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

Assessment: JAMB UTME - Mathematics - 2019 Subject: General Mathematics

Question 1 Report

4 \frac{1}{2}

3 \frac{1}{2}

7 \frac{1}{2}

5 \frac{1}{4}

Answer Details

$\int_{- 1}^{2} (2 x^{2} + x) d x$
= $[\frac{2 x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{- 1}^{2}$
= $[\frac{2 x^{3}}{3} + \frac{x^{2}}{2}]_{- 1}^{2}$
= $(\frac{2 (2)^{3}}{3} + \frac{2^{2}}{2}) - (\frac{2 (- 1)^{3}}{3} + \frac{(- 1)^{2}}{2})$
= $(\frac{16}{3} + 2) - (\frac{- 2}{3} + \frac{1}{2})$
= $\frac{22}{3} - (- \frac{1}{6})$
= $\frac{22}{3} + \frac{1}{6}$
= $\frac{15}{2}$
= $7 \frac{1}{2}$