Periodic Function Property

Step 1: Analyze the Functional Equation

We are given the functional equation:

$$f(x) + f(x + 2) = 0 \implies f(x + 2) = -f(x)$$

Replacing x with x + 2:

$$f(x + 4) = -f(x + 2) = -(-f(x)) = f(x)$$

This proves that $$f(x)$$ is a periodic function with a fundamental period of $$T = 4$$.

Step 2: Split the Main Integral

Using the linearity property of integration, we break the given integral I into two separate parts:

$$I = \int_{0}^{2026} f(x) \, dx + \int_{0}^{2026} \{x\} \, dx$$

Let $$I = I_1 + I_2$$, where:

$$I_1 = \int_{0}^{2026} f(x) \, dx$$

$$I_2 = \int_{0}^{2026} \{x\} \, dx$$

Step 3: Evaluate $$I_1$$

Since $$f(x)$$ has a period of 4, we express the upper limit 2026 in terms of the period:

$$2026 = 4 \times 506 + 2$$

Using the periodic integral property where the integral from $$0$$ to $$nT + a$$ of $$f(x) dx$$ = n times the integral from 0 to T of $$f(x) dx$$ + the integral from 0 to a of $$f(x) dx$$:

$$I_1 = 506 \int_{0}^{4} f(x) \, dx + \int_{0}^{2} f(x) \, dx$$

Let's evaluate the integral over one full period, from 0 to 4 of $$f(x) dx$$:

$$\int_{0}^{4} f(x) \, dx = \int_{0}^{2} f(x) \, dx + \int_{2}^{4} f(x) \, dx$$

For the second integral, substitute $$x = t + 24$, which gives dx = dt:

When $$x = 2, t = 0$$

When $$x = 4, t = 2$$

$$\int_{2}^{4} f(x) \, dx = \int_{0}^{2} f(t+2) \, dt = \int_{0}^{2} -f(t) \, dt = -\int_{0}^{2} f(x) \, dx$$

Substituting this back:

$$\int_{0}^{4} f(x) \, dx = \int_{0}^{2} f(x) \, dx - \int_{0}^{2} f(x) \, dx = 0$$

Thus, $$I_1$$ simplifies cleanly to:

$$I_1 = 506(0) + \int_{0}^{2} f(x) \, dx = \int_{0}^{2} f(x) \, dx$$

Given that the integral from 0 to 2 of f(x) dx = 5:

$$I_1 = 5$$

Step 4: Evaluate $$I_2$$

The fractional part function $$\{x\}$$ is periodic with a fundamental period of $$T = 1$$.

Using the periodic property:

$$I_2 = \int_{0}^{2026} \{x\} \, dx = 2026 \int_{0}^{1} \{x\} \, dx$$

In the domain where x belongs to [0, 1), \{x\} = x. Therefore:

$$\int_{0}^{1} \{x\} \, dx = \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2}$$

Substituting this value back into $$I_2$$:

$$I_2 = 2026 \times \frac{1}{2} = 1013$$

Step 5: Combine $$I_1$$ and $$I_2$$

$$I = I_1 + I_2$$

$$I = 5 + 1013 = 1018$$

Let $$I=\int_0^{2800\pi}\frac{dx}{1+3^{\sin x}}$$

The function $$f(x)=\frac{1}{1+3^{\sin x}}$$ has period $$2\pi$$

Therefore, $$I=1400\int_0^{2\pi}\frac{dx}{1+3^{\sin x}}$$

Let $$J=\int_0^{2\pi}\frac{dx}{1+3^{\sin x}}$$

Using the property

$$\int_0^{2\pi}f(x)\,dx=\int_0^{2\pi}f(2\pi-x)\,dx$$

we get

$$J=\int_0^{2\pi}\frac{dx}{1+3^{-\sin x}}$$

Adding the two expressions for $$J$$,

$$2J=\int_0^{2\pi}\left(\frac{1}{1+3^{\sin x}}+\frac{1}{1+3^{-\sin x}}\right)dx$$

Let $$t=3^{\sin x}$$

Then $$3^{-\sin x}=\frac1t$$

Hence,

$$\frac{1}{1+t}+\frac{1}{1+\frac1t}=\frac{1}{1+t}+\frac{t}{1+t}=1$$

Therefore,

$$2J=\int_0^{2\pi}1\,dx$$

$$=2\pi$$

$$J=\pi$$

Thus,

$$I=1400J$$

$$=1400\pi$$

Given,

$$I=200k\pi$$

Therefore,

$$1400\pi=200k\pi$$

$$1400=200k$$

$$k=7$$

Final Answer :

$$7$$

The fractional part function $$\{x\} = x - [x]$$ is periodic with a period of $$1$$. 

For any integer $$n$$, on the interval $$[n, n+1)$$, $$[x] = n$$, so $$\{x\} = x - n$$.
 Use the periodicity property of definite integrals.
If a function $$f(x)$$ is periodic with period $$T$$, then:
$$\int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx$$
Here, $$T = 1$$ and $$n = 100$$.
$$I = 100 \int_{0}^{1} e^{\{x\}} \, dx$$
On the interval $$[0, 1)$$, $$[x] = 0$$, so $$\{x\} = x$$.
$$I = 100 \int_{0}^{1} e^x \, dx$$
$$I = 100 \cdot \Big[ e^x \Big]_{0}^{1} = 100(e^1 - e^0) = 100(e - 1)$$