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

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?

• Can you compute the antiderivative of $1$? Or interpret this integral as the area of an extremely simple geometric region?
– user296602
Mar 7, 2016 at 20:03
• Hint: $1 = x^0$. Mar 7, 2016 at 20:05
• What do you mean "if the integration variable is canceled out"? Let $h(x)$ be the constant function $1$; you're trying to compute $\int_0^2 h(x) \, dx$; there are many standard ways to do this, with either the antiderivative or knowing the geometry.
– user296602
Mar 7, 2016 at 20:07
• @T.Bongers i just knew a very basic form of integration like $\int_0^2 (3x^3 + 13x +9) dx$ or sth similar. I was just confused by the "strange" appearance. Mar 7, 2016 at 20:15
• @all why so many dislikes? I didn't knew this form.. Mar 7, 2016 at 20:16

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 $\left[x \right] ^2_0=2-0=2$.

• For completeness, $n\neq 1$.
– Em.
Mar 7, 2016 at 20:33
• @probablyme I assume you mean $-1$ - in that case, thanks, I included it! :) Mar 7, 2016 at 21:00
• Yes, typo. Upvoted.
– Em.
Mar 7, 2016 at 21:02

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

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