Question 2

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Jambmaths question: 

Find the area of the figure bounded by the given pair of curves $y={{x}^{2}}-x+3$and y =3

Option A: 

$\tfrac{7}{6}units.sq$

Option B: 

$\tfrac{1}{6}units.sq$

Option C: 

$\tfrac{17}{6}units\text{ }sq$

Option D: 

$\tfrac{5}{6}units\text{ }sq$

Jamb Maths Solution: 

$\begin{align}  & \text{Equate the two equations to obtain the boundary of the region of the area required} \\ & {{x}^{2}}-x+3=3 \\ & {{x}^{2}}-x=0 \\ & x(x-1)=0 \\ & x=0\text{ or }x=1 \\ & \text{The bounded region is }x=0,\text{ }x=1 \\ & A=\int_{0}^{1}{({{x}^{2}}-x+3)dx}-\int_{0}^{1}{3dx} \\ & A=\left[ \tfrac{{{x}^{3}}}{3}-\tfrac{{{x}^{2}}}{2}+3x \right]_{0}^{1}-\left[ 3x \right]_{0}^{1} \\ & A=\left[ (\tfrac{1}{3}-\tfrac{1}{2}+3)-(0-0+0) \right]-\left[ 3-0 \right] \\ & A=-\frac{1}{6}\text{ Ignore the negative negative, }A=\frac{1}{6} \\\end{align}$

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