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

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

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

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

\frac{125}{6}

Solution steps

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

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

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

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

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

Simplify

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

= \frac{125}{6}

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

\int_{0}^{\frac{π}{3}} (sin (3 x)) d x

\int_{0}^{\frac{π}{3}} cos (x) - 0.5 d x

\int_{0}^{1} 12 x^{2} (1 - x) d x

\int_{0}^{3} x^{2} - 4 x + 3 d x

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