What is the integral from-1 to 0 of x^3(4x^4-2)^4 ?

The integral from-1 to 0 of x^3(4x^4-2)^4 is -4/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 0 of x^3(4x^4-2)^4

\int_{- 1}^{0} x^{3} (4 x^{4} - 2)^{4} d x

- \frac{4}{5}

Decimal Notation

- 0.8

Solution steps

\int_{- 1}^{0} x^{3} (4 x^{4} - 2)^{4} d x

Expand

x^{3} (4 x^{4} - 2)^{4} : 256 x^{19} - 512 x^{15} + 384 x^{11} - 128 x^{7} + 16 x^{3}

= \int_{- 1}^{0} 256 x^{19} - 512 x^{15} + 384 x^{11} - 128 x^{7} + 16 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_{- 1}^{0} 256 x^{19} d x - \int_{- 1}^{0} 512 x^{15} d x + \int_{- 1}^{0} 384 x^{11} d x - \int_{- 1}^{0} 128 x^{7} d x + \int_{- 1}^{0} 16 x^{3} d x

\int_{- 1}^{0} 256 x^{19} d x = - \frac{64}{5}

\int_{- 1}^{0} 512 x^{15} d x = - 32

\int_{- 1}^{0} 384 x^{11} d x = - 32

\int_{- 1}^{0} 128 x^{7} d x = - 16

\int_{- 1}^{0} 16 x^{3} d x = - 4

= - \frac{64}{5} - (- 32) - 32 - (- 16) - 4

Simplify

- \frac{64}{5} - (- 32) - 32 - (- 16) - 4 : - \frac{4}{5}

= - \frac{4}{5}

\int_{1}^{6 x} (31 + t^{2}) d t

\int_{0}^{2} 30 x d x

\int_{0}^{1} \frac{7 x}{( 2 x + 1 ) ^{3}} d x

\int_{0}^{t} (t - x)^{3} e^{x} d x

\int_{5}^{\infty} e^{- 4 x} d x

What is the integral from-1 to 0 of x^3(4x^4-2)^4 ?
The integral from-1 to 0 of x^3(4x^4-2)^4 is -4/5