Mathematical Methods for Physics and Engineering : A Comprehensive Guide

2.1 DIFFERENTIATION gradient of the function is zero but the function rises in the positivex-direction and falls in the negative ...

PRELIMINARY CALCULUS G f(x) x Figure 2.3 The graph of a functionf(x) that has a general point of inflection at the pointG. Now, ...

2.1 DIFFERENTIATION C P Q ρ θ θ+∆θ ∆θ x f(x) Figure 2.4 Two neighbouring tangents to the curvef(x) whose slopes differ by ∆θ. Th ...

PRELIMINARY CALCULUS relative to the common tangent, above or below. Thus a negative value ofρ indicates that the curve is local ...

2.1 DIFFERENTIATION
Show that the radius of curvature at the point(x, y)on the ellipse x^2 a^2 + y^2 b^2 =1 has magnitude(a^4 y ...

PRELIMINARY CALCULUS a b c f(x) x Figure 2.5 The graph of a functionf(x), showing that iff(a)=f(c)thenat one point at least betw ...

2.1 DIFFERENTIATION and hence the difference between the curve and the line is h(x)=f(x)−g(x)=f(x)−f(a)−(x−a) f(c)−f(a) c−a . Si ...

PRELIMINARY CALCULUS In each case, as might be expected, the application of Rolle’s theorem does no more than focus attention on ...

2.2 INTEGRATION a b f(x) x Figure 2.7 An integral as the area under a curve. 2.2 Integration The notion of an integral as the ar ...

PRELIMINARY CALCULUS abx 1 ξ 1 x 2 ξ 2 x 3 ξ 3 x 4 ξ 4 x 5 f(x) x Figure 2.8 The evaluation of a definite integral by subdividin ...

2.2 INTEGRATION Combining (2.23) and (2.24) withcset equal toashows that ∫b a f(x)dx=− ∫a b f(x)dx. (2.26) 2.2.2 Integration as ...

PRELIMINARY CALCULUS From the last two equations it is clear that integration can be considered as the inverse of differentiatio ...

2.2 INTEGRATION found near the end of subsection 2.1.1. A few are presented below, using the form given in (2.30): ∫ adx=ax+c, ∫ ...

PRELIMINARY CALCULUS
Evaluate the integralI= ∫ cos^4 xdx. Rewriting the integral as a power of cos^2 xand then using the double ...

2.2 INTEGRATION
Evaluate the integral I= ∫ 1 x^2 +x dx. We note that the denominator factorises to givex(x+ 1). Hence I= ∫ 1 x( ...

PRELIMINARY CALCULUS Since dt dx = 1 2 sec^2 x 2 = 1 2 ( 1+tan^2 x 2 ) = 1+t^2 2 , the required relationship is dx= 2 1+t^2 dt. ...

2.2 INTEGRATION
Evaluate the integral I= ∫ 1 x^2 +4x+7 dx. We can write the integral in the form I= ∫ 1 (x+2)^2 +3 dx. Substitu ...

PRELIMINARY CALCULUS The separation of the functions is not always so apparent, as is illustrated by the following example.
Eva ...

2.2 INTEGRATION Rearranging this expression to obtainIexplicitly and including the constant of integration we find I= eax a^2 +b ...

PRELIMINARY CALCULUS 2.2.10 Infinite and improper integrals The definition of an integral given previously does not allow for ca ...