The area of the region bounded by the curve $y = | x^{3} - 4 x^{2} + 3 x |$ and the x-axis, $0 \leq x \leq 3$ , is

\frac{37}{6}

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\frac{9}{4}

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\frac{37}{12}

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Solution

The correct option is C $\frac{37}{12}$
$y = | x^{3} - 4 x^{2} + 3 x |$
$⟹ y = | x | | x - 3 | | x - 1 |$
$For 0 \leq x \leq 1; y = (x^{3} - 4 x^{2} + 3 x)$
$For 1 \leq x \leq 3; y = - (x^{3} - 4 x^{2} + 3 x)$

$A r e a = \int_{0}^{3} y d x = \int_{0}^{1} y d x + \int_{1}^{3} y d x$
$= \int_{0}^{1} (x^{3} - 4 x^{2} + 3 x) d x + \int_{1}^{3} - (x^{3} - 4 x^{2} + 3 x) d x$

$= (\frac{x^{4}}{4} - \frac{4 x^{3}}{3} + \frac{3 x^{2}}{2})_{0}^{1} + - (\frac{x^{4}}{4} - \frac{4 x^{3}}{3} + \frac{3 x^{2}}{2})_{1}^{3}$

$= \frac{5}{12} + \frac{32}{12}$

$= \frac{37}{12}$

Suggest Corrections

Similar questions

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

x

0 \leq x \leq 3

y = x^{3}, x

x = 1, x = 4

y = 9 - x^{2}

x

x = 0

x = 3

y = x^{3}

y = x + 6

x = 0

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