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

The integral from-1 to 4 of (3x^4-4x^3-x-4) is 665/2

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

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

\frac{665}{2}

Decimal

332.5

Solution steps

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

\int_{- 1}^{4} 3 x^{4} d x = 615

\int_{- 1}^{4} 4 x^{3} d x = 255

\int_{- 1}^{4} x d x = \frac{15}{2}

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

= 615 - 255 - \frac{15}{2} - 20

Simplify

615 - 255 - \frac{15}{2} - 20 : \frac{665}{2}

= \frac{665}{2}

What is the integral from-1 to 4 of (3x^4-4x^3-x-4) ?
The integral from-1 to 4 of (3x^4-4x^3-x-4) is 665/2