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

The integral from 0 to 1 of x^2(4x^3-1)^4 is 61/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{61}{15}

Decimal

4.06666 \dots

Solution steps

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

Expand

x^{2} (4 x^{3} - 1)^{4} : 256 x^{14} - 256 x^{11} + 96 x^{8} - 16 x^{5} + x^{2}

= \int_{0}^{1} 256 x^{14} - 256 x^{11} + 96 x^{8} - 16 x^{5} + x^{2} 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}^{1} 256 x^{14} d x - \int_{0}^{1} 256 x^{11} d x + \int_{0}^{1} 96 x^{8} d x - \int_{0}^{1} 16 x^{5} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} 256 x^{14} d x = \frac{256}{15}

\int_{0}^{1} 256 x^{11} d x = \frac{64}{3}

\int_{0}^{1} 96 x^{8} d x = \frac{32}{3}

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

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

= \frac{256}{15} - \frac{64}{3} + \frac{32}{3} - \frac{8}{3} + \frac{1}{3}

Simplify

\frac{256}{15} - \frac{64}{3} + \frac{32}{3} - \frac{8}{3} + \frac{1}{3} : \frac{61}{15}

= \frac{61}{15}

\int_{0}^{1} y (y^{2} + 1)^{5} d y

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

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

\int_{- 1}^{1} \frac{1}{4 x ^{2} - 9} d x

\int_{3}^{8} \frac{x}{1 + x} d x

What is the integral from 0 to 1 of x^2(4x^3-1)^4 ?
The integral from 0 to 1 of x^2(4x^3-1)^4 is 61/15