**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

The correct answer is C.

### 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 \(\frac{32}{3} = 10\frac{2}{3}\)

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## Discussion (3)

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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 \(\frac{32}{3} = 10\frac{2}{3}\)

but how did u get 31/3