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

The integral from-2 to 1 of x(4-x^2)-(x+2) is -27/4

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

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

- \frac{27}{4}

Decimal Notation

- 6.75

Solution steps

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

Expand

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

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

\int_{- 2}^{1} x^{3} d x = - \frac{15}{4}

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

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

= - (- \frac{15}{4}) - \frac{9}{2} - 6

Simplify

- (- \frac{15}{4}) - \frac{9}{2} - 6 : - \frac{27}{4}

= - \frac{27}{4}

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

n = 1 \sum \infty \frac{1}{n ( n + 5 )}

n = 0 \sum \infty \frac{i ^{n}}{3 ^{n}}

\frac{e x}{y} = 9 x - y

\int \frac{17}{x ^{3} - 125} d x

What is the integral from-2 to 1 of x(4-x^2)-(x+2) ?
The integral from-2 to 1 of x(4-x^2)-(x+2) is -27/4