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Recitation Note 17.872 TA Jiyoon Kim Sep. 30, 2003 1 Some Integration Theorem 1. A constant can be moved through an integral sign. � cf (x)dx = c � f (x)dx 2. An integration of a sum is the sum of the integrations. � [f (x) + g(x)]dx = � f (x)dx + � g(x)dx 3. An integration of a diﬀerence is the diﬀerence of the integrations. � 2 [f (x) − g(x)]dx = � f (x)dx − � Integration Rules 1. Power Rule � 2. Exponential Rule I xn dx = � 1 xn+1 + c n+1 ex dx = ex + c 3. Exponential Rule II � 4. Logarithm Rule I f � (x)ef (x) dx = ef (x) + c � 1 dx = lnx + c x 5. Logarithm Rule II � f � (x) dx = lnf (x) + c f (x) 1 g(x)dx Examples: � 14x dx 2+5 7x � 4x dx 2) 2+1 2x � x 3) dx 2 3x + 5 1) 3 2e2x + Substitution Rule When you have integration with high power or with complicated form, you can use substitution rule. First, you put the complicated part as u and ﬁnd du du dx . Then, substitute the original equation with u and dx to solve it in u term. Example: 1) 2) � � 4 2x(x2 + 1)50 dx 6x2 (x3 + 2)99 dx Integration by Parts The integral of v with respect to u is equal to uv minus the integration of u with respect to v. � � vdu = uv − udv Here, you use the less complicated part (usually one term) as v and the rest of the� integration including integral sign and dx as� u. For example, if you 1 1 have x(x + 1) 2 dx, v should be x and u should be (x + 1) 2 dx. Then, ﬁnd dv and du to plug them in the formula. Example: 1) 2) � � xex dx ln x dx 2 5 Properties of Sigma Notation The following properties of sigma notation will help to manipulate sums: 1. n � cak = c n � ak k=1 k=1 2. n � (ak + bk ) = (ak − bk ) = ak + k=1 n � k=1 n � n � ak − n � bk n � bk k=1 3. k=1 k=1 k=1 Question Are these true? 1) � �� n d 2) dx 3) 4) 5) n � � fi (x) dx = �i=1 n � � fi (x) = i=1 ai bi = n � n �� � i=1 n � � i=1 n � fi (x)dx � � d [fi (x)] dx bi ai i=1 i=1 �n ai ai = �in=1 b i=1 bi i=1 i � n �2 n � � 2 ai = ai i=1 i=1 i=1 n � Quick Question Let x̄ denote the arithmetic average of the n numbers x1 , x2 , · · · , xn . Prove 3 that n � i=1 ¯ =0 (xi − x) 4