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WAEC Further Mathematics 2022 Paper

Question : 10

Total: 45

Evaluate

\int\limits_{-1}^{0}

(x + 1)(x - 2) dx

Solution:

Concept:

Definite integration by first expanding the product, then integrating term by term and substituting the limits.

Explanation:

First, expand the integrand:

(x + 1)(x - 2) = x^2 - x - 2

Therefore,

\int_{-1}^{0} (x + 1)(x - 2)\,dx = \int_{-1}^{0} (x^2 - x - 2)\,dx

Integrate each term:

\int (x^2 - x - 2)\,dx = \frac{x^3}{3} - \frac{x^2}{2} - 2x

Now evaluate from

-1

0

x = 0

\frac{0^3}{3} - \frac{0^2}{2} - 2(0) = 0

x = -1

\frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) = -\frac{1}{3} - \frac{1}{2} + 2

Simplify:

-\frac{1}{3} - \frac{1}{2} + 2 = \frac{-2 - 3 + 12}{6} = \frac{7}{6}

So, the definite integral is:

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

Answer:

-\frac{7}{6}

Correct option: D

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