What is the integral from-2 to 0 of pi(2x^2+4x+3)^2 ?

The integral from-2 to 0 of pi(2x^2+4x+3)^2 is pi 94/15

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integral from-2 to 0 of pi(2x^2+4x+3)^2

\int_{- 2}^{0} π (2 x^{2} + 4 x + 3)^{2} d x

π \frac{94}{15}

Decimal Notation

19.68731 \dots

Solution steps

\int_{- 2}^{0} π (2 x^{2} + 4 x + 3)^{2} d x

Take the constant out:

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

= π \cdot \int_{- 2}^{0} (2 x^{2} + 4 x + 3)^{2} d x

Expand

(2 x^{2} + 4 x + 3)^{2} : 4 x^{4} + 16 x^{3} + 28 x^{2} + 24 x + 9

= π \cdot \int_{- 2}^{0} 4 x^{4} + 16 x^{3} + 28 x^{2} + 24 x + 9 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

= π (\int_{- 2}^{0} 4 x^{4} d x + \int_{- 2}^{0} 16 x^{3} d x + \int_{- 2}^{0} 28 x^{2} d x + \int_{- 2}^{0} 24 x d x + \int_{- 2}^{0} 9 d x)

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

\int_{- 2}^{0} 16 x^{3} d x = - 64

\int_{- 2}^{0} 28 x^{2} d x = \frac{224}{3}

\int_{- 2}^{0} 24 x d x = - 48

\int_{- 2}^{0} 9 d x = 18

= π (\frac{128}{5} - 64 + \frac{224}{3} - 48 + 18)

Simplify

π (\frac{128}{5} - 64 + \frac{224}{3} - 48 + 18) : π \frac{94}{15}

= π \frac{94}{15}

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What is the integral from-2 to 0 of pi(2x^2+4x+3)^2 ?
The integral from-2 to 0 of pi(2x^2+4x+3)^2 is pi 94/15