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integration by parts with examples

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     integration by parts integration by parts Presentation Transcript

    • “Teach A Level Maths” Vol. 2: A2 Core Modules25: Integration by Parts © Christine Crisp
    • Integration by Parts Module C3 Module C4 AQA Edexcel MEI/OCR OCR"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used withpermission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
    • Integration by PartsThere is a formula for integrating some products.I’m going to show you how we get the formula butit is tricky so if you want to go directly to thesummary and examples click below. Summary
    • Integration by PartsWe develop the formula by considering how todifferentiate products.If y  uv , dy du dv where u and v are v  u both functions of x. dx dx dxSubstituting for y,  d ( uv ) du dv v u dx dx dxe.g. If y  x sin x , d ( x sin x )  sin x  1  x  cos x dx
    • Integration by PartsSo, d ( x sin x )  sin x  1  x  cos x dxIntegrating this equation, we get    d ( x sin x ) dx  sin x dx  x cos x dx dxThe l.h.s. is just the integral of a derivative, so,since integration is the reverse of differentiation,we get  x sin x  sin x dx   x cos x dxCan you see what this has to do with integrating aproduct?
    • Integration by Parts  x sin x  sin x dx   x cos x dx Here’s the product . . .if we rearrange, we get  x cos x dx  x sin x   sin x dxThe function in the integral on the l.h.s. . . . . . . is a product, but the one on the r.h.s. . . .is a simple function that we can integrate easily.
    • Integration by Parts  x sin x  sin x dx   x cos x dx Here’s the product . . . if we rearrange, we get  x cos x dx  x sin x   sin x dx  x sin x  (  cos x )  C  x sin x  cos x  C So, we’ve integrated xcos x !We need to turn this method ( called integration byparts ) into a formula.
    • Integration by Parts Example d ( x sin x ) Generalisation  sin x  x cos x dx d ( uv ) du dv Integrating: v  u dx dx dx    d ( x sin x ) dx  sin x dx  x cos x dx dx    d ( uv ) du dv dx  v dx  u dxSimplifying the l.h.s.: dx dx dx  x sinx  sinx dx   x cos x dx   du dv Rearranging: uv  v dx  u dx dx dx   x cos x dx  x sinx  sinx dx   dv du u dx  uv  v dx dx dx
    • Integration by PartsSUMMARY Integration by PartsTo integrate some products we can use the formula   dv du u dx  uv  v dx dx dx
    • Integration by Parts   dv du u dx  uv  v dx dx dxUsing this formula means that we differentiate onefactor, u to get du . . . dx
    • Integration by PartsSo,   dv du u dx  uv  v dx dx dxUsing this formula means that we differentiate onefactor, u to get du . . . dx dv , and integrate the other to get v dx
    • Integration by Parts So,   dv du u dx  uv  v dx dx dx Using this formula means that we differentiate one factor, u to get du . . . dx dv , and integrate the other to get v dx e.g. 1 Find 1  Having substituted in x dx 2 x sin the formula, notice that the completed dv st term, uv, is u  2 x andbut the sin x  2nd term still dxdifferentiatebe integrated. needs to du integrate 2 v   cos x dx ( +C comes later )
    • Integration by Parts So, dv u  2 x and  sin x dxdifferentiate integrate du 2 v   cos x dx We can now substitute into the formula   dv du u dx  uv  v dx dx dx  2 x sin x dx  2 x( cos x ) u dv u v   (  cos x ) 2 dx v du dx dx
    • Integration by Parts So, dv u  2 x and  sin x dxdifferentiate integrate du 2 v   cos x dx We can now substitute into the formula   dv du u dx  uv  v dx dx dx  2 x sin x dx  2 x( cos x )   (  cos x ) 2 dx  2 x cos x   2 cos x dx   2 x cos x  2 sin x  C The 2nd term needs integrating
    • Integration by Parts e.g. 2 Find  xe 2 x dx  dv  du u dx  uv  v dx dx dx Solution: dv u  x and  e2xdifferentiate dx 2x integrate du e 1 v dx 2 This is a compound function,  e2x  so we mustx be careful.  e2 So,  xe dx  ( x )      1 dx 2x  2     2   xe 2 x e2x xe 2 x e 2 x  2   2 dx  2  4 C
    • Integration by Parts Exercises Find 1.  xe dx 2.  x cos 3 x dx xSolutions:  xe dx  xe   e dx x x x 1.  xe  e  C x x  sin3 x   sin3 x 2.  x cos 3 x dx  ( x ) 3     3 dx x sin 3 x cos 3 x   C 3 9
    • Integration by Parts Definite Integration by PartsWith a definite integral it’s often easier to do theindefinite integral and insert the limits at the end.We’ll use the question in the exercise you have justdone to illustrate.  xe x dx  xe x   e x dx  xe x  e x  C  0 1 x xe dx   xe x x 1 e 0   1e 1 e 1   0e 0 e 0   0   1  1
    • Integration by PartsUsing Integration by PartsIntegration by parts cannot be used for every product.It works if  we can integrate one factor of the product,  the integral on the r.h.s. is easier* than the one we started with. * There is an exception but you need to learn the general rule.
    • Integration by PartsThe following exercises and examples are harderso you may want to practice more of thestraightforward questions before you tackle them.
    • Integration by Parts   dv due.g. 3 Find  x ln x dx u dx  uv  v dx dx dxSolution: What’s a possible problem? ANS: We can’t integrate ln x .Can you see what to do? dv If we let u  ln x and  x , we will need to dx differentiate ln x and integrate x. Tip: Whenever ln x appears in an integration by parts we choose to let it equal u.
    • Integration by Parts   dv du e.g. 3 Find  x ln x dx u dx  uv  v dx dx dx So, u  ln x dv x dx integratedifferentiate du 1 x2  v dx x 2 x2 x2 1   x ln x dx  2 ln x    dx 2 x The r.h.s. integral still seems to be a product! BUT . . . x cancels. 2   x x So, x ln x dx  ln x  dx 2 2 x2 x2   x ln x dx  2 ln x  4 C
    • Integration by Partse.g. 4  x 2 e  x dx   dv du u dx  uv  v dx dx dxSolution: Let dv u  x and 2  ex dx du  2x v  e  x dx   x 2 e  x dx   x 2 e  x    2 xe  x dx   x 2 e  x dx   x 2 e  x  I1  2 xe  x dx I2The integral on the r.h.s. is still a product but usingthe method again will give us a simple function.We write I   x 2e  x  I 1 2
    • Integration by Parts e.g. 4  x 2 e  x dxSolution: I1   x 2e  x  I 2 . . . . . ( 1 )I 2   2 xe  x dx dv Let u  2x and  ex du dx 2 v  e  x dx So, I 2   2 xe  x    2e  x dx   2 xe  x   x 2e  x dx   2 xe  x  2e  CSubstitute in ( 1 )  x e dx   x e 2 x 2 x x x  2 xe  2e C
    • Integration by PartsThe next example is interesting but is not essential.Click below if you want to miss it out. Omit Example   dv du e.g. 5 Find e sin x dx x u dx  uv  v dx dx dxSolution:It doesn’t look as though integration by parts willhelp since neither function in the product gets easierwhen we differentiate it.However, there’s something special about the 2functions that means the method does work.
    • Integration by Parts   dv due.g. 5 Find  e sin x dx x u dx  uv  v dx dx dx Solution: dv ue x  sin x dx du  ex v  cos x dx  e x sin x dx   e x cos x    e x cos x dx   e cos x   e x cos x dx xWe write this as: I1  e cos x  I 2 x
    • Integration by Parts   dv du e.g. 5 Find e sin x dx x u dx  uv  v dx dx dx So, I1  e x cos x  I 2 I 1   e x sin x dx and I 2  e x cos x dxwhere We next use integration by parts for I2 dv ue x  cos x du dx  ex v  sin x dx  e x cos x dx  e x sinx   e x sin x dx  I 2  e sinx  I1 x
    • Integration by Parts   dv du e.g. 5 Find e sin x dx x u dx  uv  v dx dx dx So, I1  e x cos x  I 2 I 1   e x sin x dx and I 2  e x cos x dxwhere We next use integration by parts for I2 dv ue x  cos x du dx  ex v  sin x dx  e x cos x dx  e x sinx   e x sin x dx  I 2  e sinx  I1 x
    • Integration by Parts   dv du e.g. 5 Find e sin x dx x u dx  uv  v dx dx dxSo, I1  e x cos x  I 2 . . . . . ( 1 ) I 2  e x sinx  I1 .....(2) 2 equations, 2 unknowns ( I1 and I2 ) !Substituting for I2 in ( 1 ) I1  e x cos x 
    • Integration by Parts   dv du e.g. 5 Find e sin x dx x u dx  uv  v dx dx dxSo, I1  e x cos x  I 2 . . . . . ( 1 ) I 2  e x sin x  I 1 .....(2) 2 equations, 2 unknowns ( I1 and I2 ) !Substituting for I2 in ( 1 ) I1  e x cos x  e x sin x  I 1
    • Integration by Parts   dv du e.g. 5 Find e sin x dx x u dx  uv  v dx dx dxSo, I1  e x cos x  I 2 . . . . . ( 1 ) I 2  e x sin x  I 1 .....(2) 2 equations, 2 unknowns ( I1 and I2 ) !Substituting for I2 in ( 1 ) I1  e x cos x  e x sin x  I 1  2I1  e x cos x  e x sinx  e x cos x  e x sin x  I1  C 2
    • Integration by PartsExercises x 2 1. 2. 1 ln x dx 2 sin x dx( Hint: Although 2. is not a product it can be turned into one by writing the function as 1 ln x . )
    • Integration by PartsSolutions: dv 1. x sin x dx Let u  x and  sin x 2 2 I1 du dx  2x v  cos x dx I 1   x 2 cos x    2 x cos x dx  I 1   x cos x  dv 2 2 x cos x dx . . . . . ( 1 ) I2For I2: Let u  2 x and  cos x du dx 2 v  sin x dx  I 2  2 x sin x   2 sin x dx  2 x sin x  2 cos x  CSubs. in ( 1 )   x 2 sin x dx   x 2 cos x  2 x sin x  2 cos x  C
    • Integration by Parts 2 22. 1 ln x dx  1 1 ln x dx This is an important Let u  ln x and dv application of 1 du 1 dx integration by parts  vx dx x  1   1 ln x dx  x ln x  x  dx x    C x ln x  1 dx x ln x  x   x ln x  x  2 1 ln x dx 2 So, 1   2 ln2  2    1 ln1  1   2 ln 2  1
    • Integration by Parts
    • Integration by PartsThe following slides contain repeats ofinformation on earlier slides, shown withoutcolour, so that they can be printed andphotocopied.For most purposes the slides can be printedas “Handouts” with up to 6 slides per sheet.
    • Integration by PartsSUMMARY Integration by PartsTo integrate some products we can use the formula   dv du u dx  uv  v dx dx dx
    • Integration by Parts e.g. Find  xe 2 x dx  dv du u dx  uv  v dx  dx dx Solution: dv u x and  e2xdifferentiate dx 2x integrate du e 1 v dx 2  e2x   e2x So,  xe dx  ( x )     1 dx 2x  2    2      xe 2 x e2x xe 2 x e 2 x  2   2 dx  2  4 C
    • Integration by PartsUsing Integration by PartsIntegration by parts can’t be used for every product.It works if  we can integrate one factor of the product,  the integral on the r.h.s. is easier* than the one we started with. * There is an exception but you need to learn the general rule.
    • Integration by Parts   dv du e.g. 3 Find  x ln x dx u dx  uv  v dx dx dx We can’t integrate ln x so, dv u  ln x x dx integratedifferentiate du 1 x2  v dx x 2 x2 x2 1   x ln x dx  2 ln x    dx 2 x The r.h.s. integral still seems to be a product! BUT . . . x cancels. So,  x ln x dx  x2 x2  C ln x  2 4
    • Integration by Parts This is an important2.  ln x dx   1 ln x dx application of integration by parts Let u  ln x and dv 1 dx du 1  vx dx x  1   1 ln x dx  x ln x  x  dx x  x ln x   1 dx  x ln x  x  C