MATHEMATICS FOR ENGINEERING

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1 MATHEMATICS FOR ENGINEERING INTEGRATION TUTORIAL FURTHER INTEGRATION This tutorial is essential pre-requisite material for anyone studying mechanical engineering. This tutorial uses the principle of learning by eample. The approach is practical rather than purely mathematical and may be too simple for those who prefer pure maths. Calculus is usually divided up into two parts, integration and differentiation. Each is the reverse process of the other. It is easier to eplain integration first and a numerical approach helps eplain what it is about. On completion of this tutorial you should be able to do the following. Revise Basic Integration Integrate Polynomials Integrate Trigonometric Epressions Integrate Logarithmic Epressions Integrate Eponential Epressions Integrate by Substitution Integrate by Parts List of Standard Integrals n+ n a a d + C n + d ln + C a a e d e + C a sin a d cos + C a cos a d sin + C a tan d -ln + C sin d - sin + C cos d + sin + C tan d ln d tan - + C ( ln -) + C D.J.Dunn

2 . REISION OF INTEGRATION Lets consider how to find the area under the graph of y f() +. A graph of this function looks like this. If we solve the area by use of calculus the area would be precisely solved as follows. Over the range to the area is epressed as follows. A (+ )d Carrying out the integration gives the following. Evaluating between limits we get the following. A (+ A )d + [( +.) ]. units A (+ + + )d + This is a precise answer and we will compare it with the results found in the following work. GRAPHICAL METHODS Consider the same function again and this time more grid lines are shown. COUNTING RECTANGLES A simple but crude way to find the area under the graph is to count the rectangles. Each rectangle on the graph above has an area of unit. Count them up judging the divided ones to the nearest half. You should get an answer of about units depending on how good you are at doing it. D.J.Dunn

3 MID-ORDINATE RULE The values of y corresponding to,, and so on are called the ordinates. The values of y corresponding to.5,.5,.5 and so on are called the mid-ordinates. Each column is approimately a rectangle w wide and h high. The area is approimately w h. The area under the whole graph is approimately A w h + w h + w h +w h A w(h + h + h + h ) Usually, as in this case, w Putting in the mid-ordinate values we find the following. A ( ) units Clearly if we took more strips by say making w.5, we would get a more accurate answer and in the limits as w becomes very small the answer will be the same as found by integrating. TRAPEZOIDAL RULE Consider that each strip has a straight line joining the top corners as shown. The height at the middle is not quite the same as the mid-ordinate and is the average of the two ordinates. If h is the average then h (A+B)/ h (B+ C)/ h (C+ D)/ h (D+E)/ The area of each strip is wh w(a+b)/ wh w(b+ C)/ wh w(c+ D)/ wh w(d+e)/ The total area is A (w/)[(a+b) + (B+C) + (C+D) + (D+E)] A (w/)[(a+b + B+C + C+D + D+E] A (w/)[(a + E) + (B+C+D)] Hence in our eample A (/)[(+9) +(+7+)] (/)(+6) This is slightly larger than the correct figure but again, if smaller strips are used, the answer will be accurate. The above rule may be written as follows. w A [( First + Last) + sum of the rest] D.J.Dunn

4 WORKED EXAMPLE No. Find the area under the graph of the function y sin θ between the limits and radians using integration and the trapezoidal rule. Evaluating the ordinates and mid-ordinates at intervals of /8 produces the table and graph shown. Integration A sinθ dθ [- cosθ] cos + cos + Mid-ordinates w / A w(h + h + h + h ) ( ).68 Trapezoidal Rule w / w A [( First + Last) + sum of the rest] / A [( + ) + ( ) ].896 units Note one answer is slightly large and the other slightly small. SELF ASSESSMENT EXERCISE No.. Find the area under the graph of the following functions using integration, the mid-ordinate rule and the trapezoidal rule. y between the limits and 5 y e between the limits of and 5. y sin between the limits and 8 o.. Estimate the value of the definite integral I d by Simpson's rule using four strips. What is the error in the estimate? (689. and 6.) 5 D.J.Dunn

5 In Engineering the area under the graph represents real things. For eample the area under a force distance graph represents the work done or energy used and the area under a pressure volume graph also represents work done during the compression or epansion of a gas. WORKED EXAMPLE No. The pressure (p) and volume () during a gas epansion is related by the law p. -.. Determine the work done when the volume is epanded from -6 m to -6 m. Use calculus and the trapezoidal rule to find the answer. INTEGRATION Area Work W. W. 6. [ ] ( ) ( ) [ ].69 Joules W pd but p. -.+ d TRAPEZOIDAL RULE w -6 m w W W W 5 [( First + Last) + sum of the rest] [( +.6 ) + ( )] [(.6 ) + (8.9].87 Joules -6 D.J.Dunn 5

6 SELF ASSESSMENT EXERCISE No.. The electric current charging a capacitor is related to time by the following law. I ( e -t/ ) Amps Calculate the charge Q (the area under the graph) between the limits t and calculus and the trapezoidal rule. (Graph and ordinates are calculated for you) (Answer around Coulombs) t 6 s. Use. Find the area (with units) under the following function between the limits and m using integration, the mid-ordinate rule and the trapezoidal rule. (Answers around 76.7 m ) y + m. Find the area under the following function between the limits t and t s using integration, the mid-ordinate rule and the trapezoidal rule with steps of.s. v t + e t m/s (Answers around. m). Find the area (with units) under the following function between the limits θ and θ. radian using integration, the mid-ordinate rule and the trapezoidal rule with steps of. radian T cosθ (Nm) (Answers around.96 Joules) 5. Find the area (with units) under the following function between the limits and 5 m using integration, the mid-ordinate rule and the trapezoidal rule with steps of. p ln N/m (Answers around 6. Nm or Joules) D.J.Dunn 6

7 . INTEGRATING POLYNOMIALS The rule for integrating a single polynomial is: a n n + a n + Note that no power shown against a variable (e.g. ), means and that this integrates as /. Anything raised to the power of zero is so a number on its own integrates e.g. could be written as and this integrates to /. If the polynomial is a sum each term integrates separately. WORKED EXAMPLE No. Evaluate F() ( )d F() ( F() F() [( 8 ) ( 8 6) ] F() )d SELF ASSESSMENT EXERCISE No.. Integrate the following epressions. i. y 5 / ii. y /. Evaluate the following. i. ( )d (Answer 8.67) 5 ii. ( )d (Answer 5) D.J.Dunn 7

8 . INTEGRATING TRIGONOMETRIC EXPRESSIONS The standard integrals are given in the table and used in the following eamples. WORKED EXAMPLE No. Evaluate A sin d From the list of standard integrals we see sin - cos so: A [ cos ] [ cos ] - [- cos ] - (-) - (-) WORKED EXAMPLE No. 5 Evaluate A / sin d From the list of standard integrals we see sin ½ ¼ sin so: A sin A / sin d / [( ) ( ) ]. 785 sin sin SELF ASSESSMENT EXERCISE No. Solve the following integrals. All angles are in radian... W d (Answer 5.9).. A sin( θ ) dθ (Answer ).5. A cos ( θ ) dθ (Answer.6).5. A sin ()d (Answer.55) D.J.Dunn 8

9 . INTEGRATING LOGARITHMIC AND EXPONENTIAL EXPRESSIONS The standard integrals are given in the table and used in the following eamples. WORKED EXAMPLE No. 6 Evaluate A e d From the list of standard integrals we see e a e a /a so: e A e e - WORKED EXAMPLE No. 7 Evaluate I ln() d Using the standard integral I [ { ln() } ] { ln( ) }.55 WORKED EXAMPLE No. 8 Evaluate I ln() Using the standard integrals I ln() - d I d [ { ln() } ln() ] [ { ln() } ln() ] [ { ln() } ln() ] [.8.97] I [ ] SELF ASSESSMENT EXERCISE No. 5. I { ln() e } d (-). I { sin() ln() } d (.) 5. I { e ln( )} d (). I sin () - ln() } d (.8) { D.J.Dunn 9

10 . INTEGRATION BY SUBSTITUTION A comple equation may be simplified with a substitution but it takes eperience to recognise these cases. WORKED EXAMPLE No. 9 7 Evaluate I d 8 + A suitable substitution is z 8 Differentiate to get 7 7z Substitute back into the original equation I dz + C 8 8 Substitute for z 7 I C 8 + dz d dz 8 d 8 + WORKED EXAMPLE No. The voltage c across a capacitor when it discharges through a resistance is given by dc ( S C ) T where T is a time constant and s is the voltage at t. dt Find the equation relating c with time t. Let S C Differentiate and since S is a constant we find -d C d d The equation becomes T Rearranging dt t d Integrating dt [ ln ] T The limits become clear when we substitute S C dt T d T t T t dt [ ln ( )] C S C [ ln ( ) ( ] S C ln S ) d S ln S C Take antilogs and Rearrange ln e t T C S S C S - e T t t T D.J.Dunn

11 WORKED EXAMPLE No. Find I ( ) d Substitute z and ( z)/ Differentiate to get dz -d and substitute d -dz/ Substitute to get rid of all the terms - z dz z z I ( ) d (z ) dz z z z z I + C + + C Now substitute back 6 5 ( - ) ( - ) I + C ( ) ( ) ( ) dz z - z 5 WORKED EXAMPLE No. Find I sin () θ cos( θ) dθ dsin Substitute sin(θ) and noting that () ( θ) d cos θ hence cos(θ) dθ d dθ dθ I sin () θ cos() θ dθ d sin θ I + C + C SELF ASSESSMENT EXERCISE No. 6 Solve the following integrals. d. + 5 I ( + 5) ( + 5) + C d. 6 + d. I ln( 6 + ) C ( ) + I sin ( ) + C sin. I sin () θ cos() θ θ dθ + C d 5. I 6 ln( + ) + C + D.J.Dunn

12 5. INTEGRATION BY PARTS The rule is without eplanation udv uv vdu and best shown with an eample. WORKED EXAMPLE No. Find I e d Let u and let e dv du d and v e e udv uv vdu e e d e e + C e ( ) + C WORKED EXAMPLE No. Find I e d Let u and let e e dv du d and v e e e e udv uv vdu d e d We must repeat the process for the integration of e d e Let u and dv e du d v e e e e e e d d Now put the two parts together including the constant of integration. e e e e I e d C C e + C e I [ + ] + C SELF ASSESSMENT EXERCISE No. 7 Solve the following integrals.. I ln() d 6 I cos() + sin() + e I sin() + cos() e I cos() + sin() I [ ln( ) ] + C. I sin() d ( ) C. I e cos() d. I e sin() d cos() - ( ) + C + C D.J.Dunn