Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
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Correct Answer: Option D
Explanation:
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3}Â + 2x^2 + c]\)
= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]
= \(\frac{8}{3}\) + 3 + -1 - C
= \(\frac{8}{3}\) + 2
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3}Â + 2x^2 + c]\)
= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]
= \(\frac{8}{3}\) + 3 + -1 - C
= \(\frac{8}{3}\) + 2
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)