Question about additivity of integration on intervals I was taught the following regarding integration (it's vaguely explained but my book does not go into further specifications).
If $c$ is any element of $[a,b]$:
$$\int_a^bf(x) \, dx = \int_a^cf(x)\,dx+\int_c^bf(x)\,dx$$ 
Today, during a lecture, the professor did the following:
$$\int_x^{\sin x}f(x)\,dx=\int_x^0f(x)\,dx+\int_0^{\sin x}f(x)\, dx$$
He argued that:
$$\int_x^{\sin x}f(x)\,dx=\int_x^{17000}f(x)\,dx+\int_{17000}^{\sin x}f(x)\, dx$$
is also correct (the point being that $0$ is not the only number with which the expression is correct but that any arbitrary number will do).
I am confused about why the last two expressions are true and how are they related to the property at the beginning. I don't see how there can be an interval $[x,\sin x]$ where every real number can be found.
 A: The first formula that you wrote down, which we shall call the original Additivity Theorem, can be generalized to the case when $ c \notin [a,b] $.


*

*Suppose that $ c < a \leq b $. Using the definition $ \displaystyle \int_{a}^{c} f(x) ~ d{x} \stackrel{\text{def}}{=} - \int_{c}^{a} f(x) ~ d{x} $, we see that
\begin{align}
   \int_{a}^{c} f(x) ~ d{x} + \int_{c}^{b} f(x) ~ d{x}
&= - \int_{c}^{a} f(x) ~ d{x} + \underbrace{\left[ \int_{c}^{a} f(x) ~ d{x} + \int_{a}^{b} f(x) ~ d{x} \right]}_{\text{By the original Additivity Theorem.}} \\
&= \int_{a}^{b} f(x) ~ d{x}.
\end{align}

*Suppose that $ a \leq b < c $. Using the definition $ \displaystyle \int_{c}^{b} f(x) ~ d{x} \stackrel{\text{def}}{=} - \int_{b}^{c} f(x) ~ d{x} $, we see that
\begin{align}
   \int_{a}^{c} f(x) ~ d{x} + \int_{c}^{b} f(x) ~ d{x}
&= \underbrace{\left[ \int_{a}^{b} f(x) ~ d{x} + \int_{b}^{c} f(x) ~ d{x} \right]}_{\text{By the original Additivity Theorem.}} - \int_{b}^{c} f(x) ~ d{x} \\
&= \int_{a}^{b} f(x) ~ d{x}.
\end{align}
Therefore, the Additivity Theorem holds, whether or not $ c \in [a,b] $. This means that the precise order relations among the numbers $ x $, $ \sin(x) $, $ 0 $ and $ 17,000 $ should not matter in the second and third formulae in the wording of your problem.
A: Suppose $\sin x \ge x$. Then
$$\begin{align}\int_x^{17000}f(x)\,dx+\int_{17000}^{\sin x}f(x)\, dx
&=\int_x^{\sin{x}}f(x)\,dx+\int_{\sin x}^{17000}-\int_{\sin x}^{17000}f(x)\, dx\\
&=\int_x^{\sin{x}}f(x)\,dx
\end{align}$$
using the property you mention, as well as the fact that when $a>b$ the integral $\int_a^b f(x)dx$ is defined as $-\int_b^af(x)dx$. The case $\sin x\le x$ is similar using this definition.
