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

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

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

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

\frac{81}{10}

Decimal Notation

8.1

Solution steps

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

Expand

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

= \int_{- 1}^{2} 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

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

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

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

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

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

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

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

Simplify

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

= \frac{81}{10}

x \to \infty lim (\frac{x}{0})

(t^{2} + x^{2}) d t + t x d x = 0

- cos^{2} (x)

\frac{8}{x - 2}

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