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MIT 18 01 - Integration by Parts, Reduction Formulae

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � � � Lecture 30 18.01 Fall 2006 Lecture 30: Integration by Parts, ReductionFormulae Integration by Parts Remember the product rule: (uv)� = u�v + uv� We can rewrite that as uv� = (uv)� − u�v Integrate this to get the formula for integration by parts: uv� dx = uv − u�v dx Example 1. tan−1 x dx. At first, it’s not clear how integration by parts helps. Write tan−1 x dx = tan−1 x(1 dx) = uv� dx· with u = tan−1 x and v� = 1. Therefore, 1 v = x and u� = 1 + x2 Plug all of these into the formula for integration by parts to get: 1 tan−1 x dx = uv� dx = (tan−1 x)x − 1 + x2 (x)dx = x tan−1 x − 12 ln |1 + x 2| + c Alternative Approach to Integration by Parts As above, the product rule: (uv)� = u�v + uv� can be rewritten as uv� = (uv)� − u�v This time, let’s take the definite integral: � b � b � b uv� dx = (uv)� dx − u�v dx a a a 1� � � � Lecture 30 18.01 Fall 2006 By the fundamental theorem of calculus, we can say � b �� b�b uv� dx = uv� u�v dx a −a a Another notation in the indefinite case is u dv = uv − v du This is the same because dv = v� dx = uv� dx = u dv and du = u� dx = u�v dx = vu� dx = v du⇒ ⇒ Example 2. (ln x)dx 1 u = ln x; du = dx and dv = dx; v = x x � �� � �1(ln x)dx = x ln x − x dx = x ln x − dx = x ln x − x + c x We can also use “advanced guessing” to solve this problem. We know that the derivative of something equals ln x: d (??) = ln x dx Let’s try d 1(x ln x) = ln x + x = ln x + 1 dx · x That’s almost it, but not quite. Let’s repair this guess to get: d (x ln x − x) = ln x + 1 − 1 = ln x dx Reduction Formulas (Recurrence Formulas) Example 3. (ln x)n dx Let’s try: � �1 u = (ln x)n = u� = n(ln x)n−1 ⇒ x v� = dx; v = x Plugging these into the formula for integration by parts gives us: � �� � 1 1���(ln x)ndx = x(ln x)n n(ln x)n−1 x �dx− �x Keep repeating integration by parts to get the full formula: n (n − 1) (n − 2) (n − 3) etc � → → → → Example 4. x n e x dx Let’s try: u = x n = u� = nx n−1; v� = e x = v = e x ⇒ ⇒ 2� � � � � Lecture 30 18.01 Fall 2006 Putting these into the integration by parts formula gives us: n n x x e x dx = x e nx n−1 e x dx− Repeat, going from n → (n − 1) → (n − 2) → etc. Bad news: If you change the integrals just a little bit, they become impossible to evaluate: � �2tan−1 x dx = impossible xedx = also impossible x Good news: When you can’t evaluate an integral, then � 2 xedx 1 x is an answer, not a question. This is the solution– you don’t have to integrate it! The most important thing is setting up the integral! (Once you’ve done that, you can always evaluate it numerically on a computer.) So, why bother to evaluate integrals by hand, then? Because you often get families of related integrals, such as x∞ eF (a) = dx xa 1 where you want to find how the answer depends on, say, a. 3� � Lecture 30 18.01 Fall 2006 Arc Length This is very useful to know for 18.02 (multi-variable calculus). yxdsdxdyy=f(x)Figure 1: Infinitesimal Arc Length ds dydxdsFigure 2: Zoom in on Figure 1 to see an approximate right triangle. In Figures 1 and 2, s denotes arc length and ds = the infinitesmal of arc length. ds = (dx)2 + (dy)2 = 1 + (dy/dx)2dx Integrating with respect to ds finds the length of a curve between two points (see Figure 3). To find the length of the curve between P0 and P1, evaluate: � P1 ds P0 4Lecture 30 18.01 Fall 2006 P₀P₁abFigure 3: Find length of curve between P0 and P1. We want to integrate with respect to x, not s, so we do the same algebra as above to find ds in terms of dx. (ds)2 (dx)2 (dy)2 � dy �2 = + = 1 + (dx)2 (dx)2 (dx)2 dx Therefore, �� P1 � b � �2 ds = 1 + dy dx dxP0 a Example 5: The Circle. x 2 + y 2 = 1 (see Figure 4). Figure 4: The circle in Example 1. 5� � � � � Lecture 30 18.01 Fall 2006 We want to find the length of the arc in Figure 5: aFigure 5: Arc length to be evaluated. y = 1 − x2 dy −2x 1 −x dx = √1 − x2 2= √1 − x2 � �2 ds = 1 + √1 −− xx2 dx � �2 1 + √1 −− xx2 = 1 + 1 − x2 x2 =1 −1 x− 2 x+ 2 x2 =1 − 1 x2 1 ds = dx1 − x2 � a ⏐dx ⏐a s = √1 − x2 = sin−1 x⏐ 0 = sin−1 a − sin−1 0 = sin−1 a 0 sin s = a This is illustrated in Figure 6. 6Lecture 30 18.01 Fall 2006 a11asFigure 6: s = angle in radians. Parametric Equations Example 6. x = a cos t y = a sin t Ask yourself: what’s constant? What’s varying? Here, t is variable and a is constant. Is there a relationship between x and y? Yes: x 2 + y 2 = a 2 cos2 t + a 2 sin2 t = a 2 Extra information (besides the circle): At t = 0, x = a cos 0 = a and y = a sin 0 = 0 πAt t = ,2 π π x = a cos = 0 and y = a sin = a2 2 Thus, for 0 ≤ t ≤ π/2, a quarter circle is traced counter-clockwise (Figure 7). 7� � � Lecture 30 18.01 Fall 2006 (a,0)t=0(0,a)t=π/2Figure 7: Example 6. x = a cos t, y = a sin t; the particle is moving counterclockwise. Example 7: The Ellipse See Figure 8. x = 2 sin t; y = cos t 2x+ y 2 = 1( = (2 sin t)2/4 + (cos t)2 = sin2t + cos2t = 1) 4 ⇒ (2,0)t=π/2t=0(0,1)Figure 8: Ellipse: x = 2 sin t, y = cos t (traced clockwise). …


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MIT 18 01 - Integration by Parts, Reduction Formulae

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