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Application of the mean-value theorem for integrals

  1. Let \varphi be a function with second derivative \varphi'' continuous and nonzero on an interval [a,b]. Furthermore, let m > 0 be a constant such that

        \[ \varphi'(t) \geq m \qquad \text{for all } t \in [a,b]. \]

    Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality

        \[ \left| \int_a^b \sin \varphi(t) \, dt \right| \leq \frac{4}{m}. \]

  2. If a > 0 prove that

        \[ \left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{2}{a} \qquad \text{for } x > a. \]

  1. Proof. Since we have the assumption that \varphi'(t) \geq m > 0 for all t \in [a,b] we may divide by \varphi'(t), to obtain

        \[ \left| \int_a^b \sin \varphi(t) \, dt \right| = \left| \int_a^b \frac{\sin \varphi(t)}{\varphi'(t)} \cdot \varphi'(t) \, dt \right|. \]

    Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions

        \[ f(t) = \frac{1}{\varphi'(t)} \qquad \text{and} \qquad g(t) = \varphi'(t) \sin \varphi(t). \]

    The function g is continuous since \sin \varphi(t) is a composition of continuous functions (we know \varphi(t) is continuous since it is differentiable) and \varphi'(t) is continuous (again, it is differentiable since \varphi''(t) exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that g is continuous. We also know that f meets the conditions of the theorem since it has derivative given by

        \[ f'(t) = - \frac{\varphi''(t)}{(\varphi'(t))^2}. \]

    This derivative is continuous since \varphi''(t) and \varphi'(t) are continuous and \varphi'(t) is nonzero. Furthermore, this derivative does not change since on [a,b] since \varphi''(t) is nonzero on [a,b] (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:

        \begin{align*}  \left| \int_a^b \frac{\varphi'(t) \sin \varphi(t)}{\varphi'(t)} \, dt \right| &= \left| \frac{1}{\varphi'(a)} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{\varphi'(b)} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\  &\leq \left| \frac{1}{m} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{m} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\  &\leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c + \left(-\frac{1}{m} \right) \cos \varphi(t) \Big \rvert_c^b \right| \\  & \leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c \right| + \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_c^b \right|  \\ \intertext{(by the Triangle Inequality)}  &\leq \left| -\frac{2}{m} \right| + \left| -\frac{2}{m} \right| \\  & = \frac{2}{m} + \frac{2}{m} \\  &= \frac{4}{m}. \qquad \blacksquare \end{align*}

  2. Proof. Using part (a), we take \varphi(t) = t^2, giving us

        \[ \varphi(t) = t^2 \quad \implies \quad \varphi'(t) = 2t \quad \implies \quad \varphi''(t) = 2. \]

    Thus, \varphi''(t) is continuous and never changes sign. Furthermore, \varphi' \geq m = 2a (where a is a given constant) and 2a > 0 since a> 0. Thus,

        \[ \left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{4}{2a} = \frac{2}{a} \qquad \text{for all } x > a. \qquad \blacksquare \]


  1. P. says:

    Hi, this resource was a great find!

    One thing left me stumped working on this. On the second row of the solution to part one, I initially summed the two integrals, getting 2/m rather than 4/m.

    I figured that I could use the triangle inequality; but I can’t understand if summing the two integrals gives an incorrect result. Could you explain?

    • P. says:

      On second thoughts… you can’t actually sum the two integrals when 1/(phi’) is actually evaluated at a and b. But the two integrals can still both be equal to 2.

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