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# Calc 1372 - Week 1 Math 1372

U of M

GPA 3.9

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## Popular in Calculus 2 for Engineers

## Popular in Mathematics (M)

This 4 page Class Notes was uploaded by Emily Hein on Sunday September 13, 2015. The Class Notes belongs to Math 1372 at University of Minnesota taught by Jennie Nash in Fall 2015. Since its upload, it has received 34 views. For similar materials see Calculus 2 for Engineers in Mathematics (M) at University of Minnesota.

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Date Created: 09/13/15

INTEG RATION BY PARTS Usubstitution can be thought of as undoing the chain rule for differentiation but what quotundoesquot the product rule What is the derivative offxgx assuming that bothfand g are differentiable w fxg39x fxgx So then integrating both sides we see that fxgx lfxg39xdx lgxfxdx By rearranging this equation we arrive at Ifxg39xdx fxgx Igxf39xdx This theorem is usually seen in this form ludv uv lvdu When choosing what to use as your quotuquot and your quotdvquot keep in mind that you should 0 Let quotuquot term that is simpler after derivation 0 Let quotdvquot something you can integrate Order of choices for quotuquot o Logs O Inverse trig O Polynomials o Exponentials o Trig functions Example 1 lxsinxdx xcosx lcosxdx xcosx Sinx C u x dv sinx dx One good rule of thumb after choosing your u and dv du dx v cos x quotDid I make it better Sometimes once isn t enough Example 2 2a lxzexdx xzex l2xexdx xzex 2xex lexdx xzex 2xex 26x C ux2 dvexdx ux dvexdx du2xdx vex dudx vex J J 1 2b lexodx xe l3lxolex3dx simple usub to finish problem with u x ol L uex3 dvdx 2 l du l3xTex3 v x Quick Review ldxld Znul Clnfx C u fx du f39xdx Example 3 A review of previously known integration techniques 1 Simplify algebraically 2 Usubstitution 3 Break into separate terms lm udx l 2x dx12l dx In x2 9 4arctcm g C x29 x29 x29 u x2 9 a 3 du 2xdx Recall thatl dx larctcm It would be beneficial to MEMORIZE this equation x2a2 Example 4 932ng What can we do with this since none of our previously known techniques will help Well we know that the denominator can be factored as x 3x 2 Can we somehow rewrite the integral to break up the fraction into two fractions using the factored denominator as two separate denominators This is where PARTIAL FRACTIONS come into play PARTIAL FRACTIONS m Q06 simpler fractions partial fractions that we already know how to integrate Partial fractions allow integration of any rational function fx by expressing it as a sum of Assume thatfx is proper degree of numerator Px lt degree of denominator Qx Then consider the following cases 1 Qx is a product of distinct linear factors 2x 14 2A x2 Exam le39 i p 39 x2x3 x3 2 Qx is a product of linear factors some of which are repeated D x35 41 E L E x2 x3 x32 x33 Example x2x33 x 3 Qx contains irreducible quadratic factors none of which are repeated 18x2 AxB L Example x24x3 x24 x3 4 Qx contains irreducible quadratic factors some of which are repeated 18x2 AxB CxD E Example x242x3 x24 x242 Q But what iffx is improper degree of numerator Px 2 degree of denominator Qx Then take the preliminary step of dividing Q into P by long division so thatfx x and refer back to the previous cases m 2 L Example x1 x x 2 x1 Example 4 revisited Putting it all together 2x 14 A i x2x 6dx Ix Z x3 x2 x 6gtlt11g 4 giaxxz x 6 2x 14Ax3Bx 2 x22 105A gtA 2 x 3 20 SB gtB4 2znx 24Inlx3lc

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