What is the integral from-3 to 5 of (1-x-4x^3+3x^4) ?

The integral from-3 to 5 of (1-x-4x^3+3x^4) is 7384/5

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integral from-3 to 5 of (1-x-4x^3+3x^4)

\int_{- 3}^{5} (1 - x - 4 x^{3} + 3 x^{4}) d x

\frac{7384}{5}

Decimal Notation

1476.8

Solution steps

\int_{- 3}^{5} 1 - x - 4 x^{3} + 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_{- 3}^{5} 1 d x - \int_{- 3}^{5} x d x - \int_{- 3}^{5} 4 x^{3} d x + \int_{- 3}^{5} 3 x^{4} d x

\int_{- 3}^{5} 1 d x = 8

\int_{- 3}^{5} x d x = 8

\int_{- 3}^{5} 4 x^{3} d x = 544

\int_{- 3}^{5} 3 x^{4} d x = \frac{10104}{5}

= 8 - 8 - 544 + \frac{10104}{5}

Simplify

8 - 8 - 544 + \frac{10104}{5} : \frac{7384}{5}

= \frac{7384}{5}

\int_{- 3}^{5} (x^{3} - 4 x) d x

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

\int_{4}^{8} f (t) d t

\int_{- \frac{π}{2}}^{\frac{π}{2}} 4 cos (t) d t

\int_{0}^{1} e^{x} \cdot x d x

What is the integral from-3 to 5 of (1-x-4x^3+3x^4) ?
The integral from-3 to 5 of (1-x-4x^3+3x^4) is 7384/5