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

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}

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

\int_{4}^{10} \frac{x}{4} - \frac{4}{x} d x

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

\int_{0}^{2} \frac{32}{x ^{2} + 4} d x

\int_{1}^{3} \frac{y}{( y + 1 ) ( y ^{2} + 2 )} d y

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