LESSON 4: INTEGRATION BY PARTS (I) MATH FALL 2018

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1 LESSON 4: INTEGRATION BY PARTS (I) MATH 6 FALL 8 ELLEN WELD. Integration by Parts We introduce another method for ealuating integrals called integration by parts. The key is the following : () u d = u MEMORIZE this formula. E. Suppose we want to ealuate e d. Here, u-substitution fails to produce anything useful. integration by parts. Instead, we need to use To use integration by parts, we must apply equation () which means we must identify our u, du,, d. The integrand is u d, which means once u is chosen, eerything else is d. Then, we find du by differentiating u and we find by integrating d. u = du = deriatie of u d = eerything leftoer = d For the eample gie aboe, we should take u =. This means that eerything leftoer in the integral (e d) is our d. To get our du and, we differentiate and integrate respectiely: u = du = d d = e d = e} {{ d} = e d Note. In this contet, wheneer we integrate d, we always assume the constant is. Now that we hae each part of equation (), we just plug in our alues: u d = u u e} {{ d} = d u e e d du.

2 ELLEN WELD Thus, our integral becomes e d = e original question e d = e e + C Note. We add a single C at the end of the integration by parts process. CHECK: d d (e e + C) = e Integration by parts requires attention to detail and is a difficult method to apply, but it is ery useful. The trickiest part of integration by parts is determining u. To choose an appropriate u, we use the following: L - Logarithms like ln, ln( 3 + ), etc I - Inerse trig functions (not for this class) A - Algebraic functions like, , polynomials (NO ROOTS) T - Trig functions like cos, tan, etc E - Eponential functions like e,, etc We use this list as follows: if ln appears in the integral, then we take u = ln. If not, then we let u be whateer inerse trig function is present. If neither of these, we take u to be a polynomial, etc. E. To ealuate (3 + )e d, we take u = 3 + and get the following table: u = 3 + du = (6 + ) d d = e d = e d = e Note 3. The aboe is not really a set of hard and fast rules but rather a rule of thumb for choosing u. Howeer, for this class, LIATE should be sufficient. Eamples.. Ealuate ln d Solution: By LIATE, we take u = ln. Then we get the following table: u = ln du = d d = d = d = We apply equation (), ln d = ln u ( ) ( ) ( ) d du

3 AN UNOFFICIAL GUIDE TO MATH 6 FALL 8 3 = ln = ln ( ) d d = ln 4 + C. Find cos d Solution: This is a definite integral but we still apply the same method. By LIATE, u =. So u = du = d d = cos d = cos d = sin By equation (), we write cos d = sin u = sin = sin = sin sin d sin d ( cos ) + cos = sin + cos = π ( π ) ( π ) 3 sin + cos [ }{{ 3 } 3 } sin {{ } + cos ] 3 ( ) = π 3 + 3π 3 = 6 () For definite integrals, we hae the following integration by parts formula: b a u d = u b a b a

4 4 ELLEN WELD e 3. Ealuate ln( 9 ) d Solution: By LIATE, u = ln( 9 ). Then d d ln(9 ) = d d (9 ) 9 = 98 9 = 9 Thus, u = ln( 9 ) du = 9 d d = d = d = We write e ( ) ln( 9 ) d = ln( 9 ) u = e ln( 9 ) 9 = e ln( 9 ) 9 = e ln( 9 ) 9 e e ( ) ( ) 9 d ( ) d e e = ln( 9 ) 9 4 e d ( ) e = e ln(e 9 ) 9 [ 4 e () ln( 9 ) 9 ] 4 () 9 = 9 e 9 4 e = 9 4 e Find 3 + d Solution: Sometimes you need to be ery cleer in how you choose your u een when following LIATE. If you take u = 3, then d = d. But + we don t know how to integrate this d. Instead, take u = which

5 leaes d = AN UNOFFICIAL GUIDE TO MATH 6 FALL 8 5 d. Note that + u=+ d du= d = + = ( u du = u/ du = ( + ) u / = u = +. ) u /+ So we write u = d = + d du = d = + Note that finding is itself a u-substitution problem. Thus we can write 3 d = + ( ) ( + ) ( + ) ( d) u = + + d. Where ( ) is, again, a u-substitution problem. Therefore, 3 + d = + + d = + 3 ( + ) 3/ + C Note 4. To use LIATE effectiely, choose a u that gets simpler when you differentiate and a d which you know how to integrate.. Why Integration by Parts Works Suppose we hae functions f() and g(). Then, by the product rule, [f()g()] = f ()g() + f()g (). u=+ + u= d u d = du = 3 u3/ = 3 ( + ) 3/

6 6 ELLEN WELD If we integrate both sides, then [f()g()] d = f ()g() d + f()g () d. But integration undoes differentiation by FTC, so [f()g()] d = f()g(). Therefore, [f()g()] d = f()g() f()g() = f ()g() d + f ()g() d + Now, subtract f ()g() d from both sides, f()g() f()g () d = f()g () d f()g () d f ()g() d. Finally, let to get u = f() du = f () d d = g () d = g () d = g() f() g () d = f() g() u d u g() f () d. Eamples. 3. Additional Eamples z +. Ealuate dz. e z write Solution: First, we rewrite the integral a little. Since e z z + dz = (z + ) e z = (z e z + e z ) dz = z ( ) +. = ez, we can

7 AN UNOFFICIAL GUIDE TO MATH 6 FALL 8 7 We see that can be done ia u-substitution but it is not immediately clear what z is. We turn our attention to ( ). ( ) Soling ( ): This is an integration by parts problem. By LIATE, we take u = z du = z dz d = = e z Then we simply plug this into equation (): ( ) = (z ) () = (z ) (e z ) u d u = z e z () (e z ) (z dz) z = z e z + z. ( ) Because z does not hae an obious antideriatie and u-substitution fails, we run into another situation wherein we need to use integration by parts. We focus on finding ( ). Soling ( ): By LIATE, we take u = z du = d d = = e z Then by equation (): ( ) = (z) (e z ) = (z) (e z ) u u d = ze z = ze z () = ze z + = ze z e z. (e z ) (d) du Finally, we put it all back together. z + Putting it all back together: We were originally gien the integral dz. By our work, we can now write e z

8 8 ELLEN WELD z + dz = e z z ( ) + = z e z + z ( ) + = z e z + (ze z e z ) + = z e z ze z e z e z + C = z e z ze z 3e z + C. Suppose a certain plant is growing at a rate of te t inches per day t days after it is planted. What is the height of the plant at the beginning of the third day (assuming it is planted as a seed on the first day)? Solution: Because of how we are measuring t, we make the obseration t <, t <, t < 3. Day Day Day 3 Let H(t) be the height of the plant t days after it is planted. We know that H() = (since it was planted as a seed) and we want to find H() because t = corresponds to the beginning of day 3. Note that H (t) = te t because this is the rate of change of the height of the plant. Now, the integral te t dt is an integration by parts problem. By LIATE, we choose u = t d = e t dt du = dt = e t dt = e t. So by equation (), te t dt = (t) (e t ) u (e t ) (dt) = te t e t + C = (t )e t + C. du We need to find H(t) gien our initial condition H() =. Thus, Finally, = ( )e + C = + C C =. H(t) = (t )e t +. H() = ( )e + = e +. The plant is e + inches tall at the beginning of the third day.