The area bounded by the curve y=x24+x4-12 with X- axis in 0,2

# The area bounded by the curve $y=\frac{{x}^{2}}{4}+\frac{x}{4}-\frac{1}{2}$ with $X-$ axis in $\left[0,2\right]$

1. A

$\frac{3}{4}$

2. B

$\frac{2}{3}$

3. C

$\frac{4}{3}$

4. D

3

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### Solution:

The given curve is $y=\frac{{x}^{2}}{4}+\frac{x}{4}-\frac{1}{2}$

To find the point of intersection of the abvoe curve and $x-$axis solve the equaiton $\frac{{x}^{2}}{4}+\frac{x}{4}-\frac{1}{2}=0$

it implies that

$\begin{array}{rcl}{x}^{2}+x-2& =& 0\\ {x}^{2}+2x-x-2& =& 0\\ x\left(x+2\right)-1\left(x+2\right)& =& 0\\ \left(x-1\right)\left(x+2\right)& =& 0\end{array}$

Hence, $x=1,-2$

In the given interval $\left[0,2\right]$, the curve intersect the X- axis at $x=1$

The rough sketch of the required area is as shwon below

The required area is area of shaded region.

The area of the shaded reegion is $A={\int }_{0}^{2}ydx$

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