Area bounded by the curves $y = x^{3}$ , the $x$ -axis and the ordinates $x = - 2$ and $x = 1$ is

- 9

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- \frac{15}{4}

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\frac{15}{4}

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\frac{17}{4}

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Solution

The correct option is C $\frac{17}{4}$
Required area $= - \int_{- 2}^{0} x^{3} d x + \int_{0}^{1} x^{3} d x$
$= - {[\frac{x^{4}}{4}]}_{- 2}^{0} + {[\frac{x^{4}}{4}]}_{0}^{1}$
$= - [0 - \frac{(- 2)^{4}}{4}] + \frac{1}{4}$
$= (\frac{1}{4} + 4) = \frac{17}{4}$ sq. units
Thus, the correct answer is D.

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Area bounded by the curves y=x3, the x-axis and the ordinates x=−2 and x=1 is

The correct option is C 174Required area =−∫0−2x3dx+∫10x3dx=−[x44]0−2+[x44]10=−[0−(−2)44]+14=(14+4)=174sq. unitsThus, the correct answer is D.

Area bounded by the curves $y = x^{3}$ , the $x$ -axis and the ordinates $x = - 2$ and $x = 1$ is

The correct option is C $\frac{17}{4}$
Required area $= - \int_{- 2}^{0} x^{3} d x + \int_{0}^{1} x^{3} d x$
$= - {[\frac{x^{4}}{4}]}_{- 2}^{0} + {[\frac{x^{4}}{4}]}_{0}^{1}$
$= - [0 - \frac{(- 2)^{4}}{4}] + \frac{1}{4}$
$= (\frac{1}{4} + 4) = \frac{17}{4}$ sq. units
Thus, the correct answer is D.