PureMath.pdf

# 2 z a a f x dx 0 3 z b a f x dx z c b f x dx z c a f

This preview shows pages 374–377. Sign up to view the full content.

(2) Z a a f ( x ) dx = 0. (3) Z b a f ( x ) dx + Z c b f ( x ) dx = Z c a f ( x ) dx . (4) Z b a kf ( x ) dx = k Z b a f ( x ) dx . (5) Z b a { f ( x ) + φ ( x ) } dx = Z b a f ( x ) dx + Z b a φ ( x ) dx . The reader will find it an instructive exercise to write out formal proofs of these properties, in each case giving a proof starting from ( α ) the definition by means of the integral function and ( β ) the direct definition. The following theorems are also important. (6) If f ( x ) = 0 when a 5 x 5 b , then Z b a f ( x ) dx = 0 . We have only to observe that the sum s of § 156 cannot be negative. It will be shown later ( Misc. Ex. 41) that the value of the integral cannot be zero unless * All functions mentioned in these equations are of course continuous, as the definite integral has been defined for continuous functions only.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
[VII : 160] ADDITIONAL THEOREMS IN THE CALCULUS 359 f ( x ) is always equal to zero: this may also be deduced from the second corollary of § 121 . (7) If H 5 f ( x ) 5 K when a 5 x 5 b , then H ( b - a ) 5 Z b a f ( x ) dx 5 K ( b - a ) . This follows at once if we apply (6) to f ( x ) - H and K - f ( x ). (8) Z b a f ( x ) dx = ( b - a ) f ( ξ ) , where ξ lies between a and b . This follows from (7). For we can take H to be the least and K the greatest value of f ( x ) in [ a, b ]. Then the integral is equal to η ( b - a ), where η lies between H and K . But, since f ( x ) is continuous, there must be a value of ξ for which f ( ξ ) = η ( § 100 ). If F ( x ) is the integral function, we can write the result of (8) in the form F ( b ) - F ( a ) = ( b - a ) F 0 ( ξ ) , so that (8) appears now to be only another way of stating the Mean Value Theorem of § 125 . We may call (8) the First Mean Value Theorem for Integrals . (9) The Generalised Mean Value Theorem for integrals. If φ ( x ) is positive, and H and K are defined as in (7) , then H Z b a φ ( x ) dx 5 Z b a f ( x ) φ ( x ) dx 5 K Z b a φ ( x ) dx ; and Z b a f ( x ) φ ( x ) dx = f ( ξ ) Z b a φ ( x ) dx, where ξ is defined as in (8) .
[VII : 160] ADDITIONAL THEOREMS IN THE CALCULUS 360 This follows at once by applying Theorem (6) to the integrals Z b a { f ( x ) - H } φ ( x ) dx, Z b a { K - f ( x ) } φ ( x ) dx. The reader should formulate for himself the corresponding result which holds when φ ( x ) is always negative. (10) The Fundamental Theorem of the Integral Calculus. The function F ( x ) = Z x a f ( t ) dt has a derivative equal to f ( x ) . This has been proved already in § 145 , but it is convenient to restate the result here as a formal theorem. It follows as a corollary, as was pointed out in § 157 , that F ( x ) is a continuous function of x . Examples LXV. 1. Show, by means of the direct definition of the definite integral, and equations (1)–(5) above, that (i) Z a - a φ ( x 2 ) dx = 2 Z a 0 φ ( x 2 ) dx , Z a - a ( x 2 ) dx = 0; (ii) Z 1 2 π 0 φ (cos x ) dx = Z 1 2 π 0 φ (sin x ) dx = 1 2 Z π 0 φ (sin x ) dx ; (iii) Z 0 φ (cos 2 x ) dx = m Z π 0 φ (cos 2 x ) dx , m being an integer. [The truth of these equations will appear geometrically intuitive, if the graphs of the functions under the sign of integration are sketched.] 2. Prove that Z π 0 sin nx sin x dx is equal to π or to 0 according as n is odd or or even. [Use the formula (sin nx ) / (sin x ) = 2 cos { ( n - 1) x } +2 cos { ( n - 3) x } + . . . , the last term being 1 or 2 cos x .] 3. Prove that Z π 0 sin nx cot x dx is equal to 0 or to π according as n is odd or even.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern