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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 xaxis, 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 xaxis, 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 taxis 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.

