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### Properties of the definite integral

Definite integrals obey rules similar to those for indefinite integrals. The following theorem is analogous to one for indefinite integrals.

###### Theorem (Linearity of integration)
1. If $$f$$ and $$g$$ are continuous functions on the interval $$[a,b]$$, then $\int_a^b \big( f(x) \pm g(x) \big) \; dx = \int_a^b f(x) \; dx \, \pm \int_a^b g(x) \; dx.$
2. If $$f$$ is a continuous function on $$[a,b]$$, and $$k$$ is a real constant, then> $\int_a^b kf(x) \; dx = k \int_a^b f(x) \; dx.$

The endpoints on a definite integral obey the following theorem.

Let $$f \colon [a,b] \to \mathbb{R}$$ be continuous, where $$a,b$$ are real numbers. Let $$c$$ be a real number between $$a$$ and $$b$$. Then,

$\int_a^b f(x) \; dx = \int_a^c f(x) \; dx \, + \int_c^b f(x) \; dx.$

As with many mathematical statements, it's useful to understand these two theorems both algebraically (in terms of antiderivatives) and geometrically.

For instance, we can think of the second theorem (additivity of integration) as saying geometrically that, if we consider the signed area between $$y=f(x)$$ and the $$x$$-axis from $$x=a$$ to $$x=b$$, then this signed area is equal to the sum of the signed area from $$x=a$$ to $$x=c$$ and that from $$x=c$$ to $$x=b$$. Alternatively, if we let $$F(x)$$ be an antiderivative of $$f(x)$$, we can regard the theorem as just expressing that

$F(b) - F(a) = \big(F(c) - F(a)\big) + \big(F(b) - F(c)\big).$

This piece of algebra and the fundamental theorem of calculus together give a rigorous proof of the theorem.

##### Exercise 13

Find a geometric interpretation of part (b) of the first theorem of this section (linearity of integration). You may assume the graph of $$y=f(x)$$ lies above the $$x$$-axis. Also find an interpretation in terms of antiderivatives.

We have stated the second theorem (additivity of integration) so that $$a < c < b$$. But in fact, this theorem works when $$a,b,c$$ are in any order, as long as $$f,g$$ are defined and continuous over the appropriate intervals. We just have to make sense of integrals which have their terminals 'in the wrong order'. When $$a>b$$, we define,

$\int_a^b f(x) \; dx = - \int_b^a f(x) \; dx.$

We can think of this as saying that, when $$x$$ goes backwards from $$a$$ to $$b$$, we count areas as negative. In our area estimates, $$\Delta x = \frac{1}{n} (b-a)$$ is the negative of the rectangle widths. This fits with our previous definitions, as you can show in the following exercise.

##### Exercise 14

Let $$f \colon \mathbb{R} \to \mathbb{R}$$ be a continuous function, and let $$F(x)$$ be an antiderivative of $$f(x)$$. Using the fundamental theorem of calculus, show that for any real numbers $$a,b$$ (even when $$a>b$$),

$\int_a^b f(x) \; dx = F(b) - F(a).$
##### Exercise 15

Find $$\displaystyle\int_5^0 (-2x - 3) \; dx$$.