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

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

A.

3 $\frac{2}{3}$

B.

4

C.

4 $\frac{1}{3}$

D.

4 $\frac{2}{3}$

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Correct answer is D

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

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

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

= [ $\frac{4}{3}$ + 2 + 1] - [ $\frac{-4}{3}$ +2 -1]

= $\frac{13}{3}$ + $\frac{1}{3}$

= $\frac{14}{3}$

= $4 \frac{2}{3}$

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