The area of the planar region bounded by the curve : $$ y=x^{2}+4 $$ and $$ x $$-axis and the two straight lines $$ x=-1, x=2 $$ equals $$ \ldots $$ $$ \begin{array}{llll} ext { (a) } 15 & ext { (b) } 9 & ext { (C) } 14 \frac{1}{3} & ext { (d) } 12 \frac{1}{3}\end{array} $$

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The area of the planar region bounded by the curve : $$ y=x^{2}+4 $$ and $$ x $$-axis and the two straight lines $$ x=-1, x=2 $$ equals $$ \ldots $$ $$ \begin{array}{llll}	ext { (a) } 15 & 	ext { (b) } 9 & 	ext { (C) } 14 \frac{1}{3} & 	ext { (d) } 12 \frac{1}{3}\end{array} $$

UpStudy AI · Accepted Answer

To find the area of the planar region bounded by the curve $$ y = x^2 + 4 $$, the $$ x $$-axis, and the vertical lines $$ x = -1 $$ and $$ x = 2 $$, we can set up the integral as follows:

$$
\text{Area} = \int_{-1}^{2} (x^2 + 4) \, dx
$$

Let's evaluate the integral step by step:

1. **Integrate $$ x^2 $$ from $$-1$$ to $$2$$:**
   $$
   \int_{-1}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{2} = \frac{2^3}{3} - \frac{(-1)^3}{3} = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3
   $$

2. **Integrate $$ 4 $$ from $$-1$$ to $$2$$:**
   $$
   \int_{-1}^{2} 4 \, dx = 4x \bigg|_{-1}^{2} = 4(2) - 4(-1) = 8 + 4 = 12
   $$

3. **Add the two results together:**
   $$
   \text{Total Area} = 3 + 12 = 15
   $$

Therefore, the area of the region is **15**.

**Answer:** (a) 15
The area of the region is 15. **Answer:** (a) 15

The area of the planar region bounded by the curve : and -axis
and the two straight lines equals

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The area of the planar region bounded by the curve : and -axis and the two straight lines equals