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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

- $59 \frac{1}{6}$
- $61 \frac{1}{3}$
- $40 \frac{2}{3}$
- $72$

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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.

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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