What is the integral from-2 to 0 of 9x^2(2x^3+1)^2 ?

The integral from-2 to 0 of 9x^2(2x^3+1)^2 is 1688

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integral from-2 to 0 of 9x^2(2x^3+1)^2

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 1)^{2} d x

1688

Solution steps

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 1)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{- 2}^{0} x^{2} (2 x^{3} + 1)^{2} d x

Expand

x^{2} (2 x^{3} + 1)^{2} : 4 x^{8} + 4 x^{5} + x^{2}

= 9 \cdot \int_{- 2}^{0} 4 x^{8} + 4 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

= 9 (\int_{- 2}^{0} 4 x^{8} d x + \int_{- 2}^{0} 4 x^{5} d x + \int_{- 2}^{0} x^{2} d x)

\int_{- 2}^{0} 4 x^{8} d x = \frac{2048}{9}

\int_{- 2}^{0} 4 x^{5} d x = - \frac{128}{3}

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

= 9 (\frac{2048}{9} - \frac{128}{3} + \frac{8}{3})

Simplify

9 (\frac{2048}{9} - \frac{128}{3} + \frac{8}{3}) : 1688

= 1688

\int_{- 3}^{1} ∣ x + 2 ∣ d x

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

\int_{0}^{2} 2 x 4 - x^{2} d x

\int_{1}^{e} ln (y) d y

\int_{0}^{2} e^{x} - \frac{1}{x ^{2} + 1} d x

What is the integral from-2 to 0 of 9x^2(2x^3+1)^2 ?
The integral from-2 to 0 of 9x^2(2x^3+1)^2 is 1688