Properties of Definite Integrals
Throughout, we assume $f$ and $g$ are arbitrary functions and
$a,b,c,m$ and $M$ are constants:
- $\int_a^b c \; dx = c(b-a)\\$.
- $\int_a^b [f(x) \pm g(x)] \,dx = \int_a^b f(x) \,dx
\pm \int_a^b g(x) \,dx\\$
- $\int_a^b cf(x) \,dx = c\int_a^b f(x) \,dx\\$
- $\int_b^a f(x)\, dx = -\int_a^b f(x) \,dx\\$
- $\int_a^a f(x) \,dx = 0\\$
- $\int_a^c f(x)\, dx=\int_a^b f(x)\, dx + \int_b^c
f(x)\, dx$ for any $b$.
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Items 2 and 3 are direct results of the definition of the
definite integral as a limit, since the limit of a sum (or
difference) of functions is the sum (or difference) of the limits,
and since you can pull a constant out of a limit. Item 4 is
easy to see if you think of the limit definition of the integral
-- $\Delta x$ becomes $\frac{a-b}{n}$ instead of $\frac{b-a}{n}$
and thus is negated.
DO: Sketch appropriate
graphs and convince yourself that items 1 and 5 above make
sense.
DO: Sketch a graph of
some $f$ with $a<b<c$ and convince yourself that item 6
above makes sense. Note: $b$ does not have to be
between $a$ and $c$ for item 6 to be true. This is harder
to see, but play with it if you wish.
DO: Sketch appropriate
graphs and convince yourself that the three properties below
make sense.
- If $a \le b$ and $f(x) \geq 0$ for $a \leq x \leq b$,
then $\int_a^b f(x) \,dx \geq 0$.
- If $a \le b$ and $f(x) \geq g(x)$ for $a \leq x \leq
b$, then $\int_a^b f(x)\, dx \geq \int_a^b g(x)\, dx$.
- If $a \le b$ and $m \leq f(x) \leq M$ for $a \leq x
\leq b$, then $$m(b-a) \leq \int_a^b f(x)\, dx \leq
M(b-a)$$
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In other words, we can compare $f$ to $0$, we can compare $f$ to
$g$, and we can compare $f$ to $m$ and $M$.
These properties are easy to visualize if you think about area
and graphs, as you did above. If $f(x) \ge g(x)$, then the
curve $y=f(x)$ lies above the curve $y=g(x)$, so there is more
area under the $f$ curve than under the $g$ curve.
But these properties apply even when we are dealing with
negative functions or applications that have nothing to do with
area. To see why the second property holds without thinking
of areas and graphs, notice that if $f(x) \ge g(x)$ for all $x \in
[a,b]$, then $$ f(x_i^*) \ge g(x_i^*) \quad \text{and so} \quad
f(x_i^*)\,\Delta x \ge g(x_i^*)\,\Delta x, \text{ for each }
i=1,\ldots,n. \text{ Hence, }$$ $$ {\sum_{i=1}^n
f(x_i^*)\, \Delta x} \ge {\sum_{i=1}^n g(x_i^*)\, \Delta x}.
\text{ Then by limit laws, } {\lim_{n \to \infty}\,
\sum_{i=1}^n f(x_i^*)\, \Delta x} \ge {\lim_{n \to
\infty}\,\sum_{i=1}^n g(x_i^*)\, \Delta x}, $$ and thus by the
definition of integral $$\int_a^b f(x)\, dx \ge \int_a^b g(x)\,
dx. $$
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