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Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.

∫0.52x3cosx2dx

Short Answer

Expert verified

∫0.52x3cosx2dx≈-0.342

Step by step solution

01

Step 1. Given information is: 

∫0.52x3cosx2dx

02

Step 2. Approximating the values 

FromQ58.∫0.52x3cosx2dx=∑k=0∞-1k2k!122k+322k+4-(0.5)2k+42k+4Thisimpliesthattoapproximatethevaluesofintegrals,putthevalueofk∫0.52x3cosx2dx=123·424-0.54-12!·25·626-0.56+14!·27·828-0.58=0.4980-0.1666+0.0104-0.000277+...≈-0.342

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