Home » » Find the area bounded by the curves $$y = 4 - x^2$$ and $$y = 2x + 1$$

# Find the area bounded by the curves $$y = 4 - x^2$$ and $$y = 2x + 1$$

### Question

Find the area bounded by the curves $$y = 4 - x^2$$ and $$y = 2x + 1$$

### Options

A) $$20\frac{1}{3}$$ sq. units

B) $$20\frac{2}{3}$$ sq. units

C) $$10\frac{2}{3}$$ sq. units

D) $$10\frac{1}{3}$$ sq. units

### Explanation:

$$y = 4 - x^2$$ and $$y = 2x + 1$$
$$\Rightarrow 4 - x^2 = 2x + 1$$
$$\Rightarrow x^2 + 2x - 3 = 0$$
$$(x+3)(x-1) = 0$$
Thus, $$x = -3$$ or $$x = 1.$$

Integrating $$x^2 + 2x - 3$$ from $$(-3$$ to $$1)$$ with respect to $$x$$ will give $$\frac{32}{3} = 10\frac{2}{3}$$

## Discussion (6)

• Olajubu Timileyin

I don't understand how you work it sir

• It's c

• How did you get 32/3=10⅔

• Its $$\cfrac{32}{3}$$ sq units not $$\cfrac{31}{3}$$

• $$y = 4 - x^2$$ and $$y = 2x + 1$$
$$\Rightarrow 4 - x^2 = 2x + 1$$
$$\Rightarrow x^2 + 2x - 3 = 0$$
$$(x+3)(x-1) = 0$$
Thus, $$x = -3$$ or $$x = 1.$$

Integrating $$x^2 + 2x - 3$$ from $$(-3$$ to $$1)$$ with respect to $$x$$ will give $$\frac{32}{3} = 10\frac{2}{3}$$

1. Babatunde

but how did u get 31/3