What is the integral from-4 to 1 of (4-x^2-3x) ?

The integral from-4 to 1 of (4-x^2-3x) is 125/6

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integral from-4 to 1 of (4-x^2-3x)

\int_{- 4}^{1} (4 - x^{2} - 3 x) d x

\frac{125}{6}

Solution steps

\int_{- 4}^{1} 4 - x^{2} - 3 x 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_{- 4}^{1} 4 d x - \int_{- 4}^{1} x^{2} d x - \int_{- 4}^{1} 3 x d x

\int_{- 4}^{1} 4 d x = 20

\int_{- 4}^{1} x^{2} d x = \frac{65}{3}

\int_{- 4}^{1} 3 x d x = - \frac{45}{2}

= 20 - \frac{65}{3} - (- \frac{45}{2})

Simplify

20 - \frac{65}{3} - (- \frac{45}{2}) : \frac{125}{6}

= \frac{125}{6}

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What is the integral from-4 to 1 of (4-x^2-3x) ?
The integral from-4 to 1 of (4-x^2-3x) is 125/6