# 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.

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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.

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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.

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