not my favourite area:

**Amanda-UK** · August 15th 2007, 04:46 AM

im not the biggest fan of partial fraction, doing so practice excercise to get me in the swing of things before going back to uni and have stumbled across this:

f(x) = 3
x^3 +3x^2 -10x

1) write f(x) as a sum of partial fractions
2) evaluate (intergrate f(x))

**galactus** · August 15th 2007, 05:11 AM

Start by factoring the denominator.

$x^{3}+3x^{2}-10x=x(x-2)(x+5)$

That gives:

$\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+5}=3$

Now, can you solve for A,B,C?.

**Amanda-UK** · August 15th 2007, 05:19 AM

is this correct so far:

3= A(x+5) + B(x-2)(x) + C(x)

amanda x

**topsquark** · August 15th 2007, 05:29 AM

Originally Posted by Amanda-UK

is this correct so far:

3= A(x+5) + B(x-2)(x) + C(x)

amanda x

Almost. Take it step by step and add the fractions:

$\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+5}=\frac{3}{x(x - 2)(x + 5)}$

$\frac{A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2)}{x(x - 2)(x + 5)} = \frac{3}{x(x - 2)(x + 5)}$

So
$A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2) = 3$

-Dan

**Amanda-UK** · August 15th 2007, 05:36 AM

thank you both of you. trying to get my head around intergrating it now :s

**Amanda-UK** · August 15th 2007, 07:12 AM

i have managed to get
b = 3/15 (by using 2)
c = 3/35 by using -5)

cant find the number to use to work out a :s

**topsquark** · August 15th 2007, 07:37 AM

Originally Posted by topsquark

Almost. Take it step by step and add the fractions:

$\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+5}=\frac{3}{x(x - 2)(x + 5)}$

$\frac{A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2)}{x(x - 2)(x + 5)} = \frac{3}{x(x - 2)(x + 5)}$

So
$A(x - 2)(x + 5) + Bx(x + 5) + Cx(x - 2) = 3$

-Dan

Allow me to continue, then.

$A(x^2 + 3x - 10) + B(x^2 + 5x) + C(x^2 - 2x) = 3$

$(A + B + C)x^2 + (3A + 5B - 2C)x - 10A = 3$

Thus
$A + B + C = 0$
$3A + 5B - 2C = 0$
$-10A = 3$

So
$A = -\frac{3}{10}$

This implies the other two equations are:
$B + C = \frac{3}{10}$
$5B - 2C = \frac{9}{10}$

From the first equation:
$C = \frac{3}{10} - B$

Inserting this into the second equation:
$5B - 2 \left ( \frac{3}{10} - B \right ) = \frac{9}{10}$

$B = \frac{3}{14}$

$C = \frac{3}{10} - \frac{3}{14} = \frac{3}{35}$

-Dan

**Soroban** · August 15th 2007, 07:53 AM

Hello, Amanda!

$f(x) \:=\:\frac{3}{x^3+3x^2 - 10x}$

(1) Write $f(x)$ as a sum of partial fractions

(2) Find: . $\int f(x)\,dx$

You seem to know the "fast" way for Partial Fractions.
I find it much easier, but many teachers don't know it
. . and others actively disapprove of it.

We have: . $\frac{3}{x(x-2)(x+5)} \:=\:\frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+5}$

Multiply through the LCD: . $3 \;=\;(x-2)(x+5)A + x(x+5)B + x(x-2)C$

Now select some "nice" values for $x$ . .(You'll see what I mean)

. . Let $x=5\!:\;\;3 \;=\;(\text{-}2)(5)A + 0\!\cdot\!B + 0\!\cdot\!C\quad\Rightarrow\quad A \:=\:\text{-}\frac{3}{10}$

. . Let $x = 2\!:\;\;3 \;=\;0\!\cdot\!A + 2\!\cdot\!7\!\cdot\!B + 0\!\cdot\!C \quad\Rightarrow\quad B \:=\:\frac{3}{14}$

. . Let ${\color{blue}x = \text{-}5}\!:\;\;3 \;=\;0\!\cdot\!A + 0\!\cdot\!B + (\text{-}5)(\text{-}7)C\quad\Rightarrow\quad C \:=\:\frac{3}{35}$

And we have: . $\frac{-\frac{3}{10}}{x} + \frac{\frac{3}{14}}{x-2} + \frac{\frac{3}{35}}{x+5} \;=\;\frac{3}{70}\left[-\frac{7}{x} + \frac{5}{x-2} + \frac{2}{x+5}\right]$ . (1)

Integrate: . $\frac{3}{70}\int\left[-\frac{7}{x} + \frac{5}{x-2} + \frac{2}{x+5}\right]\,dx \;= \;\frac{3}{70}\bigg[-7\ln|x|\, +\, 5\ln|x-2|\, +\, 2\ln|x+5|\bigg]\,+\,C$ . (2)

If you want to show-off, you can simplify further . . .

$\frac{3}{70}\bigg[-\ln|x^7| + \ln|x-2|^5) + \ln(x+5)^2\bigg] + C \;=\; \frac{3}{70}\ln\left|\frac{(x-2)^5(x+5)^2}{x^7}\right| + C$

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