What is the integral from-1 to 0 of y^9(5y^{10}-4)^3 ?

The integral from-1 to 0 of y^9(5y^{10}-4)^3 is 51/40

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integral from-1 to 0 of y^9(5y^{10}-4)^3

\int_{- 1}^{0} y^{9} (5 y^{10} - 4)^{3} d y

\frac{51}{40}

Decimal Notation

1.275

Solution steps

\int_{- 1}^{0} y^{9} (5 y^{10} - 4)^{3} d y

Expand

y^{9} (5 y^{10} - 4)^{3} : 125 y^{39} - 300 y^{29} + 240 y^{19} - 64 y^{9}

= \int_{- 1}^{0} 125 y^{39} - 300 y^{29} + 240 y^{19} - 64 y^{9} d y

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}^{0} 125 y^{39} d y - \int_{- 1}^{0} 300 y^{29} d y + \int_{- 1}^{0} 240 y^{19} d y - \int_{- 1}^{0} 64 y^{9} d y

\int_{- 1}^{0} 125 y^{39} d y = - \frac{25}{8}

\int_{- 1}^{0} 300 y^{29} d y = - 10

\int_{- 1}^{0} 240 y^{19} d y = - 12

\int_{- 1}^{0} 64 y^{9} d y = - \frac{32}{5}

= - \frac{25}{8} - (- 10) - 12 - (- \frac{32}{5})

Simplify

- \frac{25}{8} - (- 10) - 12 - (- \frac{32}{5}) : \frac{51}{40}

= \frac{51}{40}

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What is the integral from-1 to 0 of y^9(5y^{10}-4)^3 ?
The integral from-1 to 0 of y^9(5y^{10}-4)^3 is 51/40