What is the integral from-3 to 1 of (x^2-2)-(2x+1) ?

The integral from-3 to 1 of (x^2-2)-(2x+1) is 16/3

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integral from-3 to 1 of (x^2-2)-(2x+1)

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Simplify

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

\int_{- 3}^{1} x^{2} d x = \frac{28}{3}

\int_{- 3}^{1} 2 x d x = - 8

\int_{- 3}^{1} 3 d x = 12

= \frac{28}{3} - (- 8) - 12

Simplify

\frac{28}{3} - (- 8) - 12 : \frac{16}{3}

= \frac{16}{3}

What is the integral from-3 to 1 of (x^2-2)-(2x+1) ?
The integral from-3 to 1 of (x^2-2)-(2x+1) is 16/3