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9.11 The arc length of a curve represented by f(x), a x b, is given by L = the length of arc of one arch of a sine curve.
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1 + [ f '( x )]2 dx . Compute
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1 + f'[x] 2 x
% //N 3.8202
Mathematica returns the value of the integral as a complete elliptic integral of the second kind, represented by EllipticE[x]. We easily obtain a numerical approximation to the arc length.
9.12 The Mean Value Theorem for integrals says that if f is continuous on a closed bounded interval b [a, b], there exists a number, c, between a and b, such that f ( x ) dx = f (c)(b a). Find the value a of c that satisfies the mean value theorem for f(x) = ln x on the interval [1, 2].
SOLUTION
f[x_]= Log[x]; a = 1; b = 2;
b Solve f[x] x f[c] a), c / /Simplify (b a
4 c
%//N {{c 1.47152}}
Integral Calculus
To get a visualization of the Mean Value Theorem for integrals, consider the following plot. Observe that the area below the curve, above the x-axis, is equal to the area enclosed by the rectangle determined by c. g1 = Plot[{f[x],f[c]}, {x, a, b}, Ticks {{1, 1.2, 1.4, 1.6, 1.8, 2.0,{c, "c"}}, Automatic}]
g2 = Graphics[Line[{{2, 0},{2, f[2]}}]]; g3 = Graphics[{Dashed, Line[{{c, 0}, {c, f[c]}}]}]; Show[g1, g2, g3]
0.7 0.6 0.5 0.4 0.3 0.2 0.1 1.2 1.4 c 1.6 1.8 2.0
The area below the curve, above the x-axis, is equal to the area enclosed by the rectangle.
9.13 The work done in moving an object from a to b by a variable force, f(x), is f ( x ) dx . According a to Hooke s law, the force required to hold a spring stretched beyond its natural length is directly proportional to the displaced distance. If the natural length of a spring is 10 cm, and the force that is required to hold the spring 5 cm beyond this length is 40 Newtons, how much work is done in stretching the spring from 10 to 15 cm
SOLUTION
Hooke s law states that f(x) = kx where x represents the distance beyond the spring s natural length. Since a force of 40 Newtons is required to hold the spring 5 cm (0.05 m) beyond its natural length, 40 = 0.05k. k = 40/0.05; f[x_] = k x; work = 1.
.05 0
f[x] x
The work done is 1 Joule.
9.3 Functions Defined by Integrals
If f is continuous on [a, b], we can define a new function: F (x) =
f (t ) dt
Intuitively, if f(t) 0, F(x) represents the area bounded by f(t) and the t-axis from a to x, if x a, and the negative of this area if x < a. The (second) Fundamental Theorem of Calculus tells us that F is differentiable on (a, b) and F' (x) = f(x) for all x (a, b).
Integral Calculus
EXAMPLE 14 Let f(x) = 1/x, x > 0. The shaded area in the diagram represents F(x), assuming x 1.
f (t) = 1/t 4
1 F (x) 1 2 3 x 4 5 t
Students of calculus will recognize that F ( x ) = Mathematica knows this also. f[x_] = 1/x;
1 dt defines F ( x) to be the natural logarithm function. t
F[x_] = Integrate[f[t],{t, 1, x}, Assumptions x > 0]; F[2] Log[2] F[1/2] Log[2]
EXAMPLE 15 The continuous function f(x) = xx has an antiderivative, but it cannot be put into closed form in terms of elementary functions. However, Mathematica can deal with it as a function defined by an integral. Since all antiderivatives of f(x) differ by a constant, we define F(x) to be the antiderivative for which F(0) = 0. Let us plot this antiderivative for 0 x 4.
f[x_]= x^x; F[x_]= f[t] t;
By making the lower limit 0, we force F(0) = 0.
Plot[F[x], {x, 0, 4}]
60 50 40 30 20 10
SOLVED PROBLEMS
9.14 Let F ( x ) = esin t dt . Find F '( x ) . 1
SOLUTION
F[x_]= Exp[Sin[t]] t;
F' [x]
Sin[x]
This is in accordance with the Second Fundamental Theorem of Calculus.
Integral Calculus
9.15 Sketch, on one set of axes, the graphs of the three antiderivatives of f(x) = e sin x, 0 x 2 , for which F(0) = 0, F(1) = 0, and F(2) = 0.
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