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(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}


Learn more about this topic:

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How to Integrate Functions With Partial Fractions

from Math 104: Calculus

Chapter 13 / Lesson 9
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