What is the integral from 1 to 2 of (4x^3-(x^4)/5) ?

The integral from 1 to 2 of (4x^3-(x^4)/5) is 344/25

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integral from 1 to 2 of (4x^3-(x^4)/5)

\int_{1}^{2} (4 x^{3} - \frac{x ^{4}}{5}) d x

\frac{344}{25}

Decimal

13.76

Solution steps

\int_{1}^{2} 4 x^{3} - \frac{x ^{4}}{5} 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}^{2} 4 x^{3} d x - \int_{1}^{2} \frac{x ^{4}}{5} d x

\int_{1}^{2} 4 x^{3} d x = 15

\int_{1}^{2} \frac{x ^{4}}{5} d x = \frac{31}{25}

= 15 - \frac{31}{25}

Simplify

15 - \frac{31}{25} : \frac{344}{25}

= \frac{344}{25}

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What is the integral from 1 to 2 of (4x^3-(x^4)/5) ?
The integral from 1 to 2 of (4x^3-(x^4)/5) is 344/25