Question 39

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Jambmaths question: 

Evaluate $\int\limits_{0}^{1}{{{(2x+1)}^{2}}dx}$

Option A: 

$4\tfrac{2}{3}$

Option B: 

$4\tfrac{1}{3}$

Option C: 

$3\tfrac{2}{3}$

Option D: 

$3\tfrac{1}{2}$

Jamb Maths Solution: 

$\begin{align}  & \int\limits_{0}^{1}{{{(2x+1)}^{2}}dx} \\ & \text{Using change of variable method} \\ & \text{Let }u=2x+1 \\ & \frac{du}{dx}=2,\text{ }dx=\frac{du}{2} \\ & \int\limits_{0}^{1}{{{(2x+1)}^{2}}dx}=\int\limits_{0}^{1}{{{(u)}^{2}}\tfrac{du}{2}}=\frac{1}{2}\int\limits_{0}^{1}{{{(u)}^{2}}du} \\ & \int\limits_{0}^{1}{{{(2x+1)}^{2}}dx}=\frac{1}{2}\left[ \frac{{{u}^{3}}}{3} \right]_{0}^{1}=\frac{1}{6}\left[ {{u}^{3}} \right]_{0}^{1} \\ & \int\limits_{0}^{1}{{{(2x+1)}^{2}}dx}=\frac{1}{6}\left[ {{(2x+1)}^{3}} \right]_{0}^{1} \\ & \int\limits_{0}^{1}{{{(2x+1)}^{2}}dx}=\frac{1}{6}\left[ {{(2+1)}^{3}}-{{(0+1)}^{3}} \right] \\ & \int\limits_{0}^{1}{{{(2x+1)}^{2}}dx}=\frac{26}{6}=\frac{13}{3}=4\tfrac{1}{3} \\\end{align}$

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