257
edits
Changes
Jump to navigation
Jump to search
Line 11:
Line 11: + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Added examples and expanded article
If the function f(x) is real rather than complex, then the definite integral is also known as a Riemann integral.
If the function f(x) is real rather than complex, then the definite integral is also known as a Riemann integral.
== Solving Definite Integrals ==
Solving a definite integral usually has two main steps: [[integration]] and [[subtraction]].
== Example 1==
This is a very simple definite integral:
:<math>\int_{-3}^{5} x^2 dx</math>
=== Integration ===
<br />Using [[indefinite integral|indefinite integration]], it can be shown that:
:<math>\int x^2dx = {1 \over 3} x^3 = F(x)</math>
<br />Note that <math>F(x)</math> is the indefinite integral of <math>f(x)</math>.
=== Subtraction ===
<br />Now, plug <math>5</math> and <math>-3</math> into the new expression and subtract, as shown by the [[Fundamental Theorem of Calculus]].
: <math>F(5) = {1 \over 3}5^3 = {125 \over 3}</math>
: <math>F(-3) = {1 \over 3}(-3)^3 = {-27 \over 3}</math>
: <math>{125 \over 3} - {-27 \over 3} = {125 \over 3} + {27 \over 3} </math>
: <math>= {152 \over 3}</math>
== Example 2 ==
This is a more complex definite integral that requires [[Partial fractions in integration|partial fractions]] to solve:
: <math>\int_4^{12}\frac{3x+11}{x^2-x-6}dx </math>
=== Integration ===
<br />See the [[Partial fractions in integration]] page for how to integrate.
<br />As shown on the page mentioned above:
: <math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math>
=== Subtraction ===
This means that we can now subtract:
: <math>\left [ 4ln|12-3|-ln|12+2|+c \right ] - \left [ 4ln|4-3|-ln|4+2|+c \right ]</math>
: <math>= \left [ 4ln|9|-ln|14|+c \right ] - \left [ 4ln|1|-ln|6|+c \right ]</math>
: <math>= 4ln|9|-ln|14| - 4ln|1| + ln|6|</math>
: <math>= 4ln|9|-ln|14| + ln|6|</math>
: <math>\approx 6.02941...</math>
<br />Note the following:
* <math>ln(1) = 0</math>
* The <math>c</math> on each side cancels out because we have <math>c-c</math>
* <math>ln|6|</math> becomes positive because it was <math>-(-ln|6|)</math>
== See Also ==
== See Also ==