What is the integral from-2 to 1 of (4-x^2)-(x+2) ?

The integral from-2 to 1 of (4-x^2)-(x+2) is 9/2

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integral from-2 to 1 of (4-x^2)-(x+2)

\int_{- 2}^{1} (4 - x^{2}) - (x + 2) d x

\frac{9}{2}

Decimal Notation

4.5

Solution steps

\int_{- 2}^{1} 4 - x^{2} - (x + 2) d x

Simplify

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

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

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

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

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

= - 3 - (- \frac{3}{2}) + 6

Simplify

- 3 - (- \frac{3}{2}) + 6 : \frac{9}{2}

= \frac{9}{2}

f (x) = (5 x^{2} - 8 x) e^{x}

\frac{3}{6 x + 8}

12 - x^{2}, x^{2} - 6

y = x, y = x^{2}

x \to 4 lim (\frac{x + 2}{4 x - 16})

What is the integral from-2 to 1 of (4-x^2)-(x+2) ?
The integral from-2 to 1 of (4-x^2)-(x+2) is 9/2