Find the area under the curve $y = (x^{2} + 2)^{2} + 2 x$ between the lines $x = 0;$ $x = 2$ and the X-axis

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Solution

$y = x^{4} + 4 x^{2} + 2 x + 4$
$\frac{d y}{d x} = 4 x^{3} + 8 x + 2$
$A = \int_{0}^{2} 2 x + 4 + 4 x^{2} + x^{4} d x A = {[\frac{2 x^{2}}{2} + 4 x + \frac{4 x^{3}}{3} + \frac{x^{5}}{5}]}_{0}^{2} A = 4 + 8 + \frac{32}{3} + \frac{32}{5} A = \frac{432}{15} = 29.06$

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Similar questions

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

x

x = - 3

x = 5

y^{2} = 4 a x

X

x = 0

x = a

y = 2^{x}

y = 2 x - x^{2}

x = 0, x = 2

y = 4 - x^{2}

x = 0

x = 2

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