Evaluate ₀¹ x(x²-2)² dx

Correct Answer is : 1 161\;1/6161

Correct Answer is : 1 161\;1/6161

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WAEC Further Mathematics 2023 Paper

Question : 4

Total: 40

Evaluate

\int\limits_{0}^{1} x(x^2-2)^2 dx

Solution:

Concept:

Expand the integrand, integrate term by term, and then evaluate the definite integral at the given limits.

Explanation:

First expand the square:

(x^2 - 2)^2 = x^4 - 4x^2 + 4

Multiply by

x

to rewrite the integrand:

x(x^2 - 2)^2 = x(x^4 - 4x^2 + 4) = x^5 - 4x^3 + 4x

Now integrate term by term from

0

1

\int_{0}^{1} (x^5 - 4x^3 + 4x)\, dx = \left[ \frac{x^6}{6} - x^4 + 2x^2 \right]_{0}^{1}

Evaluate at

x=1

\frac{1^6}{6} - 1^4 + 2(1)^2 = \frac{1}{6} - 1 + 2 = \frac{1}{6} + 1 = \frac{7}{6}

Evaluate at

x=0

\frac{0^6}{6} - 0^4 + 2(0)^2 = 0

Subtract to get the definite integral:

\frac{7}{6} - 0 = \frac{7}{6} = 1\frac{1}{6}

Answer:

\;1\frac{1}{6}

(i.e.

\;\frac{7}{6}

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