4.4 Using partial fractions

Transcription

1 CHAPTER 4. INTEGRATION 43 Eample 4.9. Compute ln d. (Classic A-Level question!) ln d ln d ln d (ln ) + C. Eample 4.0. Find I sin d. I sin d sin p d p sin. 4.4 Using partial fractions Sometimes we want to compute, say, d, which we can t integrate directly. Here we must epress the integrand as a sum of partial fractions Recap: Partial fractions You can epress the function P () Q() with partial fractions if Q() factorises. For every factor of Q() (a + b) (a + b) (a + b) 3 (a + b + c) You get this partial fraction form: A (a + b) A (a + b) + A (a + b) + B (a + b) B (a + b) + A + B a + b + c C (a + b) 3 Then plug in some di erent values of to find A, B,... (or use any other method you prefer!) For the net three eamples P () will be linear and Q() will be quadratic polynomials.

2 CHAPTER 4. INTEGRATION 44 Eample 4. (Case : Denominator has two real roots) d. First things first... factorise the denominator! Hence ) 3 ( 3)( + ) A ( 3) + B A( + ) + B( 3). Let s try two di erent values of. How about...? ) 8 4B ) B, Then ) 3) 44A ) A, ( 3) d 3 + d + 3 d + + d ln 3 +ln + + C. Eample 4. (Case : Denominator has one real root). + d. Start with Let s compare coe + ( ) A + B ( ). ) A( ) + B A + B A. cients: the terms suggest that A. As for the constant terms: B A 0) A B. Therefore + d d + ln ( ) d + C.

3 CHAPTER 4. INTEGRATION 45 Eample 4.3 (Case 3: Denominator has no real roots). +5 d So we can t factorise the denominator, but we can still complete the square! thus the integral is +5( ) +4, Looks like something with (u + ), so choose ( ) +4 d. u, ) d du. Then u +5 d u +4 du u u +4 du u +4 du. Now u u +4 du ln u +4 ln ( ) +4, while for the other u-integral, try u tan ) du sec d, hence u +4 du sec 4 tan +4 d sec sec d d + C tan + C. Thus our final answer is +5 d ln +5 + tan + C. Remark 4.. If degree of P degree of Q, use long division first to get N()+ R() Q() (R for remainder!). Then use partial fractions on R() Q().

4 CHAPTER 4. INTEGRATION 46 Eample 4.4. Evaluate the indefinite integral 3 + d Do the long division first: ) 3 + d d log + C. 4.5 Some trigonometric integrals i Evaluate cos d (cos + ) d 4 sin + + C. ii Evaluate sin d ( cos )d sin + C Using integration As stated at the start of the chapter, integration is great for calculating areas under curves. Eample 4.5 (997 Eam question). Sketch the region enclosed by the curve y + and the line y and find its area. Apply the recipe for curve sketching:

5 CHAPTER 4. INTEGRATION 47 No vertical asymptotes An even function Passes through (0, ) y 6 0, and in fact y>0 for all. y! 0 as!±. For the turning points dy d ( + 0 when 0. ) Now don t forget the sketch! Figure 4.: A sketch of the curve y enclosed region is shaded in green. + (red) and the line y (yellow). The A d (Area of Rectangle) + + d tan 4 4. Eample 4.6. Question: Find the area bounded by the curve y and the

6 CHAPTER 4. INTEGRATION 48 ais between and 3. A 3 yd 3 apple apple d apple But why is the area negative? Let s draw a sketch. Figure 4.: A sketch of the curve y 6 +5 (red). The region we want to integrate over (blue) is bounded by the grey vertical lines and 3. Trouble is, the region below the ais gives a negative area! Eample 4.7. (Mechanics) A ball is thrown down from a high building with an initial velocity of 30 metres per second. Then its velocity after t seconds is given by v(t) 0t How far does the ball fall between and 3 seconds of elapsed time? The distance s(t) turns out to be the integral of the velocity, i.e. s(t) v(t)dt. Hence the distance we want is s(3) s() 3 3 v(t)dt (0t + 30)dt 5t + 30t metres.

7 CHAPTER 4. INTEGRATION 49 Eample 4.8. Find the area A of an ellipse, given by the equation a + y b, Figure 4.3: An ellipse Note from Figure 4.3 that A 4 A by symmetry. Hence for the area A, r a A 4 b 0 a d r a 4b a d, an integral that can be solved by substitution. Let and So we have 0 d sinu, ) a du a cos u r p a sin u cos u. u A 4b cos u(a cos u)du. u Reminder: In changing the variable it is also very important to change the limits, i.e. find numerical values for u and u. When a, sin u, ) u. When 0, sin u 0, ) u 0. Therefore we have A 4ab cos u du 0

8 CHAPTER 4. INTEGRATION 50 Proceeding with the integral, we get A 4ab 0 cos u du 4ab 0 + 4ab u + sin u 4 4ab 4 +0 (0 + 0) ab. Note: For a circle, a b which givws A a. cos u du 4.7 Improper integrals Often, you will come across integrals of the type a f()d. This is an improper integral, and it must be interpreted as lim b! b a f()d, if the limit eists! (If it doesn t, the integral is said to diverge). Remark 4.3. Technically, there are other kinds of improper integrals, in which I b a f()d has a problem because f() blows up at a, b or some point c in between (a <c<b). But we won t worry about them here! Eample 4.9. Consider Then I d, n >. n d lim n b! lim b! n b n d apple n b n Remark 4.4. This integral in this last eample diverges for n apple.

9 Chapter 5 Di erential Equations 5. Introduction Many problems in engineering and physical science (also biology, economics, etc.) can be reduced to solving di erential equations. Eample 5. (RLC Series Circuit). Consider the following series circuit comprised of a resistor, a capacitor and an inductor. This circuit is known as an RLC circuit. Figure 5.: An RLC Circuit where L d I dt + RdI dt + C I E (5.) I Current Flowing in a Circuit C Capacitance R Resistance L Inductance E Voltage. where C, R, L and E are constants and I is the unknown function to be found. 5

10 CHAPTER 5. DIFFERENTIAL EQUATIONS 5 An ordinary di erential equation (ODE) is a relation between a function y(),, and the derivatives dy d, d y d,etc. The order of the ODE is the order of the highest derivative in the equation. An ODE is linear if there are no products of y and its derivatives, e.g. y dy d, and no functions of y and its derivatives, such as y e y, cos y. For eample, Equation (5.) is a linear second order ode. Eample 5. (Legendre s Equation). ( )y 00 y 0 + k y 0 (k constant) is ubiquitous in problems with spherical symmetry (e.g a Hydrogen atom). It is a linear second order equation. Eample 5.3 (Radioactive decay). This is first order and linear. dr dt kr. (k constant) Eample 5.4 (Simple pendulum). d dt + g sin 0. l It is a second-order ODE. However it is non-linear, due to the sin term. Figure 5.: An simple pendulum comprised of an object with mass m attached to a string with length l. The other end of the string is attached to a ceiling. Partial di erential equations (PDEs) involve partial derivatives (see Chapter 3), such as...

11 CHAPTER 5. DIFFERENTIAL EQUATIONS 53 Eample 5.5 (Beam Equation). The Beam Equation provides a model for the load carrying and deflection properties of beams, and is given but you won t see them in this course. You ll have to wait until Maths for Engineers 3 (MATH6503) for that! 5. First order separable ODEs An ODE dy d f(), g(y). F (, y) isseparable if we can write F (, y) f()g(y) for some functions Eample 5.6. dy d y IS separable, dy d y IS NOT. Eample 5.7. Find the general solution to the ODE Separating the variables, we have 9y dy d i.e. the general solution is 9ydy 4d () 9 ydy 4 d 9 y 4 + C, 9 + y 4 which describes a family of ellipses. We can check our solution by di erentiating: K, (K C/36) i.e yy0 0 9yy Eample 5.8. Find the general solution to dy d y + +.