The area bounded by the $x$ -axis and the curve $y = 4 x - x^{2} - 3$ is

\frac{1}{3}

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\frac{2}{3}

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\frac{4}{3}

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\frac{8}{3}

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Solution

The correct option is C $\frac{4}{3}$
Graph of $y = f (x)$ is shown in above figure
so area will be equal to $\int_{1}^{3} ({- x}^{2} + 4 x - 3) d x = {[\frac{- x^{3}}{3} + 2 x^{2} - 3 x]}_{1}^{3} = [- 9 + 18 - 9] - [\frac{- 1}{3} + 2 - 3] = 0 - \frac{- 4}{3} = \frac{4}{3}$
so option C is correct

Suggest Corrections

Similar questions

y = 4 x - x^{2} - 3

y = x^{2} + x + 4

x = 1

x = 3

X

y = 4 x - x^{2} - 3

List-I	List-II
1. Area of region bounded by $y = 2 x - x^{2}$ and $x -$ axis	a. $\frac{1}{3}$
2. Area of the region ${(x, y) : x^{2} \leq y \leq \| x \|}$	b. $\frac{1}{2}$
3. Area bounded by $y = x$ and $y = x^{3}$	c. $\frac{2}{3}$
4. Area bounded by $y = x \| x \|$ , $x$ -axis and $x = - 1, x = 1$	d. $\frac{4}{3}$

1234

\frac{1}{3}

\frac{2}{3}

\frac{4}{3}

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