Definite Integrals

The definite integral $\int_{{a}}^{{b}}$f (x)dx of a function f (x) defined on the interval [a, b] is given by

$\displaystyle \int_{{a}}^{{b}}$f (x)dx = $\displaystyle \lim_{{\left\Vert P\right\Vert \rightarrow
0}}^{}$$\displaystyle \sum_{{i=1}}^{{n}}$f$\displaystyle \left(\vphantom{
\overline{x}_{i}}\right.$$\displaystyle \overline{{x}}_{{i}}^{}$$\displaystyle \left.\vphantom{
\overline{x}_{i}}\right)$Δxi

where $\overline{{x}}_{{i}}^{}$ is a point in the ith subinterval of the partition P : a = x0 < x1 < x2 < ... < xn = b of the interval [a, b] and $\left\Vert\vphantom{ P}\right.$P$\left.\vphantom{ P}\right\Vert$ = max$\left\{\vphantom{ \Delta x_{i}}\right.$Δxi$\left.\vphantom{ \Delta x_{i}}\right\}$. The sum $\sum_{{i=1}}^{{n}}$f$\left(\vphantom{ \overline{x}_{i}}\right.$$\overline{{x}}_{{i}}^{}$$\left.\vphantom{ \overline{x}_{i}}\right)$Δxi is called a Riemann sum. The function f is integrable on [a, b] if the preceding limit exists.

If f is integrable on [a, b], then

$\displaystyle \int_{{a}}^{{b}}$f (x)dx = $\displaystyle \lim_{{n\rightarrow \infty }}^{}$$\displaystyle {\frac{{b-a}}{{n}%
}}$$\displaystyle \sum_{{i=1}}^{{n}}$f$\displaystyle \left(\vphantom{ a+i\frac{b-a}{n}}\right.$a + i$\displaystyle {\frac{{b-a}}{{n}}}$$\displaystyle \left.\vphantom{ a+i\frac{b-a}{n}}\right)$

In particular, if f is continuous on [a, b], then f is integrable on [a, b].

Definite integrals can be evaluated in the same way as indefinite integrals: Leave the insertion point in the expression and choose Evaluate or Evaluate Numerically.

$\blacktriangleright$ To enter a definite integral

1.
Click itbpF0.3009in0.3009in0.0701inintegral.wmf or choose Insert + Operator and choose the integral sign.

2.
Click itbpF0.3009in0.3009in0.0701insubscrip.wmf or choose Insert + Subscript.

3.
Use the tab key to enter the upper limit of integration. (Limits of integration work the same as any other subscripts or superscripts.)

4.
Press the spacebar or the right arrow to get out of the superscript, and type the rest of the expression.

$\blacktriangleright$ Evaluate, Evaluate Numerically

$\int_{{0}}^{{1}}$x2$\sqrt{{x^{3}+1}}$dx = ${\frac{{4}}{{9}}}$$\sqrt{{2}}$ - ${\frac{{2}}{{9}}}$ = .4063171388

$\blacktriangleright$ Evaluate Numerically

$\int_{{0}}^{{1}}$x2$\sqrt{{x^{3}+1}}$dx = .4063171388

The following examples are easily handled by Scientific Notebook. Choose Evaluate or Evaluate Numerically to compute these integrals.

$\displaystyle \dint_{{0}}^{{2\pi }}$sin x dx 6pt $\displaystyle \dint_{{0}}^{{1}}$$\displaystyle \sqrt{{1-x^{2}}}$dx $\displaystyle \dint_{{1}}^{{2}}$ln x dx
$\displaystyle \dint_{{0}}^{{1}}$ln x dx 6pt $\displaystyle \dint_{{a}}^{{b}}$x ln x dx $\displaystyle \dint_{{0}}^{{1}}$ex dx
$\displaystyle \dint_{{0}}^{{\pi }}$x lnsin x dx 6pt $\displaystyle \dint_{{0}}^{{\infty
}}$e-x2 dx $\displaystyle \dint_{{1}}^{{\infty }}$$\displaystyle {\dfrac{{1}}{{x^{2}}}}$ dx
$\displaystyle \dint_{{1}}^{{\infty }}$$\displaystyle {\dfrac{{1}}{{x^{n}}}}$ dx 6pt $\displaystyle \dint_{{0}}^{{\pi /4}}$sec x dx $\displaystyle \dint_{{0}}^{{\pi /2}}$sec3x dx
$\displaystyle \dint_{{0}}^{{1}}$$\displaystyle {\dfrac{{1}}{{x^{1/3}}}}$ dx $\displaystyle \dint_{{0}}^{{\infty
}}$$\displaystyle {\dfrac{{\cos x}}{{%
\sqrt{x}}}}$ dx $\displaystyle \dint_{{0}}^{{\infty
}}$ $\displaystyle {\dfrac{{dx}}{{\left( 1+x\right) \sqrt{x}}%
}}$


Integrals involving absolute values or piecewise-defined functions can now be treated like any other function.



\begin{example}
Consider the integral $\int_{-2}^{2}\left\vert x^{2}-1\right\ve...
...t) dx=\allowbreak 4
\end{displaymath}
\medskip
\end{itemize}
\end{example}


\begin{example}
Suppose that $f$\ is defined piecewise by the expression
\beg...
...2}dx+\int_{0}^{3}xdx=\allowbreak \frac{43}{6}
\end{displaymath}
\end{example}

Note: You should use special care when working with improper integrals. Make sure answers look reasonable. Neither machines nor people are infallible, but people and machines working together can accomplish more than either can working alone or unattended.



\begin{example}
The following example illustrates one type of potential problem...
...r all $x$; thus the integral, if it exists, must be
nonnegative.
\end{example}

Note    This improper integral is examined further in the exercises at the end of this chapter.



Subsections