Integration rule for piecewise definition by interval - Revision history

Vipul: /* Definite integral version with fixed lower endpoint, variable upper endpoint; also tackles indefinite integral */

2012-07-05T22:27:49Z

Definite integral version with fixed lower endpoint, variable upper endpoint; also tackles indefinite integral

← Older revision		Revision as of 22:27, 5 July 2012
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	We make cases based on the interval in which <math>x</math> lies.		We make cases based on the interval in which <math>x</math> lies.

	Suppose <math>x < -1~~</math>. For convenience, we consider the integral in reverse, i.e., we consider <math>\int_x^0 f(t) \, dt~~</math>. The interval of integration breaks at -1, and we get:		Suppose <math>x < -1</math>. The interval of integration breaks at -1, and we get:

	<math>\~~int_x~~^0 f(t) \, dt = \~~int_x^~~{-1} f(t) \, dt + \int_{-1}^0 f(t) \, dt = \~~int_x^~~{-1} t^5 \, dt + \~~int_~~{-1}^0 t \, dt = \left[\frac{t^6}{6}\right]~~_x^~~{-1} + \left[\frac{t^2}{2}\right]_{-1}^0 = \frac{~~1 -~~ x^6}{6} + \frac{-1}{2} ~~= \frac{-1}{3} - \frac{x^6}{6}</math>~~		<math>\int_0^x f(t) \, dt = \int_{-1}^x f(t) \, dt + \int_{0}^{-1} f(t) \, dt = \int_{-1}^x t^5 \, dt + \int_0^{-1} t \, dt = \left[\frac{t^6}{6}\right]_{-1}^x + \left[\frac{t^2}{2}\right]_0^{-1} = \frac{x^6 - 1}{6} + \frac{1}{2} = \frac{1}{3} + \frac{x^6}{6}</math>

	~~Negating, we get:~~

	~~<math>\int_0^x f(t) \, dt~~ = \frac{1}{3} + \frac{x^6}{6}</math>

	Suppose now that <math>-1 \le x \le 0</math>. In this case, we can just do the integration directly, since the function has a single definition. We get:		Suppose now that <math>-1 \le x \le 0</math>. In this case, we can just do the integration directly, since the function has a single definition. We get:

Vipul: /* Indefinite integral computation */

2012-07-05T22:26:10Z

Indefinite integral computation

← Older revision		Revision as of 22:26, 5 July 2012
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	===Indefinite integral computation===		===Indefinite integral computation===

	Simply do the "definite integral version with fixed lower endpoint, variable upper endpoint" by making an arbitrary choice of fixed lower endpoint (it is usually most convenient to choose this as the left endpoint of the interval of definition of the function if such a point exists). After getting the answer, add a <math>+C</math> to it (or rather, to each of the pieces.		Simply do the "definite integral version with fixed lower endpoint, variable upper endpoint" by making an arbitrary choice of fixed lower endpoint (it is usually most convenient to choose this as the left endpoint of the interval of definition of the function if such a point exists). After getting the answer, add a <math>+C</math> to it (or rather, to each of the pieces).

	==Examples==		==Examples==

Vipul: /* Examples */

2012-07-05T22:21:40Z

Examples

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 ==Examples==
 Consider the function:
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 <math>\int_{-1}^3 f(x) \, dx = \frac{-1}{2} + \frac{1}{3} + 20 = \frac{119}{6}</math>

Vipul: Created page with "==Statement== ===Definite integration version with fixed endpoints=== The process is as follows: * Break up the interval of integration into pieces based on the pieces of d..."

2012-07-05T22:10:08Z

Created page with "==Statement== ===Definite integration version with fixed endpoints=== The process is as follows: * Break up the interval of integration into pieces based on the pieces of d..."

New page

==Statement==

===Definite integration version with fixed endpoints===

The process is as follows:

* Break up the interval of integration into pieces based on the pieces of definition of the function.
* Compute the definite integral on each piece of the corresponding function.
* Add them up.

===Definite integral version with fixed lower endpoint, variable upper endpoint===

The process is as follows:

* Make cases for the variable upper endpoint based on which interval it lands inside.
* For each case, use the "definite integration version with fixed endpoints"

===Indefinite integral computation===

Simply do the "definite integral version with fixed lower endpoint, variable upper endpoint" by making an arbitrary choice of fixed lower endpoint (it is usually most convenient to choose this as the left endpoint of the interval of definition of the function if such a point exists). After getting the answer, add a <math>+C</math> to it (or rather, to each of the pieces.

==Examples==

Consider the function:

<math>f(x) := \left\lbrace \begin{array}{rr} -5, & x \le -5 \\ x, & -5 < x \le 0 \\ x^2, & 0 < x < 1 \\ x^3, & x \ge 1 \\\end{array}\right.</math>

Suppose we are asked to calculate:

<math>\int_{-1}^3 f(x) \, dx</math>

We first break up the interval of integration into pieces based on the function definition. We note that the only points in <math>[-1,3]</math> where the function definition changes are the points 0 and 1, so we break up the interval of integration at 0 and 1. We thus get three intervals: <math>[-1,0], [0,1], and [1,3]</math>. Thus:

<math>\int_{-1}^3 f(x) \, dx = \int_{-1}^0 f(x) \, dx + \int_0^1 f(x) \, dx + \int_1^3 f(x) \, dx</math>

Within each interval, we can use the definition for that interval:

<math>\int_{-1}^3 f(x) \, dx = \int_{-1}^0 x \, dx + \int_0^1 x^2 \, dx + \int_1^3 x^3 \, dx</math>

This becomes:

<math>\int_{-1}^3 f(x) \, dx = \left[\frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^3}{3}\right]_0^1 + \left[\frac{x^4}{4}\right]_1^3</math>

This simplifies to:

<math>\int_{-1}^3 f(x) \, dx = \frac{-1}{2} + \frac{1}{3} + 20 = \frac{119}{6}</math>