Use Theorem 8.12 and the results from Exercises 41–50 to find series equal to the definite integrals in Exercises 51–60.

∫01xcos(x3)dx

Short Answer

Expert verified

∫01xcos(x3)dx=∑k=0∞-1k2k!16k+2

Step by step solution

01

Step 1. Given information is: 

∫01xcos(x3)dx

02

Step 2. Definite integral

FromQ42.Maclaurinseriesforf(x)=xcos(x3)isxcos(x)3=∑k=0∞-1k2k!x6k+1Also,F=∫fF(x)==∑k=0∞-1k2k!x6k+26k+2Addingthelimits,F(x)=∑k=0∞-1k2k!x6k+26k+201=∑k=0∞-1k2k!16k+2-06k+2=∑k=0∞-1k2k!16k+2

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