Integral of an Unknown Function

Let $I =\displaystyle\int_{0}^{2}f(x) dx$

Let $x = 2t$ implies $dx = 2dt$

Changing the limits and substituting,

$I = 2 \displaystyle\int_{0}^{1}f(2t) dt$

Since $f(t) = f(2t),$

$I = 2 \displaystyle\int_{0}^{1}f(t) dt$

Therefore,

$I = 200$

I figured the only way f(x) and be equal to f(2x) is if it was a straight line parallel to the x-axis. Thus when you double the limit, you double the value.

We can infer that $f(x)=f(\frac{x}{2})\forall x$ and therefor $f(x)=f(\frac{x}{2^n})$ for all integer $n\ge 1$ . Because of that, $\lim_{x\to 0}f(x)$ must exists and we call that limit $L$ . For all $x, y$ , we have $f(x)-f(y)=f(\frac{x}{2^n})-f(\frac{y}{2^n})=L-L=0$ . So $f(x)=L\forall x$ . Using the integral condition, we know $L=100$ and can easily calculate the later integral

We know that $\int_{ka}^{kb}f(\frac{x}{k})dx = k\int_{a}^{b}f(x)dx$ from properties of integral. We would like this $\begin{aligned}\int_{0}^{2}f(x)dx = \int_{0}^{2}f(\frac{x}{2})dx \end{aligned}$ to hold. Taking into account that $f(x) = f(2x), \forall x$ , we can proce that $f(x) = f(\frac{x}{2})$ as follows

$\\ \\ \begin{aligned} f(x) &= f(\frac{x}{2}) \\ f(4(\frac{x}{4})) &= f(2(\frac{x}{4})) \\ f(4t) &= f(2t); & t = \frac{x}{4} \\ f(2(2t)) &= f(2t) \\ f(2k) &= f(k); & k =2t \end{aligned}$

Therefore, we can claim that $\begin{aligned}\int_{0(2)}^{1(2)}f(\frac{x}{2})dx = \int_{0}^{2}f(x)dx = 2\int_{0}^{1}f(x)dx = 2(100) = 200 \end{aligned}$

Can we solve it using reflection property of integration?

Integration of f(x) from 0 to 1 represents a area and integration of the same function from 1to 2 reprents the same area as calculated before... As the graph is parallel to x axis... integration of f(x) from 0 to 2 will simply 2 times of the area we calculated at the first time... That is given 100 So the and is 200

Let $F(x)$ be the antiderivative of $f(x)$ :

$\int^1_0{f(x)}dx = F(1)-F(0)$

Okay, since $f(x) = f(2x)$ , $\int^1_0{f(x)}dx = \int^1_0{f(2x)}dx$ .

Multiplying by $\frac{2}{2}$ : $\int^1_0{f(x)}dx = \int^1_0{f(2x)}dx = \frac{1}{2}\int^1_0{f(2x)}d(2x) =\frac{1}{2}(F(2)-F(0)) =100$

Solving this equation gives the answer, 200.

This would be a horizontal line. Therefore, double the upper limits means double the integral value

Since $f(x) = f(2x) \; \forall x \in \mathbb{R}$ , it follows that $f(x) = f\left(\dfrac{x}{2^n}\right) \; \forall x \in \mathbb{R} \; \forall n \in \mathbb{N}$

By continuity, $f(x) = \displaystyle{\lim_{n \to \infty} f(x)} = \displaystyle{\lim_{n \to \infty} f\left(\dfrac{x}{2^n}\right)} = f\left(\displaystyle{\lim_{n \to \infty}\dfrac{x}{2^n}}\right) = f(0)$

So $f(x)$ is a constant function, and

$\displaystyle{\int_0^2 f(x) \,dx} = 2\displaystyle{\int_0^1 f(x) \,dx} = 2(100) = 200$

If f(x)=f(2x) then for all x values, there will be the same y value, hence a horizontal or y=c line parallel to the x axis, where c is a number. For example if x=1, f(1)=f(2), if x=2, f(2)=f(4), if x=3, f(3)=f(6) and so on. This improper integral from 0 to 1 (since the function is not defined at 0) can still be interpreted as the area from the horizontal line to the x axis from a value a approaching 0 to 1, which is 100 (kind of like a rectangular area, if you will). Going an extra unit to the right is the same as doubling the area from 0 to 1.

You can just visualize it. If f(x) = f(2x), then f(0,001) = f(0,002) = f(0,004) = ... = f(2). Hence, it is a line segment parallel to the x-axis. The area under the curve from 1 to 2 has the same magnitude of the area from 0 to 1. So it is 2 times 100 = 200.

the solution is simple integration with constant term

Let f(x) equals 100,replace it in both the equation and get the answer.