Difference between revisions of "User:Able806/integration"
| Line 56: | Line 56: | ||
===Algebraic Substitution=== | ===Algebraic Substitution=== | ||
| + | '''Integration by Algebraic Substitution''' is a [[Techniques of integration|technique]] to facilitate the integration of a rational expression by substituting a more complicated expression with a variable. | ||
| + | |||
| + | Given an integral | ||
| + | |||
| + | :<big><math>\int\frac{2x}{x^2+3}dx</math></big> | ||
| + | |||
| + | We can substitute the term :<big><math>x^2+3</math></big> with a u. Giving us | ||
| + | |||
| + | :<big><math>u=x^2+3</math></big> | ||
| + | |||
| + | We then take the derivative of u with respect to x, | ||
| + | |||
| + | :<big><math>\frac{du}{dx}x^2+3=2x</math></big> | ||
| + | |||
| + | We then set the terms equal to du, | ||
| + | |||
| + | :<big><math>du=2xdx</math></big> | ||
| + | |||
| + | Now we are ready to rewrite the integral, | ||
| + | |||
| + | :<big><math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math></big> | ||
| + | |||
| + | We can rewrite the intergal this way due to the subsitution of the x terms with the u terms. | ||
| + | |||
| + | Now we can solve the intergral in terms of u. | ||
| + | |||
| + | :<big><math>\int\frac{1}{u}du=ln|u|+c</math></big> | ||
| + | |||
| + | Now we replace u with the term :<big><math>x^2+3</math></big> to get, | ||
| + | |||
| + | :<big><math>ln|x^2+3|+c</math></big> | ||
| + | |||
| + | We can check this by taking the derivative of :<big><math>ln|x^2+3|</math></big>, | ||
| + | |||
| + | :<big><math>\frac{d}{dx}ln|x^2+3|=(\frac{1}{x^2+3})(2x)=\frac{2x}{x^2+3}</math></big> | ||
| + | |||
===Trigonometric Substitution=== | ===Trigonometric Substitution=== | ||
Revision as of 20:13, April 16, 2008
Methods of Integration
Integration by Parts
Integration by parts is a special technique to facilitate the integration of the product of two functions that otherwise lack an obvious integral. This technique utilizes the insight of the product rule.
The rule for integration by parts is stated as follows:
This rule is often useful when one function is a power of x and the other function is a trigonometric function or e raised to a power of x.
Note that it may be necessary to repeat the integration by parts several times, one for each power of x.
Partial Fractions
Integration by partial fractions is a technique to facilitate the integration of a rational expression by partial fraction decomposition.
Given an integral
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx}
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
This allows us to split the fraction in to sums by cross multiplying the denominators,
Therefore we can restate the problem,
Now we can solve for A and B by subsituting x with a value that allows the term to go to 0. For example,
We let :,
We let :,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 20=A(5)+B(0)}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=4}
We then plug in the values of A and B and get,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4}{x-3}-\frac{1}{x+2}}
Now we can solve the integral.
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c}
Algebraic Substitution
Integration by Algebraic Substitution is a technique to facilitate the integration of a rational expression by substituting a more complicated expression with a variable.
Given an integral
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{2x}{x^2+3}dx}
We can substitute the term :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2+3} with a u. Giving us
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^2+3}
We then take the derivative of u with respect to x,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{dx}x^2+3=2x}
We then set the terms equal to du,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=2xdx}
Now we are ready to rewrite the integral,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du}
We can rewrite the intergal this way due to the subsitution of the x terms with the u terms.
Now we can solve the intergral in terms of u.
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{1}{u}du=ln|u|+c}
Now we replace u with the term :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2+3} to get,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ln|x^2+3|+c}
We can check this by taking the derivative of :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ln|x^2+3|} ,
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}ln|x^2+3|=(\frac{1}{x^2+3})(2x)=\frac{2x}{x^2+3}}