What is the integral from-2 to 1 of (6-x^2)-(x+4) ?

The integral from-2 to 1 of (6-x^2)-(x+4) is 9/2

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integral from-2 to 1 of (6-x^2)-(x+4)

\int_{- 2}^{1} (6 - x^{2}) - (x + 4) d x

\frac{9}{2}

Decimal Notation

4.5

Solution steps

\int_{- 2}^{1} 6 - x^{2} - (x + 4) 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}

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

\int_{0}^{1} \frac{2}{3} (2 - x) d x

\int_{1}^{3} \frac{3 x ^{2}}{x ^{3} + 1} d x

\int_{1 5^{\circ}}^{3 0^{\circ}} x^{2} sin (5) x^{3} d x

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

What is the integral from-2 to 1 of (6-x^2)-(x+4) ?
The integral from-2 to 1 of (6-x^2)-(x+4) is 9/2