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WAEC Further Mathematics 2018 Paper

Question : 29

Total: 40

Evaluate

\int\limits_{-1}^{1} (x + 1)^{2} d x

Solution:

Concept:

Definite integrals are evaluated by finding the antiderivative and applying the limits.

Explanation:

First, expand the integrand:

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

So,

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

Integrating term by term gives:

\left[\frac{x^3}{3} + x^2 + x\right]_{-1}^{1}

At the upper limit

x=1

\frac{1^3}{3} + 1^2 + 1 = \frac{1}{3} + 1 + 1 = \frac{7}{3}

At the lower limit

x=-1

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

Subtract the lower value from the upper value:

\frac{7}{3} - \left(-\frac{1}{3}\right) = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}

Answer:

\frac{8}{3}

(Option A).

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