# Differentiation, Integration and Matrices Essay

ENGINEERING MATHEMATICS I COURSE OUTLINES PART ONE • • • • Maxima and Minima of Functions of a Single Independent Variable Tangents and Normals Differentiation Techniques of Differentiation PART TWO • Techniques of Integration: Indefinite Integrals, Integration by Parts, Definite Integrals, Improper Integrals • • Applications to Engineering Systems Introduction to Ordinary Differential Equations (ODE) and Partial Differential Equations (PDE) PART THREE • • • Properties and Evaluation of Matrices Introduction to Symmetric and Skew-symmetric Matrices Simple simultaneous Linear Equations 1 PART I

DIFFERENTIATION 1. 0 1. 1 Maxima and Minima, Tangents and Normals Gradients: The gradient of a straight line is the rate of change of y with x and is measured by = the increase in the value of y divided by the corresponding increase in the value of x. Thus, if = the tangent at that point and when the equation of the curve is known differentiation provides a method of calculating the gradient exactly. Example: Calculate the gradient of the curve Solution: 1. 2 =9 ? 4 + 5. When x = 3, =3 ? 2 + 5 ? 7 at the point (3, 71). + , = gradient of the line. The gradient of a curve at a particular point is the gradient of 74 = gradient of the curve. = ( ), Tangents and Normals: For the curve given by gives the gradient of the at that point, tangent at any point. Given the co-ordinates of a point on the curve and the value of the equation of the tangent can be easily obtained. Example: Find the equation of the tangent to the curve Solution: = + 3 + 2; =3 = + 3 + 2 when x = – 2 Therefore the required equation is = 15 + 18or + 3. When x = – 2, = 15. Also, when x = – 2, y = – 12. ? 15 ? 18 = 0. A normal is a line perpendicular to the tangent. Recall from geometry that if two lines are perpendicular then the product of their gradients is – 1.

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Example: Calculate the equation of the normal for the previous example at the points already defined. and equation of the normal is 15 + is parallel to the x-axis. Solution: The point is (– 2, – 12); the gradient of the tangent is 15. :. Gradient of the normal =? + 182 = 0. = ? 4 , Example: Find the coordinates of the points on the curve ? 3 ? 2 where the tangent 2 Solution: A line parallel to the x-axis has zero gradient. If the tangent is parallel to the axis of x then = 0 at the point of contact. =3 ? 8 ? 3 = (3 + 1)( ? 3) = 0. :. = ? 1? 3 3. These are the only values of x which give points on the curve where the gradient is 0.

Their coordinates are (? 1? 3 , ? 40? 27) and (3, – 20). 1. 3 Stationary Points A point on a curve at which = 0 is called a stationary point and the value of the function at that point is called a stationary value. At such points the tangent is parallel to the x-axis. To find the stationary points, set = 0 and face the consequence. + 15 + 18 + 7. Example: Find the stationary points of the function4 Solution: When = 12 + 30 + 18 = 6(2 + 5 + 3) = 6(x + 1) (2x + 3) = 0; 6(x + 1) (2x + 3) = 0, giving = – l or – 3/2. The stationary points are therefore (– l, 0) nd (– 3/2, 1/4) and the stationary values of the function at these points are 0 and 1/4, respectively. 1. 4 Maxima and Minima Figure 1. 1 shows a curve passing through a stationary point and reaching a maximum value at that point. As increases (from left to right) the gradient of the curve decreases from a positive value through 0 to a negative value. Figure 1. 2 shows a curve reaching a minimum at a point. As increases the gradient increases from a negative value through 0 to a positive value. Maximum and minimum points are also referred to as Turning Points (TP) because the tangent turns over at such points.

For Figures 1. 1 and 1. 2, please come to class. 1. 5 Points of Infexion A third type of stationary point is shown in Figure 1. 3, where the curve has neither a maximum nor a minimum value. This is called a point of inflexion, not a TP. For Figure 1. 3, please come to class. Note: For all stationary points, the necessary condition is that = 0. Note also that a maximum or minimum value is not meant in an absolute sense. Figure 1. 4 shows a curve which has two maxima and 1 minimum values though one of the ‘maxima’ values is greater than the other, and the minimum value is not the lowest value possible for the function.

The terms maximum and minimum apply only 3 in a local sense near the stationary point. A function can have more than one of each. They are local maximum and local minimum. For Figure 1. 4, please come to class. For a maximum or minimum, = 0 is necessary but is not sufficient to distinguish between them. ?2 + + 4. A simple test is shown in the example below. Example: Find the nature of the turning points of the function = Solution: =3 ? 4 + 1. At TP, = 0. > 3 ? 4 + 1=0, giving = 1/3 or 1. = 3/16. Now consider the value When = 1/3. When = 1/4 (a convenient value just less than 1/3), = ? 1/4. Thus 1/2 (a convenient value just greater than 1/3), = changes sign from = ? 0. 23. Thus positive to negative when Now consider the value = 1/3. This means that y is a maximum. The value of y is 112/27. = 1. When = 0. 9, =? 0. 17. When = 1. 1, changes sign from negative to positive when is 4. = 1. This means thatb y is a minimum. The value of y A second procedure for distinguishing between maximum and minimum values is the following: (a) At a TP giving maximum value (b) At a TP giving minimum value (c) If at a TP “( “( “( use our original criterion for distinguishing between maximum and minimum values.

For the example just considered, When = 1, “( ) = 0, no conclusions can be drawn using the above argument and we have to “( ) &gt; 0. ) &lt; 0. ) = 2 (positive). Thus ) = 6 ? 4. When = 1/3, “( ) = ? 2 (negative). = 1/3 gives a maximum and = 1 gives a minimum for y. Example 1: Find the maximum and minimum values and the points of inflexion of = ? 6 + 9 + 1. Solution: & & = 6 ? 12. When =3 ? 12 + 9 = 3( ? 1)( ? 3). = 1, & = ? 6 &lt; 0 and When & = 0 when = 3, & & = 1 3. = + 6 &gt; 0. When = 1, y has a maximum value of 5 and when 4 = 3, y has a minimum value of 1. & (2, 3). & = 6 ? 12is 0 when x = 2 and changes sign.

Therefore there is just one point of inflexion at Example 2: The difference between two numbers is’, what are they if their product is to be a minimum? Solution: Let the two numbers be x and y and their product be p. ? ? For a minimum value , =? . * :. ( = ? ‘ . If their product is to be a minimum, it means that &) & ) &gt; 0. &) & = 0. > = ‘ and ( = ) . * = 2 ? ‘ or = . =2 &gt; 0; their product is a minimum. The two numbers are ? 3 + 4 = 0. and = * Example 3: Find the equations of the tangent and normal to the curve point (1, 2). Solution: Differentiate with respect to x: 2 2 +4? 3 ? +( = ? 6 ( ? ) ). + ? 3 ? 3 (2 ) + + 4 = 3 at the defined by: = 0. At the point (1, 2), ? 2=? ?2= + + The equation of the normal at (1, 2) is ( ? 1) or 4 ? 11 + 3 = 0. ( ? 1) or 11 + 4 ? 26 = 0. =? . The equation of the tangent is Example 4: A closed rectangular tank is to be made to contain 9 m3 of water. The length must be twice the breadth and the total surface area must be a minimum. Find the dimensions of the tank. minimised is the surface area (A). , = 4 ,=4 + 0 Solution: Let the breadth be x m, then the length 2x m and the height h m. The quantity to be + 6 ?. But . = 2 1 and 1 3/2, the surface area is a minimum. 8 ? 0 & . For a TP, = 0; :. x = 3/2. Since ? or 9 = 2 &1 & = 8+ ?, i. e. ? = + 2 / = 24 when x = & . :. Dimensions of the tank are = 1. 5 ? 3 ? 2 , giving a surface area of 27 . 5 1. 6 Differentiation Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. Calculus is a subject that falls into two parts: (i) differential calculus (or differentiation) and (ii) integral calculus (or integration). Differentiation is used in calculations involving velocity and acceleration, rates of change and maximum and minimum values of curves.

Differentiation of 4 = 567 by the general rule 8 From differentiation by first principles, a general rule for differentiating ‘ are constants. This rule is: if y = ‘ , then = ‘9 8: ; When differentiating, results can be expressed in a number of ways. For example: (i) If y = 3×2, then dy/dx = 6x, (ii) If f(x) = 3×2, then (3 ) = 6x. 3×2 is 6x, (iv) The derivative of 3×2 is 6x, and (v) ? + ( )= 6x, (iii) The differential coefficient of or, if ( ) = ‘ 8 8 emerges where a and n ;( , then ) = ‘9 8: . Problem: Using the general rule, differentiate the following with respect to x: (a) y = 5×7 (b) y = 3vx (c) y = 4/x2 (d) y = + 4v +7 =’? , @ /@? =– ‘=;gt;9’? (where ‘ is a constant) and if – ‘=&gt;9(‘? + B) (where ‘ and B are constants). (where ‘ is a constant) and if = =;gt;9(‘? + B), @ /@? = ‘ =(‘? + B) (where ‘ and B are constants). Similarly, If If = =;gt;9? , then @ /@? = =?. It may also be shown that: If = =;gt;9’? , @ /@? = ‘ =’? Differentiation of Sine and Cosine Functions = =? , C? [email protected] /@? =– =&gt;9?. It may also be shown that: If = = =(‘? + B), @ /@? = Problem: An alternating voltage is given by: v = 100 sin 200t volts, where t is the time in seconds.

Calculate the rate of change of voltage when (a) t = 0. 005 s and (b) t = 0. 01 s. Differentiation of F56 and ln ax Problems: (1) If (C) = + If = ln , then @ /@ = 1/ . It may also be shown that if = ln’ , then @ /@ = 1/ . I JK If = D , then @ /@ = D . It may also be shown that if = D * , then @ /@ = ‘D * . , find ? (C) 6 (2) Evaluate (3) Given (? ) = 5 ln 2? ? 4 ln 3? , determine when = 1, given = 3D + ? I 2L ; (? ). + 8 ln 5 . Give the answer correct to 3 s. f. 1. 7 1. 7. 1 Techniques of Differentiation Differentiation of Common Functions coefficients of the separate terms.

Thus, if ( ) = (( ) + M( ) ? ( ), (where f, p, q, and r are functions), then ;( 1. The differential coefficient of a sum or difference is the sum or difference of the differential ) = (; ( ) + M ; ( ) ? =5 + ; ( ). 2. In general, the differential coefficient of a constant is always zero. Problems: (1) Differentiate (2) (a) Differentiate correct to 4 s. f. 1. 7. 2 When Differentiation of a Product = QR, and Q and R are both functions of , then =Q S T = N; + 2 ln 2? ? 2( =5? + 3=;gt;92? ) ? I 2O (b) Evaluate +4 ? ; + v ? 3 with respect to . N when ? = , P +R . This is known as the product rule.

Note that the differential coefficient of a product is not obtained by merely differentiating each term and multiplying the two answers together. The product rule formula must be used when differentiating products. Example: Find the differential coefficient of y = 3×2 sin 2x. [Answer: Problems: (1) Differentiate: y = x3 cos 3x ln x. (2) Determine the rate of change of voltage, given that v = 5t sin 2t volts, when t = 0. 2 s. =6 ( =2 + =;gt;92 )]. 1. 7. 3 If Differentiation of a Quotient T = S , and Q and R are both functions of , then = S UV UW :T UL UL S; . This is known as the quotient rule.

Note that the differential coefficient is not obtained by merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient formula must be used when differentiating quotients. Examples: (1) Find the differential coefficient of (2) Determine the gradient of the curve = ; [+ +XY8 Z . [Answer: v = + J (5 =5 ? 4=;gt;95 )] at the point (v3, ). Answer: ? 7 1. 7. 4 = Function of a Function T It is often easier to make a substitution before differentiating. If Examples: (1) If y = (3x ? 1)9; ? T is a function of then: . This is known as the ‘function of a function’ rule (or sometimes the chain rule). 27(3 ? 1)]. = 36C’9 3 =D 3 ]. (2) Differentiate y = 3 tan4 3x. [Answer: 1. 7. 5 Successive differentiation ; When a function or and is written as ( ). If the expression is differentiated again, the second differential coefficient is obtained & & = ( )is differentiated with respect to (pronounced dee two by dee the differential coefficient is written as ;; 2 squared) or ). By successive differentiation further higher derivatives such as = 2 D: & & & ( ) (pronounced 2 and Z double–dash Z may be obtained. Example: Given Solution: 18 D : Substituting values into & = 2 D: show that +6 & (i. . a product). Hence & +6 Problem: Evaluate = 0. Thus when = 2 D : , N& + 9 = 0 gives: 18 D : = 4=D 2? . & = ? 6 D : +9 = 0 + 2D : when ? = 0 given +6 + 9 = 0. ? 12D : and ? 36 D : & & = 18 D : + 12D : ? 12D : + 1. 7. 6 sec : Differentiation of Inverse Trigonometric Functions = cosec : = sin: , y = cos : , The inverse trigonometric functions: = cos , and = tan , = cot , correspond respectively to the trigonometric functions: = sec and = cosec . = tan: , = cot : = sin , , = The inverse trigonometric functions do not fall into any of the categories (product or quotient) so far considered.

Therefore new techniques are involved to differentiate them. It can be shown that if = ( ), then = tan: = f( ) and , = tan and = ? Example: If Problem: Differentiate the remaining inverse trigonometric functions. = =D = = 1 + C’9 ? . =1+ .? = ? = . [ ; 8 1. 7. 7 Let Differentiation of Implicit Functions be defined as an implicit function of to obtain .? 2 ( by the equation )+ ( )= the equation with respect to Notice that ( )= ( is a function of )? =2 which is itself a function of . Using the function of a function rule, + =;gt;9 = 3. = : Examples: (1) Find ?2 + 2 2 @ @ + = if @ @ 0. ? = : ; + + 2 = 0 so that . =? . (1). ? + = 1. Differentiate each term of ( ) + 2 = 0. (2) Find the gradient of the curve ?3 +3 [email protected] @ = 0. ? ; = + [hiX : ; ; . The gradient at = 2 at the point (1, 1) on the curve. = 1, = 1 is ? 1. 0. 0000004? . Determine the rate of change of length, in mm/°C, when the temperature is (a) 100 °C and (b) 400 °C. 1. The length j metres of a certain metal rod at temperature ? °C is given by: j = 1 + 0. 00005? + 1. 8 Some Applications of Differentiation 2. The luminous intensity I candelas of a lamp at varying voltage V is given by: I = 4 ? 10:+ V2.

Determine the voltage at which the light is increasing at a rate of 0. 6 candelas per volt. 3. Newton’s law of cooling is given by ? = ? D :kl , where the excess of temperature at zero time is ? °C and at time t seconds is ? °C. Determine the rate of change of temperature after 40 s, given that ? = 16 °C and k =? 0. 03. 4. The displacement s cm of the end of a stiff spring at time t seconds is given by: s = aD :kl sin 2mft. Determine the velocity of the end of the spring after 1 s, if a = 2, k = 0. 9 and f = 5. 5. The distance x metres moved by a car in a time t seconds is given by: x = 3t3 – 2t2 + 4t – 1.

Determine the velocity and acceleration when (a) t = 0 and (b) t = 1. 5 s. 6. Supplies are dropped from a helicopter and the distance fallen in a time t seconds is given by x = gt2, where g = 9. 8 m/s2. Determine the velocity and acceleration of the supplies after it has fallen for 2 seconds. 7. The distance x metres travelled by a vehicle in time t seconds after the brakes are applied is given by = 20C ? C . Determine (a) the speed of the vehicle (in km/h) at the instant the brakes are applied, and (b) the distance the car travelled before it stopped. 9 8. A rectangular area is formed having a perimeter of 40 cm.

Determine the length and breadth of the rectangle if it is to enclose the maximum possible area. 9. A rectangular sheet of metal having dimensions 20 cm by 12 cm has squares removed from each of the four corners and the sides bent upwards to form an open box. Determine the maximum possible volume of the box. 10. Determine the height and radius of a cylinder of volume 200 cm3 which has the least surface area. 10 PART 2 INTEGRAL CALCULUS AND INTRODUCTION TO ODE AND PDE 2. 0 Indefinite Integrals When differentiating a function we solve the problem of finding the derivative of a given function.

Now we proceed to the inverse problem: given the derivative of a function, to find this function. This is called INTEGRATION. whose derivative is the given function:n ; ( ) = ( ). Let us consider some examples elucidating the relationship between the given function and its integral. Let y = x 2 . For which function does x 2 serve as its derivatives? Obviously for Therefore the function Indeed, the derivative 2 2 2 Definition: An integral (a primitive) of a given function f(x) in a given interval is any function F(x) 😮 p = 2 ; is the integral of x 2 . However, the function x 2 possesses other integrals. 5, of 2 2 is also equal to x2 . Hence, any function of the form between the functions x2 and ? 100 and, generally, of 2 of the latter and the latter is the integral of the former. Similarly, for x;gt;0, any function of the form ln + q, for any arbitrary constant, C is that the former is the derivative + q is an integral of x2 . Thus, the connection 2 + q, where C is an arbitary constant, + C is an integral of , any function of the form Sinx + C is an integral of Cosx , etc. These examples indicate that a given function has infinitely many integrals because the constant C can take on arbitrary values. Therefore ? ( x )dx = f ( x ) + C . This is the indefinite integral of g(x). C is an arbitrary constant called the constant of integration and must always be added any time we integrate a function. Theorem: Every continuous function possesses an infinite number of integrals, any two of them only differing by a constant. Definition: The operation of finding integrals is called indefinite integration and the collection of all integrals of a given function f(x) is called the indefinite integral of f(x) and is denoted by the symbol 11 ? f (x )dx .

2. 1 The function f(x) is called the Integrand, the expression f ( x )dx is the element of ntegration and the variable x is the variable of integration. The Rules of Integration Theorem 1: The integral of a sum or difference of a finite number of functions is equal to the sum or the difference of the integrals of these functions: ? (u ± v ± ……………… ± w)dx = ? udx ± ? vdx ± …………. ± ? wdx, Where u, v, ……….. , w are functions of the independent variable x. Theorem 2: A constant factor in the integrand can be taken outside the sign of indefinite integration: ? kf (x )dx = k ? f (x )dx Examples: 1. (k= constant) ? (x 3 + tan x + e x dx = ) x4 ? log cos x + e x + C 4 . ? (2 sin x ? 3 cos x )dx = 2? sin dx ? 3? cos xdx = ? 2 cos x ? 3sin x + C Note: Though every intermediate integration contributes an arbitrary constant term to the final result we write only one such term since the sum of arbitrary constants is also an arbitrary constant. Theorem 3: Invariance of integration formulas Let ? f (x )dx = F (x ) + C be any given integration formula and U = g (x ) any function possessing continuous derivative. Then 2 ? f (u )du = F (u ) + C e. g. ? 2 xe x dx. observing that 2xdx is the differential d x 2 we write the integral as ( ) ? 2 xe x2 dx = ? x d e x = ? eu du = e x + C ; Where u = x 2 . 2 2 2 ( ) Theorem 4: The addition of a constant to the variable makes no difference to the form of the result. Theorem 5: Multiplying the variable by a constant makes no difference to the form of the result but we have to divide by the constant. 12 Theorem 6: The integral of a fraction whose numerator is the derivative of its denominator is the logarithm of the denominator. In some cases a constant factor has to be inserted to make the numerator exactly equal to the derivative of the denominator. In some cases also, we multiply by minus sign.

At times we have to separate the integrand f ( x ) into Examples: rs ( ) r( ) . 1. ? sin 5xdx = ? 5 cos 5x + C ? e ?3 x 1 2. 1 dx = ? e ? 3 + C 3 100 3. ? (2 x ? 1) ? (2 x ? 1) dx = 1 (2 x ? 1)101 + C 202 Note: 100 dx = 1 100 ? (2 x ? 1) d (2 x ? 1). 2 100 Let 2 x ? 1 = u;Q ? (2 x ? 1) dx = 1 100 1 u101 1 u du = +C = (2 x ? 1)101 + C ? 2 2 101 202 4. ? x x +1 2 dx = 1 2x 2 ? x2 + 1 dx = x + 1 + C 2 1 5. du 1 (3 x ? 1) 1 d (3 x ? 1) 1 = ln(3x ? 1) + C. = ? dx = ? ? 3x ? 1 3 3x ? 1 3 3x ? 1 3 6. ?1+ x x 2 dx = 1 2 xdx 1 2 ? 1 + x 2 = 2 ln 1 + x + C 2 ( ) 7. e x ? e? x x ? x ? e x + e ? dx = ln e + e + C ( ) 8. ? 2 x ? 1 dx = ? ?1. 5 + 2 x ? 1 ? dx = 1. 5x + 1. 75 ln(2 x ? 1) + C ? ? ? 1+ x x 4 3x + 2 ? 3 . 5 ? 9. dx Note that xdx = 1 d x2 2 ( ) 13 1 d x2 ? 1 + x4 = 2 ? 1 + x2 xdx ( ) ( ) 2 1 = arctan x 2 + C 2 ( ) 10. ?x 2 1 ? x? a? 1 1 d (x a ) 1 1 ? x a – 1 ? dx dx = 2? = ? = . ln? 2 2 2 ? x a + 1 ? + C = 2a ln? x + a ? + C ? ?a a ? x? a (x a ) ? 1 a 2 ? ? ? ? ? ? ? 1 ? a? dx = 1 x arctan + C a a 11. ?x 2 +a 2 An integral of the type ?x 2 du can readily be reduced to the above integrals. The denominator + px + q should be represented as a sum or a difference of squares.

Example: ?x 2 dx dx 1 x+2 =? = arctan +C 2 + 4x + 8 2 (x + 2) + 4 2 dx . Observe that the derivative of the denominator is 2 x + 4 we write the numerator 12. ? x+3 x + 4x + 8 2 in the form 1 (2 x + 4) + 1 and split the given integral into two integrals reducible to standard ones: 2 ?x 13. 2 x+3 1 2x + 4 dx 1 1 x+2 dx = ? 2 dx + ? 2 +C = ln x 2 + 4 x + 8 + arctan + 4x + 8 2 x + 4x + 8 x + 4x + 8 2 2 2 ( ) ? Sinx cos xdx = 2 ? sin 2 xdx = 4 ? sin 2 xdx(2 x ) = ? 4 cos 2 x + C 1 1 1 1 2 sin x + C 2 Alternatively: ? sin x cos xdx = ? sin xd (sin x ) = 1 Or ? sin x cos xdx = ? ? cos xd (cos x ) = ? os2 x + C 2 It may seem that for one and the same integral we have obtained three different expressions 1 1 1 ? cos 2 x + C , sin 2 x + C and ? cos 2 x + C . However, it can be easily checked that the difference 4 2 2 between any two of them is constant. Therefore each of these expressions describes the set of all integrals of =;gt;9 = . 14 14. The integrals ? sin 2 xdx and ? cos 2 xdx can be found using the formulas sin 2 x = 1 + cos 2 x 1 ? cos 2 x and cos2 x = 2 2 1 1? sin 2 x ? ? (1 + cos 2 x )dx = 2 ? x + 2 ? + C 2 ? ? 3 ? cos 15. 2 xdx = ? cos xdx = ? cos 2 xd (sin x ) = ? (1 ? sin 2 x )d (sin x ) = ? (sin x ) ? ? sin 2 xd (sin x ) 1 = sin x ? sin 3 x + C 3 2. 2 2. 2. 1 Integration by Parts and by Change of Variable Integration by Parts The method of integration by parts is implied by the formula for differentiating a product of two functions. Let u(x) and v(x) be functions of x possessing continuous derivatives. We have d (uv ) = udv + vdu or udv = d (uv) ? vdu Integrate both sides of the above equation to get ? udv = ? d (uv ) ? ? vdu i. e. ? udv = uv ? ? vdu. This is the formula of integrating by parts. Examples: 1. ? xe x dx. Put: u = x, du = dx; dv = e x dx, v = e x . Substitute into the formula to obta xe dx = xe ? ? e dx = e (x ? 1) + C x x x x 2. ? xlnxdx = dx 1 x2 x2 , dv = xdx , v = x 2 lnx ? + C . Let u = ln x, du = x 2 2 4 3. 4. ? log xdx = x log x ? x + C. Verify! ?x ? x 2 cos xdx. Put: u = x 2 , du = 2 xdx , dv = cos xdx , v = sin x cos xdx = x 2 sin x ? 2 ? x sin xdx 2 The last integral is again found with the aid of integration by parts. 15 ?x 2 cos xdx = x 2 sin x + 2 x cos x ? 2 sin x + C Repeated integration by parts makes it possible to find the integrals. ?x m sin xdx , ? x m cos xdx , ?x m x e dx where m is a positive integer, and hence the integral x ? P(x )sin xdx , ? P(x )cos xdx , ?

P(x )e dx. Repeated integration by parts sometimes leads to the original integral, and then we either arrive at a useless identity (if the repeated integration has been done improperly) or at an equality that enables us to find the integral required. Let us consider an example demonstrating this technique. ?e x cos xdx. Put: u = cos x , du = sin xdx , dv = e x dx , v = e x ? ? e x cos xdx = e x cos x + ? e x sin xdx………………………. (1) Let us again apply integration by parts. Put: u = sin x , du = cos xdx ; dv = e x dv , v = e x ? ? e x sin xdx = e x sin x ? ? e x cos xdx…………………….. (2 ) .

Put (2) in (1) and face the consequence. ?e x cos xdx = e x cos x + e x sin x ? ? e x cos xdx ? 2 ? e x cos xdx = e x cos x + e x sin x ? ? e x cos xdx = ex (sin x + cos x ) + C 2 2. 2. 2 Integration by Change of Variable (by Substitution) to another A commonly used device in integration is to change the independent variable, say oneQ, provided that the relationship between x and u is known. 1 Examples: 1. ? x 2 3 4 ? 3×3 dx . Let: u = 4 ? 3×3 , du = ? 9 x 2 dx and x 2 dx = ? du 9 ? ? x 2 3 4 ? 3×3 dx = ? 13 u du = ? 9 Q+? + q = ? 1 4 ? 3x 3 12 ( ) 3 4 ? 3x 3 + C 2. ? ? ex + 1 ? 1? ?+C = ln? ? ex +1 +1? ex +1 ? ? dx 16 3 1 1 3 3. x 5e x dx = e y ( y ? 1) + C = e x x 3 ? 1 + C 3 3 ( ) 1? 1 ? 4. ? 1 ? x 2 dx = ? cos 2 udu = ? u + sin 2u ? + C 2? 2 ? = 1 arcsin x + x 1 ? x 2 + C 2 ( ) Notes: If the integrand involves: i. iii. a 2 ? x 2 , try x = a sec? a 2 ? x 2 , try x = a sin ? iv. ii. v. a 2 + x 2 , try x = a tan ? e f ( x ) , put u = f(x) a + x, put = a + x cos5 x cos 3 x 5. ? sin x cos xdx = ? +C 5 3 3 2 6. 16 ? x 2 x 2 dx = ? 1 x 16 ? x 2 ? sin ? 1 + C x 4 2. 2. 3 Integration of Rational Algebraic Fractions We have so far introduced the basic ideas and rules of integration using simple integrals obtained from the inverse of differential coefficients.

The aim of this section is to treat the more complicated integrals (which cannot be integrated by use of the general rule). We try to resolve them into simpler forms readily integrable or recognizable as standard integrals. Note that no student will progress happily with Part 2 of this course without proper investment of time and effort because ability to integrate comes readily only with experience and constant practice. Denominator of the First Degree A. If the numerator is of the same degree as or higher degree than the denominator, we first divide out. 2x ? 8×2 1 ? ? 1 ? dx = x ? x 2 ? og(1 + 4 x ) + C dx = ? ?1 ? 2 x ? Examples: 1. ? ? ? 1 + 4x 1 + 4x ? 4 ? x3 1 ? ? x3 x2 ? dx = dx = ? ? x 2 + x + 1 + 2. ? + + x + log( x ? 1) + C ? x ? 1 x ? 1? 3 2 ? ? B. Denominator of the second degree and which does not revolve into rational factors. Here we shall discuss two cases: 17 i. ii. When the numerator is constant. When the numerator is a linear expression in x. i. Consider ? ax k 2 + bx + c dx. This can always be put in the form k dx ? (x + ? )2 ± ? 2 a We shall consider the case +” ? 2 ” only. ? 5×2 + 6 x +1 2 (x + ? ) k dx k 1 2 ? (x + ? )2 + ? = a x ? arctan ? + C a 2. Examples: 1. ? x = 5 x + arctan x + C ? 2x 2 5 2x ? 1 dx = arctan +C 5 ? 2 x + 13 ii. If the numerator is a linear expression in x we put it equal to k times (the derivative of the denominator) plus l, where k, l can be determined by inspection. The integral now splits into 2 parts. Examples: 1. 5x + 1 ?x x+7 1 2x dx 1 7 x dx = ? 2 dx + 7 ? 2 = log x 2 + 16 + arctan + C 2 + 16 2 x + 16 x + 16 2 4 4 ( ) 2. ? = (6 x ? 12) 5 dx dx = ? 2 dx + 11? 2 6 3 x ? 12 x + 13 3 x ? 12 x + 13 3 x ? 12 x + 13 2 5 11 5 11 dx = log 3x 2 ? 12 x + 13 + arctan 3 ( x ? 2) + C log 3 x 2 ? 12 x + 13 + ? 2 6 3 x ? 4 x + 13 3 6 3 ( ) ( ) C.

Denominator which revolves into rational factors of the first and second degrees. If the denominator factorises we use the technique of partial fractions to express the integrand in a form suitable for integration. Examples: 1. ?x x? 3 3 ?x dx . Notice that x 3 ? x = x( x ? 1)( x + 1) Put x? 3 A B C = + + . Solving gives: A = 3, B = ? 1, C = ? 2 3 x ? x x x ? 1 x +1 ? ? x? 3 dx dx dx x3 dx = 3? ? ? ? 2? = 3lnx ? ln( x ? 1) ? 2ln( x + 1) + C = ln? 3 ? (x – 1)( x + 1)2 ? + C ? x ? x x x ? 1 x +1 ? ? x? 5 x ? 3x + 4 3 2 ?? 2. ? dx . x 3 ? 3 x 2 + 4 = ( x + 1)( x ? 2 ) 2 Set x? 5 x ? 3x + 4 3 2 = A B C 2 2 + + A = ? , B = , C = ? 2 ? x + 1 x ? 2 (x ? 2) 3 3 18 ?? 2 2 1 x ? 5 2 dx 2 dx dx = ? ( x + 1) + ln( x ? 2) + +C dx = ? ? + ? ?? 2 2 3 3 x? 2 x ? 3x + 4 3 x +1 3 x ? 2 (x ? 2) 3 2. 3 Definite Integrals There are various problems leading to the notion of the definite integrals: Determining the area of a plane figure, computing the work of a variable force, finding the distance travelled by a body with a given velocity, and many other problems. Recall that an arbitrary constant can always be added any time we integrate a function. If conditions are stipulated such that the arbitrary constant can be evaluated the result is a definite integral.

If the function f(x) is continuous in the closed interval [a, b], its integral sum tends to a definite limit as the length of the greatest subinterval (the interval of continuity is divided into subintervals) tends to zero provided that the limit exists. 2. 3. 1 Some Properties of the Definite Integral a. The integral of a sum of a finite number of functions is equal to the sum of the integrals of these functions. ? (u + v + ………… + w )dx = ? udx + ? vdx + ……….. + ? wdx b b b b a a a a b. A constant factor in the integral can be taken outside the integral sign. ? cudx = c ? udx a a b b c.

If the upper and the lower limits of the definite integral are interchanged the integral only changes its sign: ? f ( x )dx = ? ? f ( x )dx b a a b d. If the interval of integration [a, b] is split into two parts [a, c] and [c, b] then ? f (x )dx = ? f (x )dx + ? f (x )dx b c b a a c Examples: 1. ?e 2 3 3x 3 1 1 dx = e3 x = e 9 + e 6 ? 2600 2 3 3 ( ) 2. ?? ? 3 1 1 ? 3 cos 3 xdx = sin 3x ? 6 = ? 6 3 3 3. A particle starts with an initial speed of 30m/s. Its acceleration at any time t is (18 + 2t) m/s2. Find the speed at the end of 6 seconds and the distance travelled in that time. 4. Find the equation of the curve whose gradient is 2 x 2 ? and which passes through the point ( ? 1, 0). Solution: 2 x3 dy = 2 x 2 ? 1 or ? dy = ? (2 x 2 ? 1)dx i. e. y = ? x+C 3 dx 19 But the curve passes through the point ( ? 1, 0). ?C = 1 2 x3 1 ? y = ? x+ 3 3 3 5. Find the area bounded by the parabola y = 6 ? x ? x 2 and the x-axis. Theory: To find the area between the curve y = f(x) and the x-axis, we first find the indefinite integral ? ydx or ? f ( x )dx . We then substitute the upper and lower limits in the indefinite integral and subtract the two results. The notation adopted is ? f (x )dx b a or ? b a ydx . Now let us solve the above Example 5.

The curve crosses the x-axis when y = 0, i. e. 6 ? x ? x 2 = 0 giving x = ? 3 or 2 ? Required area = ? 6 ? x ? x 2 dx = ? 3 2 ( ) 6. Calculate the area between the curve = (6 ? ) and the line = 5. Solution: Represent the curve and the line in a diagram and shade the required area. The abscissae of the equation (6 ? ) = 5, i. e. x 2 ? 6 x + 5 = 0 , giving = 1 ? Area = ? x(6 ? x )dx ? area ABCD = 5 1 125 sq. units 6 the points, say, A and B where the quadratic graph meets the line y = 5 are given by the solutions of 5 32 sq. units . (Please come to class if you wish to see the 3 area I am talking about). 7.

Find the area between the curve y = x3 , the axis of y and the lines = 1, = 8. Solution: Area = ? xdy = ? y1/ 3 dy = 1 1 8 8 45 sq. units 4 2. 3. 2 The volume swept out when the area enclosed by the curve = ( ), the x-axis and the ordinates given by: Volume of revolution = ? ?y 2 dx ……………………….. (1) a b Volume of a Solid of Revolution = ‘, = t is rotated through 2? radians or one complete revolution about OX (i. e. the x- axis) is When any portion of the area contained between the curve and the y-axis is rotated about the y-axis the volume swept out is given by: Volume of revolution = ? ?x 2 dy ………………………… (2) c d 0 Solution: Volume = ? ?y 2 dx = ? ? (x 4 + 2 x 2 + 1)dx = 3 3 2 2 and the ordinates at = 2and = 3 is rotated through 2? radians about the x-axis. Examples: 1. Find the volume swept out when the area between the parabola y = x 2 + 1 , the x-axis centre and both measured in the same direction. Assume that ? &gt; ? . 2. Find the volume cut from a sphere of radius a by two parallel planes distances ? , ? from the 838? cubic units 15 Solution: A sphere is generated by the rotation through 360o of a semi-circle about its bounding diameter. Take the centre of the circle as origin and the bounding diameter as the x-axis.

Then by equation (1), required volume = ? ?y 2 dx h1 h2 The equation of the semi circle is x 2 + y 2 = a 2 ? Required volume = ? ? h2 h1 (a 2 ? x 2 dx = ) ? 3 (h2 ? h1 )[3a 2 ? (h22 + h1h2 + h12 )] 2. 4 Improper Integrals Up to now, when speaking of definite integrals we assumed that the intervals of integration was finite and closed and that the integrand was continuous. It is this particular case to which the existence theorem for the definite integrals applies. However, it often becomes necessary to extend the definition of the definite integral to an infinite interval of integration or to an unbounded integrand function.

At this point we shall consider the case of infinite limits (the infinite integrals). Definition: The improper integral u* of u* ( )@ w v equation (*) is meaningless and the improper integral u* Examples: 1. u D : @ = ? D : v ? 1 ? 0 If the limit exists the improper integral u* as l > ? provided that this limit exists:u* v ( )@ of the function f (x) over the interval [a,? ) is the limit ( )@ is said to be convergent. If the limit does not exist v v ( )@ = limw>v u* ( )@ w (*) ( )@ is said to be divergent. ? = 0 +1 = 1. Therefore ? e ? x dx is convergent and equal to 1. 0 2. ? dx ? ln x 1 = ?. The integral is divergent because the integral ln x tends to infinity as x > ? x 21 3. ? cos xdx is divergent since the integral 0 ? ? cos xdx = sin x = sin l has no limit as l > ? (it 0 0 l l oscillates between – 1 and +1). 4. ? dx ? ? = arc tan x ?? = ? (? ? 2) = ? 2 ?? 1 + x 2 ? ? ? ?? 5. ? e x dx = e x ?? =?? 0=? 6. ? x 1 dx = ln 1 + x 2 ?? 1 + x 2 2 ? ( ) ? ?? Here both f (-? ) and f (? ) are equal to infinity, and the integral is divergent. 7. ? ? a 1 1 1 dx = ? 3 x 2 x2 ? ? = a 1 2a 2 8. ? ? a dx x ? m+1 = for m ? 1 xm ? m + 1 a If m &gt;1 the exponent 1 ? is negative, and the integral tends to zero as x>? ; therefore the integral is convergent. If m 0, and the integral tends to infinity as x >? , i. e. the integral is divergent. If m = 1, the integral is divergent. 2. 5 Some Applications of Integral Calculus to Engineering Systems Natural Law of Growth and Decay (NLGD) In nature and society it often happens, under ideal conditions, that the rate of growth or decay (negative growth) of a natural substance or a social matter (e. g. population) at time t, is proportional to the amount A (t) at that time. In mathematical language, A? (t ) = KA(t ) , i. . A? ? KA = 0 , where K is the constant of proportionality.

This is of course a first order homogeneous OLDE whose general solution is given by theorem to be A(t ) = Ce ? kt , where C is any constant. If the value of A(t) at t = 0 (i. e. initial value) is known or given, then C=A(0)=A0, and so the unique solution for that particular substance at time t is A(t ) = A0 e ? kt . This gives the amount A at any time t. The formula A(t ) = A0 e ? kt is called the Natural Law of Growth and Decay (NLGD). Radioactive substances, bacteria cultures, population of nations, etc, all normally obey this law. 2 Examples: 1. Find the curve through the point (1, 1) in the xy-plane having at each of its points the slope – . Solution: y? = ? 1 y c ? y = . When x = 1, y = 1 ? c = 1 ? The curve is y = x x x 2. Newton’s Law of Cooling: A copper ball is heated to a temperature of 100oC. Then at time t = 0 it is placed in water which is maintained at a temperature of 30oC. At the end of 3 minutes the temperature of the ball is reduced to 70oC. Find the time at which the temperature of the ball is reduced to 31oC. Solution: dT = ? k (T ? 30) . Separate variables and integrate. T (t ) = Ce ? kt + 30 dt 7 When C = 0, y = 100 ? T (0) = 100 ? T (t ) = 70e ? kt + 30 . But T (3) = 70, k = ln = 0. 1865 3 4 ? When y = 31oC, C = 22. 78 &gt;9=. EXERCISES 3. Velocity of escape from the earth: Find the minimum initial velocity of a projectile which is fired in radial direction from the earth and is supposed to escape from the earth. Neglect the air resistance and the gravitational pull of other celestial bodies. 4. Skydiver: Suppose that a skydiver falls from rest towards the earth and the parachute opens at an instant, call it t = 0, when the skydiver’s speed is v (0) = V0 = 10. 0 m/s.

Find the speed v (t) of the skydiver at any later time t. Does v (t) increase indefinitely? 5. Radioactivity, exponential decay: Experiment shows that a radioactive substance decomposes at a rate proportional to the amount present. Starting with a given amount of substance, say 2 grams, at a certain time say, t = 0, what can be said about the amount available at a later time? 6. Flow of water through orifices (Torricelli’s law): A cylinder tank 1. 50 m high stands on its circular base of diameter 1. 00 m and is initially filled with water. At the bottom of the tank there is a hole of diameter 1. 0 cm, which is opened at some instant, so that the water starts draining under the T (t ) = 70e ? 0. 1865t + 30 . 23 influence of gravity. Find the height h(t) of the water in the tank at any time t. Find the times at which the tank is one- half full, one-quarter full, and empty. 7. How much of radioactive Polonium (Po) is stilling remaining after t days, if the initial amount is 10 mg and if Po has a half-life of 140 days? How much is still left after 1 year (360 days)? 8. On a flat private road, a driver turned off the engine of his car and allowed