Z 1n x n x n 0 1n x n 1n 1 x n 1 1n 1 x nn now to get

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1.e2x2:Pn=0(2x2)nn!=Pn=0(1)n2nx2nn!2.x ·cos(x)=Pn=0(1)nx2n+1(2n)!3. cos(x2)=Pn=0(1)nx4n(2n)!4.x ·cos(12x2)=Pn=0(1)nx4n+122n·(2n)!Use the MacLaurin series expansions to compute the following integrals:1.Rsin(x2)dx=Pn=0(1)nx4n+3(2n+1)!(4n+3)2.Rln(1 +x2)dx:This requires two steps.The first one to find the MacLaurin series ofln(1 +x2) and then the second one to compute the integral.In total we getRln(1 +x2)dx=Pn=1(1)n−1x2n+1(2n+1).3.Re3xdx=Pn=03n·xn2+1n!·(n2+1)=Pn=02·3n·xn2+1n!·(n+2)What is the general formula of the MacLaurin series? What is the general forumla for the TaylorSeries at a pointa?Please refer to the textbook.Partial DerivativesFind all second partial derivatives offat the give points.1.f(x, y) =xysin(y2) at (3,2).2.f(x, y) =y53x2yat (4,1).3.f(x, y) =x(x2+y)2at (1,1).4.f(x, y) =xln(yx) at (1,4).1. Findfzyx=∂f∂x∂y∂zoff(x, y, z) = 3xyz+x2y3z7.

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