my integration variable is canceled to $\int_0^2 1 dx $ - what is that?

Question

i have functions: $f(x)$ and $g(x) = f(x) +1)$. I want to calculate the area between those two functions in $[0;2]$.

Therefore, my integral is $\int_0^2 g(x) - f(x) dx $, which results in $\int_0^2 f(x) + 1 - f(x) dx $, and in the end is $f(x)$ cut out and the $1$ remains. Therefore my question:

What is

$$\int_0^2 1 dx $$

EDIT: To address @TBongers comment: Yes i can. I would estimate the geometrical area as 2, but i need to justify my estimation somehow and my actual approach is not very reasonable. But my question is actually specific for that situation. What happensi f the integration variable is canceled out?

Answer 1

Remember the general rule for antiderivatives of polynomial terms:

$$\int x^n=\frac{x^{n+1}}{1+n}+C \;, \;\;n\not=-1$$

This applies here as well, only now $n=0$, so you just get $x+C$. Evaluating this from $0$ to $2$ gives $\left[x \right] ^2_0=2-0=2$.

Answer 2

$$\int_0^2 1dx=\int_0^2 x^0 dx$$ Knowing that $$\int x^n dx=\frac{x^{n+1}}{n+1}+c$$ for $n\ne-1$, we have, $$\int_0^2 1dx=\int_0^2 x^0 dx=\frac{x^{0+1}}{0+1}=x|_0^2=2$$

Answer 3

Hint: Since $\frac{d}{dx}x=1$, what can you then say about the indefinite integral $\int 1 dx$? How does this help you with your calculation?

Further: What does the region under $y=1$ look like? What is the "length" of this shape? Here, geometry will suffice