CHAPTER 4. INTEGRATION 68. Previously, we chose an antiderivative which is correct for the given integrand 1/x 2. However, 6= 1 dx x x 2 if x =0.

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1 CHAPTER 4. INTEGRATION 68 Previously, we chose an antierivative which is correct for the given integran /. However, recall 6 if 0. That is F 0 () f() oesn t hol for apple apple. We have to be sure the function is well efine over the entire interval over which we integrate. Eample 4.8. Consier the function f(). Then the integral over the interval [, ] is apple Figure 4.8: Integrating to fin the shae area uner the curve y on the interval [, ]. Have two shae regions boune by the curve an the -ais. In this case, the area cancels out. The shae area is actually given by apple apple Inefinite integrals So far we know f() F ()+C, F 0 () f(). What function F can we i erentiate to get f? Powers of : n n + n+ + C, (n 6 ),

2 CHAPTER 4. INTEGRATION 69 since n+ n. n + For n, ln + C, since for >0 an for <0, (ln ), (ln ) (ln( )). Special rule: Let us consier the erivative of the logarithm of some general function f(), i.e. This implies that [ln(f())] f() [f()] f 0 () f(). f 0 () ln(f()) + C, f() where C is some arbitrary constant of integration. Hence, if we can calculate integrals by inspection if the integran takes the form f 0 ()/f() by the above formula. Eample 4.9. Consier the the following integral: +5 I Now, if we choose f() +5 +, then f 0 () + 5. So, if we i erentiate ln(f()), in this case we have ln( +5 + ) , by the chain rule. Thus, we know the integral must be I ln( +5 + ) + C, where C is some arbitrary constant of integration. Eample 4.0. Consier the following integral: I +. Now, if we choose f() + then f 0 (). However, the numerator of the integran is. Not to worry, as we can simply re-write or manipulate the initial integral as follows: I Since / is a constant, which we are able to take out of the integral sign, we nee not worry about this an can procee with the integration using what we have learnt above, giving I ln( + ) + C.

3 CHAPTER 4. INTEGRATION 70 To check, we i erentiate the above epression, so which is correct! I apple ln( + ) + C +, This special case is an eample of a metho calle substitution, an is not limite to integrals which give you logarithms. Nevertheless, it is a goo sighter for what s to follow. Together with inspection, it can be etene for other function by choosing suitable substitutions (i.e. f()). Trigonometric functions: cos sin + C, sin cos + C, since ( since (sin ) cos, cos ) sin. Eponential function: e e + C, since (e )e. As with i erentiation, we also have a sum rule for integration, that is (f()+g()) (f) + g(). (4.4) In other wors, the antierivative of a sum is the sum of the antierivatives. Also, Kf() K f(), (4.5) where K is a constant, i.e. a constant can be taken outsie of the integration sign. Eample 4.. ( ) C. NOTE: Do not forget the constant C when the integral is inefinite! Eample 4.. ( )( 4 + ) C.

4 CHAPTER 4. INTEGRATION 7 Eample C. 4.. Substitution We can use substitution to convert a complicate integral into a simple one. Eample 4.4. Consier the inefinite integral with integran ( + ) 00. We make the substitution u + ) u i.e. u. So we calculate the integral as follows: ( + ) 00 u 00 u u 00 u 0 ( + )0 + C. We can check the result by performing the following i erentiation: apple 0 ( + )0 + C 0 0 ( + )00 ( + ) 00, Eample 4.5. Suppose we have the integran ( + ) 50. Let us try the substitution So we have u +, so u ) u i.e. u. ( + ) 50 (u )u 50 u u 5 u u 50 u 5 u5 5 u5 + C 5 ( + )5 5 ( + )5 + C apple 5 ( + )5 5 ( + )5 + C ( + ) 5 ( + ) 50 ( + ) 50,

5 CHAPTER 4. INTEGRATION 7 Eample 4.6. Consier the integran /( ln ). Let us try u ln ) u i.e. u. Thus, we calculate the integral as ln u u u u ln u + C ln ln + C. (ln ln ) ln, Eample 4.7. Consier the integran /( + p ). Let us try the substitution Di erentiating we have u + p ) p u. u ) u (u )u. Therefore, we calculate the integral as + p u (u ) u u u u u u u ln u + C ( + p ) ln + p + C. ( + p ) ln + p + C p p ( + p ) +p p ( + p ) Eample 4.8. Consier the integran sin( + ). Let us try the substitution So integrate as follows: u + ) u i.e. u. sin + sin uu cos u + C cos( + ) + C. apple cos( + ) + C + sin( + ) sin( + ). + p.

6 CHAPTER 4. INTEGRATION 7 Eample 4.9. Consier the integran sin /. Let us try the substitution u So we calculate the integral as follows: ) u i.e. u. sin sin u ( ) u sin uu cos u + C cos + C. apple cos + C sin sin, How o we fin a suitable substitution? Usually by observation an using our knowlege of i erentiation. Or we put If we have then we can write u f(), For instance, in the case then we know that (ln ) So we write the integral as ln Therefore we know u ln will work! u f 0 () ) u f 0 (). [f()] f 0 (), [f()] f 0 (). ln, i.e. (ln ). (ln ) ln (ln ). ln Similarly we know, i.e. so the integral from eample 4.9 can be written as sin sin, ) u.

7 CHAPTER 4. INTEGRATION 74 Also, from eample 4.4, we have ( + ) 00 since ( + ) ( + ) 00 ( + ), ) ( + ) ) let u +, which is the same as what we trie earlier. En Lecture 7.