Area bounded by curve $y = x^{3}, x - a x i s$ axis and ordinates x=1 and x=4, is

$64 s q . u n i t$

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$27 s q . u n i t$

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\frac{127}{4} s q . u n i t

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$\frac{255}{4} s q . u n i t$

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Solution

The correct option is D
$\frac{255}{4} s q . u n i t$

$R e q u i r e d a r e a = \int_{1}^{4} x^{3} d x = {[\frac{x^{4}}{4}]}_{1}^{4} = \frac{255}{4} s q . u n i t .$

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Similar questions

y = x^{2} + x + 4

x = 1

x = 3

y = x^{3}, x

x = 1, x = 4

y = x^{3}

x = - 2

x = 1

Area bounded by curve $y = x^{3}, x - a x i s$ axis and ordinates x=1 and x=4, is

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