What is the integral from-3 to 4 of (2x^5-3x^4+4x^3) ?

The integral from-3 to 4 of (2x^5-3x^4+4x^3) is 8057/15

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integral from-3 to 4 of (2x^5-3x^4+4x^3)

\int_{- 3}^{4} (2 x^{5} - 3 x^{4} + 4 x^{3}) d x

\frac{8057}{15}

Decimal

537.13333 \dots

Solution steps

\int_{- 3}^{4} 2 x^{5} - 3 x^{4} + 4 x^{3} 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}^{4} 2 x^{5} d x - \int_{- 3}^{4} 3 x^{4} d x + \int_{- 3}^{4} 4 x^{3} d x

\int_{- 3}^{4} 2 x^{5} d x = \frac{3367}{3}

\int_{- 3}^{4} 3 x^{4} d x = \frac{3801}{5}

\int_{- 3}^{4} 4 x^{3} d x = 175

= \frac{3367}{3} - \frac{3801}{5} + 175

Simplify

\frac{3367}{3} - \frac{3801}{5} + 175 : \frac{8057}{15}

= \frac{8057}{15}

What is the integral from-3 to 4 of (2x^5-3x^4+4x^3) ?
The integral from-3 to 4 of (2x^5-3x^4+4x^3) is 8057/15