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Notice that we nd a speci c antiderivative, x 3 /3 + x 4 /4, and then add the arbitrary constant C. Rules 1 through 4 enable us to compute the antiderivative of any polynomial. EXAMPLE
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1 3x 5 x 4 + 7x 2 + x 3 2 dx = 3 = x6 6 1 x5 2 5 +7 x3 3 + x2 3x + C 2
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The next rule will prove to be extremely useful. RULE 5. (Quick Formula I). (g(x))r g (x) dx = (g(x))r+1 +C r+1 for r = 1
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The power chain rule implies that Dx (g(x))r+1 r+1 = 1 1 Dx (g(x)r+1 ) = (r + 1)(g(x))r g (x) = (g(x))r g (x) r+1 r+1
which yields quick formula I. EXAMPLES
(a) (b)
7 8 1 1 2 1 2 x + 5 x dx = x +5 +C 2 8 2
1 2x 5 dx = 2
(2x 5)1/2 (2) dx =
1 (2x 5)3/2 1 + C = (2x 5)3/2 + C 3 2 3
ANTIDERIVATIVES
[CHAP. 29
RULE 6. (i) Find
(Substitution Method). Deferring the general formulation and justi cation to Problem 29.18, we illustrate the method by three examples. x 2 cos x 3 dx. Let x 3 = u. Then, by Section 21.2, the differential of u is given by du = Dx (x 3 ) dx = 3x 2 dx or x 2 dx = 1 du 3
Now substitute u for x 3 and
du for x 2 dx, 1 1 cos u du = 3 3 cos u du = 1 1 sin u + C = sin x 3 + C 3 3
x 2 cos x 3 dx = (ii) Find
(x 2 + 3x 5)3 (2x + 3) dx. Let u = x 2 + 3x 5, du = (2x + 3) dx. Then (x 2 + 3x 5)3 (2x + 3) dx = u3 du = u4 1 + C = (x 2 + 3x 5)4 + C 4 4
(iii) Find
sin2 x cos x dx. Let u = sin x. Then du = cos x dx, and sin2 x cos x dx = u2 du = sin3 x u3 +C = +C 3 3
Notice that Quick Formula I (Rule 5) is a special case of Rule 6, corresponding to the substitution u = g(x). The beauty of quick formula I is that, when it is applicable, it allows us to avoid the bother of going through the substitution process.
Solved Problems
29.1 Find the following antiderivatives: (a) ( 3 x 5x 2 ) dx (b) (4x + x 5 2) dx (c) (x 2 sec2 x) dx
( 3 x 5x 2 ) dx = = =
2 x + 3x 2 dx x
(x 1/3 5x 2 ) dx x 4/3
x3 3
[by Rules 2 4]
3 4/3 5 3 x x +C 4 3 x2 2 x 7/2
(4x +
x 5 2) dx =
(4x + x 5/2 2) dx = 4
2x + C
2 = 2x 2 + x 7/2 2x + C 7 (c) (x 2 sec2 x) dx = x 2 dx sec2 x dx = x3 tan x + C 3
CHAP. 29] 2 x + 3x 2 dx = x
ANTIDERIVATIVES
2 + 3x x
dx = 2
x 1/2 dx + 3
x dx
[by Rules 2 4]
x2 3 x 1/2 = 2 1 + 3 + C = 4 x + x2 + C 2 2
29.2 Find the following antiderivatives: (a) (2x 3 x)4 (6x 2 1) dx (b)
5x 2 1 x dx
(a) Notice that Dx (2x 3 x) = 6x 2 1. So, by Quick Formula I, (2x 3 x)4 (6x 2 1) dx = (b) Observe that Dx (5x 2 1) = 10x. Then, by Rules 2 and 3,
1 3 (2x x)5 + C 5
5x 2 1 x dx = = =
(5x 2 1)1/3 x dx = 1 (5x 2 1)4/3 +C 4 10
1 10
(5x 2 1)1/3 10x dx [by Quick Formula I]
3 (5x 2 1)4/3 + C 40 3 3 (5x 2 1)4 + C = 40
(For manipulations of rational powers, review Section 15.2.)
29.3 Use the substitution method to evaluate: sin x (a) dx (b) x sec2 (3x 2 1) dx x
(a) Let u = x. Then,
x 2 x + 2 dx
Hence,
1 1 du = Dx ( x) dx = Dx (x 1/2 ) dx = x 1/2 dx = dx 2 2 x sin x dx = 2 sin u du = 2 cos u + C = 2 cos x + C x
(b) Let u = 3x 2 1. Then du = 6x dx, and 1 x sec 2 3x 2 1 dx = 6 sec2 u du = 1 1 tan u + C = tan (3x 2 1) + C 6 6
(c) Let u = x + 2. Then du = dx and x = u 2. Hence, x 2 x + 2 dx = = = The substitution u = u 2
u du =
u2 4u + 4 u1/2 du [by ur us = ur+s ]
u5/2 4u3/2 + 4u1/2 du
2 7/2 8 5/2 8 3/2 2 8 8 7/2 5/2 3/2 u x+2 x+2 x+2 u + u +C = + +C 7 5 3 7 5 3
x + 2 would also work.
ANTIDERIVATIVES
[CHAP. 29
29.4 Prove Theorem 29.1.
Let a and b be any two numbers in I . By the mean-value theorem (Theorem 17.2), there is a number c between a and b, and therefore in I , such that F (c) = F(b) F (a) b a
But by hypothesis, F (c) = 0; hence, F(b) F(a) = 0, or F(b) = F(a).
29.5 A rocket is shot straight up into the air with an initial velocity of 256 feet per second. (a) When does it reach its maximum height (b) What is its maximum height (c) When does it hit the ground again (d) What is its speed when it hits the ground
In free-fall problems, v = a dt and s = per second per second (when up is positive), v= s= v dt because, by de nition, a = dv/dt and v = ds/dt. Since a = 32 feet
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