Understanding the Problem

We are asked to find the derivative of the function \(\phi(a)\) with respect to \(a\), where \(\phi(a)\) is defined as a definite integral:

\(\phi(a) = \int_a^{a+100\pi} |\sin x|dx\)

The integrand is the absolute value of the sine function, \(|\sin x|\). The limits of integration depend on the variable \(a\). We need to find \(\phi'(a)\).

Method 1: Using the Leibniz Integral Rule

The Leibniz integral rule allows us to differentiate an integral whose limits are functions of the differentiation variable. The rule states:

\(\frac{d}{da} \int_{p(a)}^{q(a)} f(x, a) dx = f(q(a), a) \cdot q'(a) - f(p(a), a) \cdot p'(a) + \int_{p(a)}^{q(a)} \frac{\partial}{\partial a} f(x, a) dx\)

In our case:

• The integrand is \(f(x) = |\sin x|\). Note that the integrand does not depend on \(a\).
• The lower limit is \(p(a) = a\).
• The upper limit is \(q(a) = a+100\pi\).

We find the derivatives of the limits:

• \(p'(a) = \frac{d}{da}(a) = 1\)
• \(q'(a) = \frac{d}{da}(a+100\pi) = 1\)

And the partial derivative of the integrand with respect to \(a\):

• \(\frac{\partial}{\partial a} |\sin x| = 0\) (since \(|\sin x|\) does not depend on \(a\))

Now, apply the Leibniz rule:

\(\phi'(a) = |\sin(a+100\pi)| \cdot (1) - |\sin(a)| \cdot (1) + \int_a^{a+100\pi} 0 dx\)

We know that the sine function has a period of \(2\pi\). Therefore, \(\sin(a+100\pi) = \sin(a)\) because \(100\pi\) is an integer multiple of \(2\pi\). This also means \(|\sin(a+100\pi)| = |\sin(a)|\).

Substituting this back:

\(\phi'(a) = |\sin(a)| \cdot 1 - |\sin(a)| \cdot 1 + 0\)

\(\phi'(a) = |\sin(a)| - |\sin(a)|\)

\(\phi'(a) = 0\)

Method 2: Using Properties of Periodic Functions

Let's analyze the integrand \(f(x) = |\sin x|\). This function is periodic.

Periodicity of \(|\sin x|\): The function \(\sin x\) has a period of \(2\pi\). The function \(|\sin x|\) repeats its values every \(\pi\) units. For example:

• \(|\sin 0| = 0\), \(|\sin \pi| = 0\), \(|\sin 2\pi| = 0\)
• \(|\sin (\pi/2)| = 1\), \(|\sin (3\pi/2)| = |-1| = 1\)
• So, the fundamental period of \(|\sin x|\) is \(\pi\).

Integral over a multiple of the period: The integral of a periodic function over an interval whose length is an integer multiple of its period is constant and does not depend on the starting point of the interval.

In our function \(\phi(a) = \int_a^{a+100\pi} |\sin x|dx\), the length of the integration interval is \((a+100\pi) - a = 100\pi\).

Since the period of \(|\sin x|\) is \(\pi\), the interval length \(100\pi\) is exactly \(100\) times the period.

Therefore, the value of the integral \(\int_a^{a+100\pi} |\sin x|dx\) is constant for all values of \(a\).

Let's calculate the value of the integral over one period, say from \(0\) to \(\pi\):

\(\int_0^{\pi} |\sin x| dx = \int_0^{\pi} \sin x dx \quad (\text{since } \sin x \ge 0 \text{ for } x \in [0, \pi])\)

\(= [-\cos x]_0^{\pi} = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2\)

The integral over the interval \([a, a+100\pi]\) can be thought of as summing the integrals over 100 periods of length \(\pi\).

\(\phi(a) = \int_a^{a+100\pi} |\sin x|dx = 100 \times \int_0^{\pi} |\sin x| dx = 100 \times 2 = 200\)

So, \(\phi(a) = 200\). This means \(\phi(a)\) is a constant function.

The derivative of a constant function is always zero.

\(\phi'(a) = \frac{d}{da}(200) = 0\)

Conclusion

Both methods show that the derivative \(\phi'(a)\) is equal to 0.

Final Answer: The final answer is \(\boxed{0}\)