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

The integral from-1 to 2 of x^2(x^3+1)^2 is 81

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

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

81

Solution steps

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

Expand

x^{2} (x^{3} + 1)^{2} : x^{8} + 2 x^{5} + x^{2}

= \int_{- 1}^{2} x^{8} + 2 x^{5} + x^{2} 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^{8} d x + \int_{- 1}^{2} 2 x^{5} d x + \int_{- 1}^{2} x^{2} d x

\int_{- 1}^{2} x^{8} d x = 57

\int_{- 1}^{2} 2 x^{5} d x = 21

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

= 57 + 21 + 3

Simplify

= 81

\frac{d y}{d x} = 100 x (y + 2)

y = 3 e^{x} cos (x), (0, 3)

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0

y^{^{'}} = \frac{y}{x} + 2 x + 1

f (x) = \frac{1}{2} x^{8}

What is the integral from-1 to 2 of x^2(x^3+1)^2 ?
The integral from-1 to 2 of x^2(x^3+1)^2 is 81