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THE DEFINITE INTEGRAL
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EXAMPLE Suppose that f (x) 0 for all x in [a, b]. The graph of f along with its mirror image, the graph of f is shown
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in Fig. 30-5. Since f (x) 0,
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f (x) dx = area B
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Fig. 30-5
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But by symmetry, area B = area A; and by Theorem 30.2, with c = 1,
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f (x) dx =
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f (x) dx
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It follows that
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f (x) dx = (area A)
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In other words, the de nite integral of a nonpositive function is the negative of the area above the graph of the function and below the x-axis.
Theorem 30.3: If f and g are integrable on [a, b], then so are f + g and f g, and
(f (x) g(x)) dx =
f (x) dx
g(x) dx
Again, this property is implied by the corresponding property of the approximating sums,
n n n
[P(i) Q(i)] =
i=1 i=1
P(i)
Q(i)
Theorem 30.4: If a < c < b and if f is integrable on [a, c] and on [c, b], then f is integrable on [a, b], and
f (x) dx =
f (x) dx +
f (x) dx
For f (x) 0, the theorem is obvious: the area under the graph from a to b must be the sum of the areas from a to c and from c to b.
THE DEFINITE INTEGRAL
[CHAP. 30
EXAMPLE Theorem 30.4 yields a geometric interpretation for the de nite integral when the graph of f has the appearance
shown in Fig. 30-6. Here,
b a c1 c2 c3 c2 c4 c3 b
f (x) dx =
f (x) dx+
a c1
f (x) dx +
f (x) dx +
f (x) dx +
f (x)dx
= A1 A2 + A3 A4 + A5 That is, the de nite integral may be considered a total area, in which areas above the x-axis are counted as positive, and areas below the x-axis are counted as negative. Thus, we can infer from Fig. 27-2(b) that 2
sin x dx = 0
because the positive area from 0 to is just canceled by the negative area from to 2 .
Fig. 30-6
Arbitrary Limits of Integration In de ning a f (x) dx, we have assumed that the limits of integration a and b are such that a < b. Extend the de nition as follows:
f (x) dx = 0.
(2) If a > b, let
f (x) dx =
f (x) dx (with the de nite integral on the right falling under the original de nition).
Under this extended de nition, interchanging the limits of integration in any de nite integral reverses the algebraic sign of the integral. Moreover, the equations of Theorems 30.2, 30.3, and 30.4 now hold for arbitrary limits of integration a, b, and c.
Solved Problems
30.1 Show that
1 dx = b a.
For any subdivision a = x0 < x1 < x2 < < xn 1 < xn = b of [a, b], the approximating sum (30.1) is
n i=1 f (xi ) i x = n ix i=1
[since f (x) = 1 for all x]
= (x1 x0 ) + (x2 x1 ) + (x3 x2 ) + + (xn xn 1 ) = xn x0 = b a
CHAP. 30]
THE DEFINITE INTEGRAL
Since every approximating sum is equal to b a,
1 dx = b a
b As an alternative, intuitive proof, note that a 1 dx is equal to the area of a rectangle with base of length b a and height 1, since the graph of the constant function 1 is the line y = 1. This area is (b a)(l) = b a (see Fig. 30-7).
Fig. 30-7
30.2 Calculate
x 2 dx. [You may assume the formula 12 + 22 + + n2 =
n(n + 1)(2n + 1) , which is established 6
i x = 1/n. In the ith subinterval
in Problem 30.12(a, ii).]
Divide the interval [0, 1] into n equal parts, as indicated in Fig. 30-8, making each [(i 1)/n, i/n], let xi be the right endpoint i/n. Then (30.1) becomes
n i=1 f (xi ) i x = n i=1 n 1 i 21 = 3 i2 n n n i=1
1 n+1 1 n(n + 1)(2n + 1) = = 3 6 6 n n 1 1 1 = 1+ 2+ 6 n n
2n + 1 n
Fig. 30-8
We can make the subdivision ner and ner by letting n approach in nity. Then,
x 2 dx = lim
n 6
1 1 (1)(2) = 6 3
THE DEFINITE INTEGRAL
[CHAP. 30
This kind of direct calculation of a de nite integral is possible only for the very simplest functions f (x). A much more powerful method will be explained in 31.
30.3 Let f (x) and g(x) be integrable on [a, b]. (a) If f (x) 0 on [a, b], show that
f (x) dx 0
(b) If f (x) g(x) on [a, b], show that
f (x) dx
g(x) dx
(c) If m f (x) M on [a, b], show that m(b a)
f (x) dx M(b a)
(a) The de nite integral, being the area under the graph of f , cannot be negative. More fundamentally, every approximating sum (30.1) is nonnegative, since f (xi ) 0 and i x > 0. Hence (as shown in Problem 9.10), the limiting value of the approximating sums is also nonnegative. (b) Because g(x) f (x) 0 on [a, b],
b a b a
(g(x) f (x)) dx 0 [by (a)]
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