Definite And Indefinite Integrals

Step 1: Check if the integrand is odd or even.
Let . . So, is an odd function.
Step 2: Apply the property of definite integrals for odd functions.
For an odd function , . Here, , so .

Step 1: Differentiate the given integral equation.
We are given . Differentiating both sides with respect to using the Fundamental Theorem of Calculus and the chain rule: $ f(\cos x) x \in (0, \frac{\pi}{2}) \sin x \neq 0 \cos x \neq 0 -\sin x \cos x f\left(\frac{1}{\sqrt{2}}\right) f \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}} x = \frac{\pi}{4} (0, \frac{\pi}{2}) \cos x = \frac{1}{\sqrt{2}} f(\cos x) f\left(\frac{1}{\sqrt{2}}\right) = \boxed{2\sqrt{2}}$.

Step 1: Identify the periodicity of the integrand.
Let . The term represents the fractional part of , which is periodic with a period of 1. Therefore, is also periodic with a period .
Step 2: Use the property of integrals of periodic functions.
For a periodic function with period , we have , where is an integer. In our case, and . $ 0 \le x<1 \lfloor x \rfloor = 0 x - \lfloor x \rfloor = x \boxed{1000(e - 1)}$.

Step 1: Rewrite the equations.
and .
Step 2: Find intersection points.
Setting . Points are and .
Step 3: Set up the integral for the area.
Area .
Step 4: Evaluate the integral using symmetry.
Area .

Step 1: Split the integral based on the absolute value We are asked to evaluate: First, break the absolute values into pieces based on the points where the expressions inside the absolute values change sign: - changes sign at , - changes sign at . Thus, we split the integral into two parts: Step 2: Evaluate the integrals Now evaluate each part: For : For : Thus, the value of the integral is:

Step 1: Substituting using symmetry Let: Using the property: Applying transformations and simplifications, we get: Step 2: Solving for