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by the rst mean value theorem for integrals (Page 93). Then if x is any point interior to a; b , F 0 x lim
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since f is continuous. If x a or x b, we use right- or left-hand limits, respectively, and the result holds in these cases as well.
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5.11. Prove the fundamental theorem of the calculus, Part 2 (Pages 94 and 95).
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By Problem 5.10, if F x is any function whose derivative is f x , we can write x F x f t dt c
where c is any constant (see last line of Problem 22, 4). b b Since F a c, it follows that F b f t dt F a or f t dt F b F a .
5.12. If f x is continuous in a; b , prove that F x
f t dt is continuous in a; b .
If x is any point interior to a; b , then as in Problem 5.10,
lim F x h F x lim h f  0
and F x is continuous. If x a and x b, we use right- and left-hand limits, respectively, to show that F x is continuous at x a and x b.
CHAP. 5]
INTEGRALS
Another method: By Problem 5.10 and Problem 4.3, 4, it follows that F 0 x exists and so F x must be continuous.
CHANGE OF VARIABLES AND SPECIAL METHODS OF INTEGRATION 5.13. Prove the result (7), Page 95, for changing the variable of integration.
Let F x x
f x dx and G t
f fg t g g 0 t dt, where x g t .
Then dF f x dx, dG f fg t g g 0 t dt. Since dx g 0 t dt, it follows that f x dx f fg t g g 0 t dt so that dF x dG t , from which F x G t c. Now when x a, t or F a G c. But F a G 0, so that c 0. Hence F x G t . Since x b when t , we have b f x dx f fg t g g 0 t dt
as required.
5.14. Evaluate: a x 2 sin x2 4x 6 dx b cot ln x dx x
dx p x 2 3 x 1 2 x tanh 21 x dx
1=p 2
x sin 1 x2 p dx 1 x4
d
f
x dx p 2 x 1 x
(a) Method 1: Let x2 4x 6 u. Then 2x 4 dx du, x 2 dx 1 du and the integral becomes 2 1 1 1 2 sin u du cos u c cos x 4x 6 c 2 2 2 Method 2: 1 1 sin x2 4x 6 d x2 4x 6 cos x2 4x 6 c x 2 sin x2 4x 6 dx 2 2 (b) Let ln x u. Then dx =x du and the integral becomes cot u du ln j sin uj c ln j sin ln x j c dx dx dx dx p p p q 2 2 x 2 3 x 6 x x 6 x x 25=4 x 1 2
2 1 2
c Method 1:
Letting x u, this becomes   du u 2x 1 p sin 1 c sin 1 c 5=2 5 25=4 u2 Then 1       dx 2x 1 1  sin 1 1 sin 1 3 p sin 1  5 5 5 1 x 2 3 x 1 sin 1 :2 sin 1 :6
INTEGRALS
[CHAP. 5
Method 2: Let x 1 u as in Method 1. Now when x 1, u 3; and when x 1, u 1. Thus 2 2 2 by Formula 25, Page 96. 1 dx p x 2 3 x 1 1 dx q 1 25=4 x 1 2 2 1=2  du u 1=2  p sin 1 5=2  3=2 3=2 25=4 u2
sin1 :2 sin 1 :6 (d) Let 21 x u. Then 21 x ln 2 dx du and 2 x dx du , so that the integral becomes 2 ln 2
1 1 tanh u du ln cosh 21 x c 2 ln 2 2 ln 2
Let sin 1 x2 u.
1 2x dx Then du q 2x dx p and the integral becomes 1 x4 2 2 1 x 1 1 1 u du u2 c sin 1 x2 2 c 2 4 4
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