integral de-4 a 1 de-x^2-3x+4

Solução

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

\frac{125}{6}

Passos da solução

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

Aplicar a regra da soma:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

Simplificar

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

= \frac{125}{6}

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