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Added some of the instructions for performing integration using parti…
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…al fraction decomposition.
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Tom Malone committed Oct 22, 2015
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84 changes: 77 additions & 7 deletions 07_Integration_Techniques/.ipynb_checkpoints/Partial_Fraction_Decomposition-checkpoint.ipynb
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Expand Up @@ -21,16 +21,86 @@
"\n",
"To integration the partial fractions, just integrate each fraction individually and the add the integrals together.\n",
"\n",
"$$\\int \\frac{5x-3}{x^2-2x-3}dx = \\int \\frac{2}{x+1}dx + \\int \\frac{3}{x-3}dx$$"
"$$\\int \\frac{5x-3}{x^2-2x-3}dx = \\int \\frac{2}{x+1}dx + \\int \\frac{3}{x-3}dx$$\n",
"\n",
"### Steps\n",
"\n",
"To use the method to integrate $\\int \\frac{5x-3}{x^2-2x-3}dx, follow these steps:\n",
"\n",
"<table>\n",
"<tr>\n",
" <td style='width: 40%;'>\n",
" __Step 1__\n",
" Break the rational function into its constituent parts by splitting the denominator into its factors. \n",
" </td>\n",
" <td style='width: 60%;'>\n",
" $$\\frac{5x-3}{x^2-2x-3} = \\frac{5x-3}{(x+1)(x-3)}$$\n",
" </td>\n",
"</tr>\n",
"<tr>\n",
" <td style='width: 40%;'>\n",
" __Step 2__\n",
" Use $A$ and $B$ (and successive capital constants as needed) for the numerators, for now.\n",
" </td>\n",
" <td style='width: 60%;'>\n",
" $$\\frac{5x-3}{(x+1)(x-3)} = \\frac{A}{x+1} + \\frac{B}{x-3}$$\n",
" </td>\n",
"</tr>\n",
"<tr>\n",
" <td colspan='2'>\n",
" __Step 3__\n",
" Before you can find $A$ and $B$, clear the equation of fractions by multiplying both sides by the common denominator, $(x+1)(x-3)$ <br/><br/>\n",
" \n",
" $$(x+1)(x-3) \\left( \\frac{5x-3}{(x+1)(x-3)} \\right) = \\left[ \\frac{A}{x+1} + \\frac{B}{x-3} \\right] \\cdot (x+1)(x-3)$$\n",
" <br/>\n",
" $$5x-3 = A(x-3) + B(x+1)$$\n",
" </td>\n",
"</tr>\n",
"\n",
"<tr>\n",
" <td colspan='2'>\n",
" __Step 4__\n",
" To find $A$ and $B$, substitute an appropriate value into (first) the $(x+1)$ term to remove it from the equation (i.e. to make it equal to $0$), then do the same for $(x-3)$.<br/><br/>\n",
" \n",
" <table class='no-border' style='width: 100%'>\n",
" <tr>\n",
" <td>\n",
" Subtitute $x = -1$:\n",
" $$5(-1)-3 = A[(-1)-3] + B[(-1)+1]$$<br/>\n",
" \n",
" $$-5-3 = A(-4) + B(0)$$<br/>\n",
"\n",
" $$-4A = -8$$<br/>\n",
"\n",
" $$A = \\frac{-8}{-4}$$<br/>\n",
"\n",
" $$A = 2$$\n",
" </td>\n",
" \n",
" <td>\n",
" Substitute $x = 3$\n",
" $$5(3)-3 = A[(3)-3] + B[(3)+1]$$<br/>\n",
" \n",
" $$15-3 = A(0) + B(4)$$<br/>\n",
" \n",
" $$4B = 12$$<br/>\n",
" \n",
" $$B = \\frac{12}{4}$$<br/>\n",
" \n",
" $$B = 3$$\n",
" </td>\n",
" </tr>\n",
" </table>\n",
" </td>\n",
"</tr>\n",
"</table>\n",
"\n",
"$$$$"
]
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84 changes: 77 additions & 7 deletions 07_Integration_Techniques/Partial_Fraction_Decomposition.ipynb
View file
Original file line number Diff line number Diff line change
Expand Up @@ -21,16 +21,86 @@
"\n",
"To integration the partial fractions, just integrate each fraction individually and the add the integrals together.\n",
"\n",
"$$\\int \\frac{5x-3}{x^2-2x-3}dx = \\int \\frac{2}{x+1}dx + \\int \\frac{3}{x-3}dx$$"
"$$\\int \\frac{5x-3}{x^2-2x-3}dx = \\int \\frac{2}{x+1}dx + \\int \\frac{3}{x-3}dx$$\n",
"\n",
"### Steps\n",
"\n",
"To use the method to integrate $\\int \\frac{5x-3}{x^2-2x-3}dx, follow these steps:\n",
"\n",
"<table>\n",
"<tr>\n",
" <td style='width: 40%;'>\n",
" __Step 1__\n",
" Break the rational function into its constituent parts by splitting the denominator into its factors. \n",
" </td>\n",
" <td style='width: 60%;'>\n",
" $$\\frac{5x-3}{x^2-2x-3} = \\frac{5x-3}{(x+1)(x-3)}$$\n",
" </td>\n",
"</tr>\n",
"<tr>\n",
" <td style='width: 40%;'>\n",
" __Step 2__\n",
" Use $A$ and $B$ (and successive capital constants as needed) for the numerators, for now.\n",
" </td>\n",
" <td style='width: 60%;'>\n",
" $$\\frac{5x-3}{(x+1)(x-3)} = \\frac{A}{x+1} + \\frac{B}{x-3}$$\n",
" </td>\n",
"</tr>\n",
"<tr>\n",
" <td colspan='2'>\n",
" __Step 3__\n",
" Before you can find $A$ and $B$, clear the equation of fractions by multiplying both sides by the common denominator, $(x+1)(x-3)$ <br/><br/>\n",
" \n",
" $$(x+1)(x-3) \\left( \\frac{5x-3}{(x+1)(x-3)} \\right) = \\left[ \\frac{A}{x+1} + \\frac{B}{x-3} \\right] \\cdot (x+1)(x-3)$$\n",
" <br/>\n",
" $$5x-3 = A(x-3) + B(x+1)$$\n",
" </td>\n",
"</tr>\n",
"\n",
"<tr>\n",
" <td colspan='2'>\n",
" __Step 4__\n",
" To find $A$ and $B$, substitute an appropriate value into (first) the $(x+1)$ term to remove it from the equation (i.e. to make it equal to $0$), then do the same for $(x-3)$.<br/><br/>\n",
" \n",
" <table class='no-border' style='width: 100%'>\n",
" <tr>\n",
" <td>\n",
" Subtitute $x = -1$:\n",
" $$5(-1)-3 = A[(-1)-3] + B[(-1)+1]$$<br/>\n",
" \n",
" $$-5-3 = A(-4) + B(0)$$<br/>\n",
"\n",
" $$-4A = -8$$<br/>\n",
"\n",
" $$A = \\frac{-8}{-4}$$<br/>\n",
"\n",
" $$A = 2$$\n",
" </td>\n",
" \n",
" <td>\n",
" Substitute $x = 3$\n",
" $$5(3)-3 = A[(3)-3] + B[(3)+1]$$<br/>\n",
" \n",
" $$15-3 = A(0) + B(4)$$<br/>\n",
" \n",
" $$4B = 12$$<br/>\n",
" \n",
" $$B = \\frac{12}{4}$$<br/>\n",
" \n",
" $$B = 3$$\n",
" </td>\n",
" </tr>\n",
" </table>\n",
" </td>\n",
"</tr>\n",
"</table>\n",
"\n",
"$$$$"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"cell_type": "markdown",
"metadata": {},
"source": []
}
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