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

The integral from-1 to 2 of (9-x^2)-(x+1) is 39/2

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

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

\frac{39}{2}

Decimal Notation

19.5

Solution steps

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

Simplify

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

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

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

\int_{- 1}^{2} 8 d x = 24

= - 3 - \frac{3}{2} + 24

Simplify

- 3 - \frac{3}{2} + 24 : \frac{39}{2}

= \frac{39}{2}

\frac{d}{d x} (x^{n + 1})

\frac{d}{d x} (\frac{( x - 1 )}{x})

\frac{\partial}{\partial x} (x^{3} - 3 x y)

\frac{\partial}{\partial y} (3 x + 5 y - 1)

\frac{d}{d x} (ln (x^{2} x^{2} + 1))

What is the integral from-1 to 2 of (9-x^2)-(x+1) ?
The integral from-1 to 2 of (9-x^2)-(x+1) is 39/2