Jambmaths question:
Integrate $\int_{-1}^{2}{(2{{x}^{2}}+x)dx}$
Option A:
$4\tfrac{1}{2}$
Option B:
$3\tfrac{1}{2}$
Option C:
$7\tfrac{1}{2}$
Option D:
$5\tfrac{1}{4}$
Jamb Maths Solution:
$\begin{align} & \int_{-1}^{2}{(2{{x}^{2}}+x)dx}=\left[ \frac{2{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2} \right]_{-1}^{2} \\ & \int_{-1}^{2}{(2{{x}^{2}}+x)dx}=\left( \frac{16}{3}+\frac{4}{2} \right)-\left( \frac{-2}{3}+\frac{1}{2} \right) \\ & \int_{-1}^{2}{(2{{x}^{2}}+x)dx}=\left( \frac{32+12}{6} \right)-\left( \frac{-4+3}{6} \right) \\ & \int_{-1}^{2}{(2{{x}^{2}}+x)dx}=\frac{44}{6}+\frac{1}{6}=\frac{45}{6}=\frac{15}{2}=7\tfrac{1}{2} \\\end{align}$
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