Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)

  • A 3\(\frac{2}{3}\)
  • B 4
  • C 4\(\frac{1}{3}\)
  • D 4\(\frac{2}{3}\)
jamb 1994mathematics

The correct answer is D. 4\(\frac{2}{3}\)

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

= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)

= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + c]\)

= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]

= \(\frac{8}{3}\) + 3 + -1 - C

= \(\frac{8}{3}\) + 2

= \(\frac{14}{3}\)

= \(4 \frac{2}{3}\)

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