Evaluate ∫1−1 ∫ − 1 1 (2x + 1)2dx

Question 1 Report

Evaluate $\int_{- 1}^{1}$ (2x + 1)2dx

\frac{2}{3}

\frac{1}{3}

\frac{1}{3}

Answer Details

$\int_{- 1}^{1}$ (2x + 1)2dx
= $\int_{- 1}^{1}$ (4x2 + 4x + 1)dx
= $\int_{- 1}^{1}$ [ $\frac{4 x^{3}}{3}$ + 2x2 + c]
= [ $\frac{4}{3}$ + 3 + c] - [4 + $\frac{1}{3}$ + c]
= $\frac{8}{3}$ + 3 + -1 - C
= $\frac{8}{3}$ + 2
= $\frac{14}{3}$
= $\frac{42}{3}$

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Integration

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