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4. (a) Express $$\frac{25}{x^2(2x + 1)}$$ in partial fractions - Edexcel - A-Level Maths: Pure - Question 7 - 2014 - Paper 8

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4. (a) Express $$\frac{25}{x^2(2x + 1)}$$ in partial fractions. (b) Use calculus to find the exact volume of the solid of revolution generated, giving your ans... show full transcript

Worked Solution & Example Answer:4. (a) Express $$\frac{25}{x^2(2x + 1)}$$ in partial fractions - Edexcel - A-Level Maths: Pure - Question 7 - 2014 - Paper 8

Step 1

Express \(\frac{25}{x^2(2x + 1)}\) in partial fractions.

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Answer

To express the given fraction in partial fractions, we can assume it takes the following form:

25x2(2x+1)=Ax+Bx2+C2x+1\frac{25}{x^2(2x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x + 1}

Next, we multiply through by the denominator, resulting in:

25=Ax(2x+1)+B(2x+1)+Cx225 = A x (2x + 1) + B (2x + 1) + C x^2

Expanding both sides gives:

25=A(2x2+x)+B(2x+1)+Cx225 = A(2x^2 + x) + B(2x + 1) + C x^2

To find the coefficients, we need to collect terms:

25=(2A+C)x2+(A+2B)x+B25 = (2A + C)x^2 + (A + 2B)x + B

Now, equate coefficients from both sides:

  • For the constant term:
    • B=25B = 25
  • For the coefficient of (x):
    • A+2B=0A+50=0A=50A + 2B = 0 \Rightarrow A + 50 = 0 \Rightarrow A = -50
  • For the coefficient of (x^2):
    • 2A+C=02(50)+C=0C=1002A + C = 0 \Rightarrow 2(-50) + C = 0 \Rightarrow C = 100

Thus, the partial fraction decomposition is:

25x2(2x+1)=50x+25x2+1002x+1\frac{25}{x^2(2x + 1)} = \frac{-50}{x} + \frac{25}{x^2} + \frac{100}{2x + 1}

Step 2

Use calculus to find the exact volume of the solid of revolution generated.

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Answer

The volume (V) of the solid generated by rotating region (R) around the x-axis can be computed using the formula:

V=14π[f(x)]2dxV = \int_{1}^{4} \pi [f(x)]^2 \,dx

Here, the function defined by the curve (C) is:

f(x)=52x+1f(x) = \frac{5}{\sqrt{2x + 1}}

Thus, we have:

V=π14(52x+1)2dxV = \pi \int_{1}^{4} \left(\frac{5}{\sqrt{2x + 1}}\right)^2 \,dx

This simplifies to:

V=π14252x+1dxV = \pi \int_{1}^{4} \frac{25}{2x + 1} \,dx

Computing the integral, we can separate the terms:

V=25π1412x+1dxV = 25\pi \int_{1}^{4} \frac{1}{2x + 1} \,dx

Using the substitution (u = 2x + 1) yields:

du=2dxdx=du2du = 2 \,dx \Rightarrow dx = \frac{du}{2}

Changing the limits accordingly, when (x=1), (u=3) and when (x=4), (u=9). Thus, the entry becomes:

V=25π2391uduV = \frac{25\pi}{2} \int_{3}^{9} \frac{1}{u} \,du

Evaluating the integral gives:

V=25π2[lnu]39=25π2(ln(9)ln(3))=25π2ln(3)V = \frac{25\pi}{2} \left[\ln|u|\right]_{3}^{9} = \frac{25\pi}{2} (\ln(9) - \ln(3)) = \frac{25\pi}{2} \ln(3)

Finally, we can express the volume in the required form:

75π2(31)\frac{75\pi}{2} \ell(\frac{3}{1})
where:
a = 0, b = \frac{75\pi}{2}, c = 3.

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