Calc Integral: |sinx-cos2x| from 0 to \pi/2

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SUMMARY

The integral of |sin(x) - cos(2x)| from 0 to π/2 requires splitting the integral into two parts based on the intersection points of the functions sin(x) and cos(2x). The correct intervals are from 0 to π/6, where cos(2x) is greater, and from π/6 to π/2, where sin(x) is greater. The user correctly identified the need to split the integral but struggled with evaluating the individual integrals. Proper evaluation techniques must be applied to obtain the correct result.

PREREQUISITES
  • Understanding of definite integrals
  • Knowledge of trigonometric functions and their graphs
  • Familiarity with the concept of absolute value in integrals
  • Ability to evaluate integrals using standard techniques
NEXT STEPS
  • Review techniques for evaluating definite integrals involving absolute values
  • Study the properties of trigonometric functions, specifically sin(x) and cos(2x)
  • Practice solving integrals with piecewise functions
  • Learn about graphical methods for determining intersection points of functions
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Students studying calculus, particularly those focusing on integral calculus and absolute value functions, as well as educators teaching these concepts.

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\int^{\pi/2}_{0}|sinx-cos2x| dx
Hi guys i can't seem to get the correct answer for this. What i did was draw the graph of sinx and cos2x and for 0 to \pi/6 i use cos2x-sinx since cos2x is higher. From \pi/6 to \pi/2 i use sinx-cos2x. However i am not able to get the same answer as my book can someone tell me is this the way to do integrals involving the absolute value?
 
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You've split the integral up correctly. Your problem must be in evaluating the two integrals.
 

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