What is the integral from 0 to 2 of (x^3-1)^2 ?

The integral from 0 to 2 of (x^3-1)^2 is 86/7

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integral from 0 to 2 of (x^3-1)^2

\int_{0}^{2} (x^{3} - 1)^{2} d x

\frac{86}{7}

Decimal

12.28571 \dots

Solution steps

\int_{0}^{2} (x^{3} - 1)^{2} d x

Expand

(x^{3} - 1)^{2} : x^{6} - 2 x^{3} + 1

= \int_{0}^{2} x^{6} - 2 x^{3} + 1 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^{6} d x - \int_{0}^{2} 2 x^{3} d x + \int_{0}^{2} 1 d x

\int_{0}^{2} x^{6} d x = \frac{128}{7}

\int_{0}^{2} 2 x^{3} d x = 8

\int_{0}^{2} 1 d x = 2

= \frac{128}{7} - 8 + 2

Simplify

\frac{128}{7} - 8 + 2 : \frac{86}{7}

= \frac{86}{7}

y^{^{'}} = y - 2 x - 1

x + 2 y x^{2} + 1 y^{^{'}} = 0

\int_{0}^{0.5} f (2 t) d t

x \to - 1 lim (x^{2} - x + 1)

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

What is the integral from 0 to 2 of (x^3-1)^2 ?
The integral from 0 to 2 of (x^3-1)^2 is 86/7