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

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

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

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

\frac{125}{6}

Decimal

20.83333 \dots

Solution steps

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

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

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

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

= \frac{45}{2} + 20 - \frac{65}{3}

Simplify

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

= \frac{125}{6}

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