Evaluate \(\int_{-1}^{0} (x+1)(x-2) \mathrm{d}x\)
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Correct Answer: Option D
Explanation:
Expanding \((x+1)(x-2) = x^{2} - 2x + x - 2 = x^{2} - x - 2\)
\(\int_{-1}^{0} (x^{2} - x - 2) \mathrm{d}x = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x]_{-1}^{0}\)
= \([\frac{0}{3} - \frac{0}{2} - 2\times0 - (\frac{-1^{3}}{3} - \frac{-1^{2}}{2} - 2\times-1)]\)
= \(0 + \frac{1}{3} + \frac{1}{2} - 2 = \frac{-7}{6}\)
Note: This can also be solved using integration by parts.
\(\int uv \mathrm{d}x = u\int v \mathrm{d}x - \int u'(\int v \mathrm{d}x)\mathrm{d}x\).
Expanding \((x+1)(x-2) = x^{2} - 2x + x - 2 = x^{2} - x - 2\)
\(\int_{-1}^{0} (x^{2} - x - 2) \mathrm{d}x = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x]_{-1}^{0}\)
= \([\frac{0}{3} - \frac{0}{2} - 2\times0 - (\frac{-1^{3}}{3} - \frac{-1^{2}}{2} - 2\times-1)]\)
= \(0 + \frac{1}{3} + \frac{1}{2} - 2 = \frac{-7}{6}\)
Note: This can also be solved using integration by parts.
\(\int uv \mathrm{d}x = u\int v \mathrm{d}x - \int u'(\int v \mathrm{d}x)\mathrm{d}x\).