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WAEC Further Mathematics 2010 Paper

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Question : 14

Total: 40

Evaluate

\int\limits_{1}^{2} \frac{4}{x^{3}} \, dx

$-1\frac{1}{2}$
$-\frac{17}{16}$
$\frac{15}{16}$
$1\frac{1}{2}$

Solution:

Concept:

Use the power rule of integration after rewriting

\frac{4}{x^3}

as

4x^{-3}

.

Explanation:

Rewrite the integrand:

\int_{1}^{2} \frac{4}{x^3}\,dx = \int_{1}^{2} 4x^{-3}\,dx

Integrate using the power rule,

\int x^n\,dx = \frac{x^{n+1}}{n+1}

for

n \neq -1

:

4 \cdot \frac{x^{-3+1}}{-3+1} = 4 \cdot \frac{x^{-2}}{-2} = -2x^{-2}

Write this as

-\frac{2}{x^2}

and apply the limits

1

to

2

:

\left[-\frac{2}{x^2}\right]_{1}^{2} = -\frac{2}{2^2} - \left(-\frac{2}{1^2}\right)

= -\frac{2}{4} + 2 = -\frac{1}{2} + 2 = 1\frac{1}{2}

Answer:

The value of the integral is

1\frac{1}{2}

, which is option D.

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