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The area under the curve $x^2+3x-4$ in the positive quadrant and bounded by the line $x=5$ is equal to

1. $59 \frac{1}{6}$
2. $61 \frac{1}{3}$
3. $40 \frac{2}{3}$
4. $72$
in Geometry
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## 1 Answer

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The function  $y = x^{2} + 3x - 4$ intersects the x-axis at $\left ( 1,0 \right )$

We will take only first quadrant into consideration. So, the intersection point of the given equation and the line $x = 5$  is the point $\left ( 5,36 \right )$. (put $x = 5$ in the quadratic equation and $y$ turns out be 36).

So, the answer to this question is the area under the curve of the given equation and the line  $x = 5$  i.e. from $x = 1$  to $x = 5$ .

This can easily be calculated using integration.

$\int_{1}^{5} x^{2} + 3x - 4$   $dx$   = $\frac{368}{6}$   =  $61 \frac{1}{3}$

Option B is the correct answer.
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