What is the integral from 0 to 2 of (x(64-(x+3)^3)) ?

The integral from 0 to 2 of (x(64-(x+3)^3)) is -202/5

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integral from 0 to 2 of (x(64-(x+3)^3))

\int_{0}^{2} (x (64 - (x + 3)^{3})) d x

- \frac{202}{5}

Decimal Notation

- 40.4

Solution steps

\int_{0}^{2} x (64 - (x + 3)^{3}) d x

Expand

x (64 - (x + 3)^{3}) : - x^{4} - 9 x^{3} - 27 x^{2} + 37 x

= \int_{0}^{2} - x^{4} - 9 x^{3} - 27 x^{2} + 37 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{2} x^{4} d x - \int_{0}^{2} 9 x^{3} d x - \int_{0}^{2} 27 x^{2} d x + \int_{0}^{2} 37 x d x

\int_{0}^{2} x^{4} d x = \frac{32}{5}

\int_{0}^{2} 9 x^{3} d x = 36

\int_{0}^{2} 27 x^{2} d x = 72

\int_{0}^{2} 37 x d x = 74

= - \frac{32}{5} - 36 - 72 + 74

Simplify

- \frac{32}{5} - 36 - 72 + 74 : - \frac{202}{5}

= - \frac{202}{5}

\frac{d}{d x} ((x^{2} - x)^{3})

\frac{d}{d x} (\frac{1}{3} (x^{2} + 2)^{\frac{3}{2}} - 5)

\frac{d}{d x} (\frac{1}{sec ^{2} ( x )})

f (x) = \frac{2}{x + 1}, at x = 7

x \to 0 + lim (\frac{12}{x ^{2}})

What is the integral from 0 to 2 of (x(64-(x+3)^3)) ?
The integral from 0 to 2 of (x(64-(x+3)^3)) is -202/5