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

The integral from-1 to 2 of (x^3+1)^2 is 405/14

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

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

\frac{405}{14}

Decimal

28.92857 \dots

Solution steps

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

Expand

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

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

\int_{- 1}^{2} x^{6} d x = \frac{129}{7}

\int_{- 1}^{2} 2 x^{3} d x = \frac{15}{2}

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

= \frac{129}{7} + \frac{15}{2} + 3

Simplify

\frac{129}{7} + \frac{15}{2} + 3 : \frac{405}{14}

= \frac{405}{14}

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What is the integral from-1 to 2 of (x^3+1)^2 ?
The integral from-1 to 2 of (x^3+1)^2 is 405/14