The area of the region bounded by $y = x + 6$ , $y = x^{3}$ in sq units is

10

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12

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\frac{1}{13}

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\frac{1}{12}

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Solution

The correct option is A $10$
$\begin{matrix} W e d r a w t h e f i g u r e y = x^{3} a n d y = x + 6 T h e r e q u i r e d a r e a A = \int_{0}^{2} (y_{2} - y_{1}) d x . A = \int_{0}^{2} (x + 6) d x - \int_{0}^{2} x^{3} d x . A = {[\frac{x^{2}}{2} + 6 x]}_{0}^{2} - {[\frac{x^{4}}{4}]}_{0}^{2} A = [\frac{2^{2}}{2} - 0 + 6 \times 2 - 0] = [14 - 4] = 10 T h e r e q u i r e d a r e a = 10 s q . u n i t s H e n c e, t h e o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$

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The area between the curves y = x² and y = x³ is:

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