# (a) Find the partial fraction decomposition of \frac{1}{(x-3)(x+1)}, and hence evaluate...

## Question:

(a) Find the partial fraction decomposition of {eq}\frac{1}{(x-3)(x+1)} {/eq}, and hence evaluate

{eq}\int^2_1\frac{1}{(x-3)(x+1)}dx {/eq}.

(b) Determine the general solution of the differential equation,

{eq}\frac{d^2x}{dt^2}-5\frac{dx}{dt}-24x=0, {/eq}

and find the particular solution that satisfies the initial conditions

{eq}x(0) = 1 {/eq} and {eq}\frac{dx}{dt}(0) = 5 {/eq}.

(c) Find the volume of the solid that is generated when the area enclosed by the curve y = sin(2x) and the x-axis on the interval {eq}\left [ 0, \frac{\pi}{2} \right ] {/eq} is rotated about the x-axis. [Use the disk method.]

## Partial Fraction Decomposition

Partial fraction decomposition is a technique in calculus that is used to express a rational term as a sum of its partial fractions. It is especially used in integrals of rational function since the resulting partial fractions are easier to integrate compared to its original form.

## Answer and Explanation:

**(a)**

Let's get the partial fraction decomposition of

{eq}\displaystyle \frac{1}{(x-3)(x+1)} {/eq}

Let's begin by expressing it as sum of its partial fraction whose denominators are the factors of denominators with unknown coefficients.

{eq}\displaystyle \frac{1}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1} {/eq}

Multiply both sides by the denominator of the left side.

{eq}\displaystyle (x-3)(x+1)\left [ \frac{1}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1} \right ](x-3)(x+1) \\ \displaystyle 1 = A(x+1) + B(x-3) \\ \displaystyle 1 = Ax + A + Bx - 3B \\ \displaystyle 1 = Ax + Bx + A - 3B \\ \displaystyle 1 = (A + B)x + (A - 3B) {/eq}

Equate the coefficients.

{eq}A + B = 0 \\ A - 3B = 1 {/eq}

Solve the unknown coefficients.

{eq}A = -B \\ (-B) - 3B = 1 \\ -4B = 1 \\ B = \dfrac{1}{-4} \\ B = -\dfrac{1}{4} \\ A = -\left ( -\dfrac{1}{4} \right ) \\ A = \dfrac{1}{4} {/eq}

The partial fraction of the given rational term is

{eq}\displaystyle \frac{1}{(x-3)(x+1)} = \frac{1}{4(x-3)} - \frac{1}{4(x+1)} {/eq}

The integral becomes

{eq}\displaystyle \int^2_1\frac{1}{(x-3)(x+1)}dx = \int^2_1 \left ( \frac{1}{4(x-3)} - \frac{1}{4(x+1)} \right )\,dx {/eq}

Temporarily remove the limits of integration and solve the indefinite integral.

{eq}\displaystyle = \int \left ( \frac{1}{4(x-3)} - \frac{1}{4(x+1)} \right )\,dx {/eq}

Apply sum rule.

{eq}\displaystyle = \int \frac{1}{4(x-3)}\,dx - \int \frac{1}{4(x+1)} \,dx \\ \displaystyle = \frac{1}{4}\int \frac{1}{(x-3)}\,dx - \frac{1}{4}\int \frac{1}{(x+1)} \,dx {/eq}

Apply u-substitution for each term.

{eq}u = x - 3 \\ du = dx \\ v = x + 1 \\ dv = dx {/eq}

The integral becomes

{eq}\displaystyle = \frac{1}{4}\int \frac{1}{u}\,du - \frac{1}{4}\int \frac{1}{v} \,dv \\ {/eq}

These are common integrals.

{eq}\displaystyle = \frac{1}{4}\ln|u| - \frac{1}{4}\ln|v| {/eq}

Substitute back the values of {eq}u {/eq} and {eq}v {/eq}.

{eq}\displaystyle = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| {/eq}

Bring back the limits of integration and evaluate it.

{eq}\displaystyle = \left (\frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| \right )\Bigg|_1^2 \\ \displaystyle = \left (\frac{1}{4}\ln|2-3| - \frac{1}{4}\ln|2+1| \right ) - \left (\frac{1}{4}\ln|1-3| - \frac{1}{4}\ln|1+1| \right ) \\ \displaystyle = \left (\frac{1}{4}\ln|-1| - \frac{1}{4}\ln|3| \right ) - \left (\frac{1}{4}\ln|-2| - \frac{1}{4}\ln|2| \right ) \\ \displaystyle = \left (\frac{1}{4}\ln(1) - \frac{1}{4}\ln(3) \right ) - \left (\frac{1}{4}\ln(2) - \frac{1}{4}\ln(2) \right ) \\ \displaystyle = \left (\frac{1}{4}(0) - \frac{1}{4}\ln(3) \right ) - \left (0 \right ) \\ \displaystyle = 0 - \frac{1}{4}\ln(3) - 0 {/eq}

The integral is equal to

{eq}\displaystyle = \boldsymbol{-\frac{1}{4}\ln(3)} {/eq}

**(b)**

To find the general solution of the given differential equation

{eq}\displaystyle \frac{d^2x}{dt^2}-5\frac{dx}{dt}-24x=0 {/eq}

Assume the solution to be

{eq}x(t) = e^{rt} {/eq}

The first and second derivative of this is

{eq}\displaystyle \frac{dx}{dt} = re^{rt} \\ \displaystyle \frac{d^2x}{dt^2} = r^2e^{rt} {/eq}

Substitute these to the differential equation.

{eq}\displaystyle (r^2e^{rt})-5(re^{rt})-24(e^{rt})=0 \\ \displaystyle r^2e^{rt} - 5re^{rt} - 24e^{rt} = 0 \\ \displaystyle e^{rt}(r^2 - 5r - 24) = 0 \\ \displaystyle \frac{e^{rt}(r^2 - 5r - 24)}{e^{rt}} = \frac{0}{e^{rt}} {/eq}

The characteristic equation of the differential equation is

{eq}r^2 - 5r - 24 = 0 {/eq}

The roots of this are computed using quadratic formula.

{eq}r = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ r = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-24)}}{2(1)} \\ r = \dfrac{5 \pm \sqrt{25 + 96}}{2} \\ r = \dfrac{5 \pm \sqrt{121}}{2} \\ r = \dfrac{5 \pm 11}{2} \\ r = \dfrac{5 + 11}{2}, r = \dfrac{5 - 11}{2} \\ r = \dfrac{16}{2}, r = \dfrac{-6}{2} \\ r = 8, r = -3 {/eq}

The general solution is therefore equal to

{eq}\boldsymbol{x(t) = C_1e^{8t} + C_2e^{-3t}} {/eq}

where {eq}C_1 {/eq} and {eq}C_2 {/eq} are arbitrary coefficients. We need to find these values to get the particular solution. We can get these using the given conditions.

{eq}x(0) = C_1e^{8(0)} + C_2e^{-3(0)} = 1 \\ C_1e^{0} + C_2e^{0} = 1 \\ C_1(1) + C_2(1) = 1 \\ C_1 + C_2 = 1 \\ C_1 = 1 - C_2 {/eq}

The second condition involves the first derivative of the general solution.

{eq}\displaystyle \frac{dx}{dt} = 8C_1e^{8t} - 3C_2e^{-3t} \\ \displaystyle \frac{dx}{dt}(0) = 8C_1e^{8(0)} - 3C_2e^{-3(0)} = 5 \\ \displaystyle 8C_1e^{0} - 3C_2e^{0} = 5 \\ \displaystyle 8C_1(1) - 3C_2(1) = 5 \\ \displaystyle 8C_1 - 3C_2 = 5 {/eq}

Substitute the expression that we got from the first condition.

{eq}\displaystyle 8(1 - C_2) - 3C_2 = 5 \\ \displaystyle 8 - 8C_2 - 3C_2 = 5 \\ \displaystyle - 8C_2 - 3C_2 = 5 - 8 \\ \displaystyle - 11C_2 = -3 \\ \displaystyle C_2 = -\frac{3}{-11} \\ \displaystyle C_2 = \frac{3}{11} \\ \displaystyle C_1 = 1 - C_2 = 1 - \frac{3}{11} \\ \displaystyle C_1 = \frac{11}{11} - \frac{3}{11} \\ \displaystyle C_1 = \frac{8}{11} {/eq}

Hence, the particular solution of the differential equation is

{eq}\boldsymbol{x(t) = \dfrac{8}{11}e^{8t} + \dfrac{3}{11}e^{-3t}} {/eq}