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Integral Calculus
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Other options, which control more precisely how the algorithm should be implemented, are available, but will not be discussed here. These options are useful for integrals involving pathological functions 1 1000 such as sin 1 dx or e x dx. The interested reader should consult the Mathematica Documentation .0001 1000 x Center for details. b The sequence N[Integrate[f[x], {x, a, b}]] or f[x] x //N evaluates the integral, a whenever possible, by first finding the antiderivative and then using the Fundamental Theorem of Calculus. If this is impossible, NIntegrate[f[x], {x, a, b}] is called automatically.
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EXAMPLE 5 To evaluate
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x e x sin x dx we input
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Integrate[x Exp[x] Sin[x], {x, 0, 1}] 1( 1 + Sin[1]) 2 As an alternate representation, we can use the Basic Math Input palette.
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1( 1 + Sin[1]) 2
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If a numerical approximation is desired, we can type
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Sin[x] x //N
Integrate[x Exp[x] Sin[x], {x, 0, 1}] //N
0.643678 Here, the antiderivative of the function x e x sin x was computed and then evaluated from 0 to 1. If a strictly numerical procedure is preferred, we can use Nintegrate. NIntegrate[x Exp[x] Sin[x], {x, 0, 1}] 0.643678
EXAMPLE 6 Obtain an approximation to
sin (sin x ) dx accurate to (a) 6 significant digits and (b) 20 significant digits.
Sin[Sin[x] x //N ]
1 ] (b) N Sin[Sin[x] x,20 0
Mathematica automatically adjusts WorkingPrecision and PrecisionGoal to achieve the desired result.
0.43060610312069060491
Mathematica can handle certain improper integrals. An improper integral of type I is an integral with one or two infinite limits of integration. We define
f ( x ) dx = lim f ( x ) dx and
t a
provided the limits exist. Such an integral is said to be convergent. If both verge, we define
EXAMPLE 7
f ( x ) dx =
f ( x ) dx +
b
f ( x ) dx = lim
f ( x ) dx and
t t
f ( x ) dx
f ( x ) dx con-
f ( x ) dx .
-x x
EXAMPLE 8
x x
This integral is divergent.
Integrate idiv : Integral of x does not converge on {0, }.
x x
Integral Calculus
EXAMPLE 9
1 x 1 + x2
The value of a type I improper integral may depend upon the values of parameters within the integrand. The option Assumptions allows the specification of conditions to be imposed upon these parameters.
Assumptions conditions specifies conditions to be applied to parameters within the integral.
EXAMPLE 10
x n dx converges if n < 1 and diverges otherwise.
Integrate[xn, {x, 1, }, Assumptions n < 1] 1 1+ n Integrate[xn, {x, 1, }, Assumptions n 1]
Integrate idiv : Integral of xn does not converge on {1, }.
Integrate[xn, {x, 1, }, Assumptions n 1]
An improper integral of type II is an integral whose integrand is discontinuous on the interval of integration. If f is continuous on [a, b) but not at b, we define ous on (a, b] but not at a we define convergent. If f has a discontinuity at c (a, b) and both
f ( x ) dx = lim f ( x ) dx . If the limit exists, we say the integral is +
t a t
f ( x ) dx = lim f ( x ) dx , and if f is continu
t b a
f ( x ) dx =
f ( x ) dx + f ( x ) dx .
f ( x ) dx and
f ( x ) dx are convergent, then
EXAMPLE 11
Log[x] x
EXAMPLE 12 Integrate idiv : Integral of 1 does not converge on { 2, 3}. x
1 x or x
Integrate[1/x, {x, 2, 3}]
1 x x
Because of the discontinuity at 0, the integral of Example 12 is improper. If we break up the integral 0 dx 3 dx + into the sum of two integrals, , each integral, evaluated separately, diverges. However, if we 2 x 0 x consider the limits simultaneously,
t 0 + 3 dx t dx 3 t lim 2 x + t x = tlim ln | x | 2 + ln x t 0 +
= lim [ ln t ln 2 + ln 3 ln t ] +
t 0
= ln 3 ln 2 = ln 3 2
This number is called the Cauchy Principal Value. The option PrincipalValue instructs Integrate to compute the Cauchy Principal Value of an integral.
PrincipalValue True specifies that the Cauchy Principal Value of an integral is to be determined.
Integral Calculus
EXAMPLE 13
Integrate[1/x, {x, 2, 3}, PrincipalValue True] Log 3 2
Compare with the result of Example 12.
SOLVED PROBLEMS
9.8 Compute the area bounded by the curves f(x) = 1 x2 and g(x) = x4 3x2.
SOLUTION
f[x_] = 1 x2; g[x_] = x4 3x2; Plot[{f[x], g[x]}, {x, 2, 2}]
4 3 2 1 2 1 1 2 3 1 2
First we must find the points of intersection of the two curves. intersectionpoints = Solve[f[x] g[x]]
{{x
1 + 2 , x 1 + 2 , x 1 + 2 , x 1 + 2
{a, b, c, d} = x /. intersectionpoints 1 + 2, 1 + 2, 1 + 2, 1 + 2
The points of intersection correspond to the real solutions of this equation c and d.
(f[x] g[x]) x 8 1 + 2 (4+ 2) 15
% //N 4.48665
/ /Simplify
9.9 The volume of the solid of revolution obtained by rotating about the x-axis the area bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b is [ f ( x )]2 dx . Compute the volume of the a sphere obtained if the semicircle y = r 2 x 2 , r x r, is rotated about the x-axis.
SOLUTION
y y= r2 x2
y = r2 x 2 o
y2 x
r r x
4 r3 3
Integral Calculus
9.10 Compute the volume of a frustum of a cone with height h and radii r and R, and use this to derive the formula for the volume of a cone of radius R and height h.
SOLUTION
Position the frustum as shown in the diagram. The frustum is generated by rotating about the x-axis the region bounded by the line segment connecting (0, r) and (h, R), the x-axis, and the vertical lines x = r and x = R. The equaR r tion of the line segment is y = h x + r . The volume is h y 2 dx .
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