What is the integral from-1 to 0 of (2x+1)^4(x+1) ?

The integral from-1 to 0 of (2x+1)^4(x+1) is 1/10

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integral from-1 to 0 of (2x+1)^4(x+1)

\int_{- 1}^{0} (2 x + 1)^{4} (x + 1) d x

\frac{1}{10}

Decimal

0.1

Solution steps

\int_{- 1}^{0} (2 x + 1)^{4} (x + 1) d x

Expand

(2 x + 1)^{4} (x + 1) : 16 x^{5} + 48 x^{4} + 56 x^{3} + 32 x^{2} + 9 x + 1

= \int_{- 1}^{0} 16 x^{5} + 48 x^{4} + 56 x^{3} + 32 x^{2} + 9 x + 1 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}^{0} 16 x^{5} d x + \int_{- 1}^{0} 48 x^{4} d x + \int_{- 1}^{0} 56 x^{3} d x + \int_{- 1}^{0} 32 x^{2} d x + \int_{- 1}^{0} 9 x d x + \int_{- 1}^{0} 1 d x

\int_{- 1}^{0} 16 x^{5} d x = - \frac{8}{3}

\int_{- 1}^{0} 48 x^{4} d x = \frac{48}{5}

\int_{- 1}^{0} 56 x^{3} d x = - 14

\int_{- 1}^{0} 32 x^{2} d x = \frac{32}{3}

\int_{- 1}^{0} 9 x d x = - \frac{9}{2}

\int_{- 1}^{0} 1 d x = 1

= - \frac{8}{3} + \frac{48}{5} - 14 + \frac{32}{3} - \frac{9}{2} + 1

Simplify

- \frac{8}{3} + \frac{48}{5} - 14 + \frac{32}{3} - \frac{9}{2} + 1 : \frac{1}{10}

= \frac{1}{10}

What is the integral from-1 to 0 of (2x+1)^4(x+1) ?
The integral from-1 to 0 of (2x+1)^4(x+1) is 1/10