Second mean value theorem in Bonnet's form

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SUMMARY

The discussion focuses on applying the Second Mean Value Theorem in Bonnet's form to demonstrate the existence of a point \( p \) in the interval \([0, p]\) such that the integral \(\int_0^p e^{-x} \cos x \, dx = \sin p\). The Mean Value Theorem for Integrals is utilized, where the function \( f(x) = e^{-x} \cos x \) is evaluated. The conclusion is that there exists a \( c \) in the interval \((0, p)\) satisfying the equation, confirming the theorem's application in this context.

PREREQUISITES
  • Understanding of the Second Mean Value Theorem in Bonnet's form
  • Knowledge of integral calculus, specifically integration of exponential and trigonometric functions
  • Familiarity with the concept of definite integrals
  • Basic understanding of continuity and differentiability of functions
NEXT STEPS
  • Study the properties of the Second Mean Value Theorem in detail
  • Explore applications of the Mean Value Theorem for Integrals in various contexts
  • Learn about the integration techniques for functions involving \( e^{-x} \) and \( \cos x \)
  • Investigate the implications of the existence of points in intervals as described by the Mean Value Theorem
USEFUL FOR

Mathematicians, calculus students, and educators looking to deepen their understanding of integral theorems and their applications in real analysis.

Suvadip
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Using second mean value theorem in Bonnet's form show that there exists a
$$p $$in $$[a,b]$$ such that
$$\int_a^b e^{-x}cos x dx =sin ~p$$

I know the theorem but how to show this using that theorem .
 
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We can solve this by using Mean Value Theorem for Integrals in Bonnet's Form. Let f(x) = e^{-x}cos x, a = 0 and b = p.By the Mean Value Theorem for Integrals in Bonnet's Form, there exists c ∈ (0, p) such that\int_0^p e^{-x}cos x dx = f(c) (p - 0) = e^{-c}cos c (p - 0) = sin p Therefore, there exists c ∈ (0, p) such that \int_0^p e^{-x}cos x dx = sin p
 

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