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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 \(\frac{31}{3} = 10\frac{1}{3}\)


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Dicussion (1)

  • \(y = 4 - x^2\) 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 \(\frac{31}{3} = 10\frac{1}{3}\)

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