit as n ! 1 and each xk ! 0, we have the required result.
2 5.8. Prove that lim
n!1 0
sin nx dx 0. x2 n 2
 2    sin nx Then lim  dx 0, and so the required result follows.   n!1 0 x2 n2
 2   2  2    sin nx  sin nx dx 2    dx @ dx @ 2    2 2 2 2 2 n 0 x n 0 x n 0 n
INTEGRALS
[CHAP. 5
MEAN VALUE THEOREMS FOR INTEGRALS 5.9. Given the right triangle pictured in Fig. 5-6: (a) Find the average value of h. (b) At what point does this average value occur (c) Determine the average value of f x sin 1 x; 0 @ x @ 1. 2 (d) Determine the average value of f x cos2 x; 0 @ x @
(a) h x
(Use integration by parts.)  . 2
Fig. 5-6
H x. According to the mean value theorem for integrals, B the average value of the function h on the interval 0; B is 1 BH H x dx A B 0 B 2
(b) The point, , at which the average value of h occurs may be obtained by equating f  with that average H H B value, i.e.,  . Thus,  . B 2 2
FUNDAMENTAL THEOREM OF THE CALCULUS x 5.10. If F x f t dt where f x is continuous in a; b , prove that F 0 x f x .