JAMB - Mathematics (1994 - No. 34)
Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
3\(\frac{2}{3}\)
4
4\(\frac{1}{3}\)
4\(\frac{2}{3}\)
Explanation
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)
= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]
= \(\frac{13}{3}\) + \(\frac{1}{3}\)
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)
= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]
= \(\frac{13}{3}\) + \(\frac{1}{3}\)
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
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