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

The integral from-2 to 1 of pi(x^2+x-2)^2 is pi 81/10

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

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

π \frac{81}{10}

Decimal Notation

25.44690 \dots

Solution steps

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

Expand

(x^{2} + x - 2)^{2} : x^{4} + 2 x^{3} - 3 x^{2} - 4 x + 4

= π \cdot \int_{- 2}^{1} x^{4} + 2 x^{3} - 3 x^{2} - 4 x + 4 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}^{1} x^{4} d x + \int_{- 2}^{1} 2 x^{3} d x - \int_{- 2}^{1} 3 x^{2} d x - \int_{- 2}^{1} 4 x d x + \int_{- 2}^{1} 4 d x)

\int_{- 2}^{1} x^{4} d x = \frac{33}{5}

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

\int_{- 2}^{1} 3 x^{2} d x = 9

\int_{- 2}^{1} 4 x d x = - 6

\int_{- 2}^{1} 4 d x = 12

= π (\frac{33}{5} - \frac{15}{2} - 9 - (- 6) + 12)

Simplify

π (\frac{33}{5} - \frac{15}{2} - 9 - (- 6) + 12) : π \frac{81}{10}

= π \frac{81}{10}

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What is the integral from-2 to 1 of pi(x^2+x-2)^2 ?
The integral from-2 to 1 of pi(x^2+x-2)^2 is pi 81/10