5.2: Definite Integrals

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below are two videos showing worked examples.

Problems

If $f(x)$ is a continuous function or a function with finitely many jump discontinuities on an interval $[a,b]$ , then the definite integral, denoted $\displaystyle \int _a^b f(x)dx$ represents the signed area under $y = f(x)$ on the interval $[a,b]$ , and is defined to be

$\int _a^b f(x)dx = \lim _{n \to \infty } \sum _{j=1}^n f(x_j^*)\Delta x.$

Consider the limit

$\lim _{n \to \infty } \sum _{j=1}^n (1-x_j^*)\Delta x.$ This represents a definite integral $\displaystyle \int _0^3 f(x)dx$ for some function $f(x)$ .

The function $f(x) = \answer {1-x}$ .
Evaluate the definite integral using geometry. $\int _0^3 f(x)\,dx = \answer {-3/2}.$
Remember that if a graph goes below the $x$ -axis, the contribution to the definite integral is negative.

Consider a function $f(x)$ whose graph is shown below.

Evaluate:

$\displaystyle \int _{-2}^0 f(x)dx = \answer {\pi }$ .
$\displaystyle \int _{-2}^2 f(x)dx = \answer {2\pi }$ .
$\displaystyle \int _2^3 f(x)dx = \answer {-1}$ .
$\displaystyle \int _3^0 f(x)dx = \answer {1-\pi }$ .
Use the property $\displaystyle \int _a^b f(x)dx = -\int _b^a f(x)dx$ .
$\displaystyle \int _0^2 (2f(x)-1)dx = \answer {2\pi -2}$ .
Use the property $\displaystyle \int _a^b (kf(x)\pm g(x))dx = k\int _a^b f(x)dx \pm \int _a^b g(x)dx$ .

If $f(x)$ is a continuous function or a function with finitely many jump discontinuities on $[a,b]$ , and $m$ and $M$ are the minimum and maximum values of $f(x)$ on $[a,b]$ , respectively, then

$m(b-a) \leq \int _a^b f(x)dx \leq M(b-a).$

Consider $f(x) = e^{-x}$ on the interval $[0,2]$ .

The maximum and minimum values of $f(x)$ on $[0,2]$ are, respectively, $M = \answer {1}$ and $m = \answer {e^{-2}}$ .
Using the property described in the theorem, we can say $\answer {2e^{-2}} \leq \int _0^2 e^{-x}\,dx \leq \answer {2}.$

Suppose $\displaystyle \int _0^a f(x)dx = a+2$ . Then

$\displaystyle \int _0^3 f(x)dx = \answer {5}$ .
$\displaystyle \int _5^0 2f(x)\,dx = \answer {-14}$ .
$\displaystyle \int _1^4 f(x)\,dx = \answer {3}$ .

Consider the graph of $f(x)$ below.

Let $\displaystyle g(x) = \int _{-2}^x f(t)\,dt$ for $-2 < x < 4$ .

The value of $g(-2) =\answer {0}$ .
The value of $g(1) = \answer {9/2}$ .
The value of $g(3) = \answer {5/2}$ .
On which interval in $(-2,4)$ is $g(x)$ increasing? In interval notation: $(\answer {-2},\answer {1})$ .
On which interval in $(-2,4)$ is $g(x)$ decreasing? In interval notation: $(\answer {1},\answer {4})$ .
The point $(1,9/2)$ is a local
of $g(x)$ .

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