What is the integral from 4 to 0 of (2x^5+3x^4-2x^3) ?

The integral from 4 to 0 of (2x^5+3x^4-2x^3) is -27776/15

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integral from 4 to 0 of (2x^5+3x^4-2x^3)

\int_{4}^{0} (2 x^{5} + 3 x^{4} - 2 x^{3}) d x

- \frac{27776}{15}

Decimal

- 1851.73333 \dots

Solution steps

\int_{4}^{0} 2 x^{5} + 3 x^{4} - 2 x^{3} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{4} 2 x^{5} + 3 x^{4} - 2 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_{0}^{4} 2 x^{5} d x + \int_{0}^{4} 3 x^{4} d x - \int_{0}^{4} 2 x^{3} d x)

\int_{0}^{4} 2 x^{5} d x = \frac{4096}{3}

\int_{0}^{4} 3 x^{4} d x = \frac{3072}{5}

\int_{0}^{4} 2 x^{3} d x = 128

= - (\frac{4096}{3} + \frac{3072}{5} - 128)

Join

\frac{4096}{3} + \frac{3072}{5} - 128 : \frac{27776}{15}

= - \frac{27776}{15}

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\int_{1}^{2} (\frac{4}{x} - e^{- x}) d x

What is the integral from 4 to 0 of (2x^5+3x^4-2x^3) ?
The integral from 4 to 0 of (2x^5+3x^4-2x^3) is -27776/15