Definite Integral of Both Sides

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SUMMARY

The discussion focuses on the process of integrating both sides of equations in calculus, specifically using definite integrals with the same starting and ending points. The user seeks clarification on how to approach this integration method and shares their attempts at integrating specific equations, including dy = 3 sin 2x dx and df = -2 cos (x/2) dx. Feedback from other participants confirms the correctness of the first two integrations while questioning the clarity of the third integration involving dP = A sin^2 (kx) dx.

PREREQUISITES
  • Understanding of definite integrals in calculus
  • Familiarity with basic integration techniques
  • Knowledge of trigonometric functions and their derivatives
  • Ability to differentiate functions to verify integration results
NEXT STEPS
  • Study the properties of definite integrals in calculus
  • Learn advanced integration techniques, such as integration by parts
  • Explore the application of trigonometric identities in integration
  • Practice verifying integration results through differentiation
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Students and educators in mathematics, particularly those studying calculus and integration techniques, as well as anyone seeking to improve their understanding of definite integrals and trigonometric functions.

SherwinS
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Homework Statement


"Integrate both sides of these equations from some starting point to some ending point (such as t0 to t): r dt = d[A], C dt = dH and dw = (nRT/V) dV


Homework Equations



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The Attempt at a Solution



I'm have trouble understanding the concept of taking the definite integral of both sides using the same starting points for both sides. If someone could guide me on how to do one of these, I'll be able to finally do the rest. Thanks!

As a side note, can someone check my integration of these? :)
dy = 3 sin 2x dx ---> -3/2 cos 2x + C
df = -2 cos (x/2) dx ---> 4 sin (x/2) + C
dP = A sin^2 (kx) dx ---> A/2 (x - 1/2ksin (2kx)) + C
 
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If you differentiate your three solutions then you could check them yourself.

The first two look okay. The third looks ambiguous. Is the sin in the numerator or denominator?

Maybe you can show your attempt before someone shows you how.
 
Sorry, 1/2k is the coefficient, sin is in the numerator. Yes I've differentiated them already and come up with the right result, but I always feel it's better to have another pair of eyes glaze over them. :)
 

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