What is the integral from 0 to 1 of x^6(2+2x^7)^4 ?

The integral from 0 to 1 of x^6(2+2x^7)^4 is 496/35

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^6(2+2x^7)^4

\int_{0}^{1} x^{6} (2 + 2 x^{7})^{4} d x

\frac{496}{35}

Decimal

14.17142 \dots

Solution steps

\int_{0}^{1} x^{6} (2 + 2 x^{7})^{4} d x

Expand

x^{6} (2 + 2 x^{7})^{4} : 16 x^{6} + 64 x^{13} + 96 x^{20} + 64 x^{27} + 16 x^{34}

= \int_{0}^{1} 16 x^{6} + 64 x^{13} + 96 x^{20} + 64 x^{27} + 16 x^{34} 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} 16 x^{6} d x + \int_{0}^{1} 64 x^{13} d x + \int_{0}^{1} 96 x^{20} d x + \int_{0}^{1} 64 x^{27} d x + \int_{0}^{1} 16 x^{34} d x

\int_{0}^{1} 16 x^{6} d x = \frac{16}{7}

\int_{0}^{1} 64 x^{13} d x = \frac{32}{7}

\int_{0}^{1} 96 x^{20} d x = \frac{32}{7}

\int_{0}^{1} 64 x^{27} d x = \frac{16}{7}

\int_{0}^{1} 16 x^{34} d x = \frac{16}{35}

= \frac{16}{7} + \frac{32}{7} + \frac{32}{7} + \frac{16}{7} + \frac{16}{35}

Simplify

\frac{16}{7} + \frac{32}{7} + \frac{32}{7} + \frac{16}{7} + \frac{16}{35} : \frac{496}{35}

= \frac{496}{35}

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

\int_{- 1}^{1} \frac{1}{v} d v

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

\int_{- 2}^{2} (19 - ∣ x ∣^{3}) d x

\int_{0}^{5} \frac{x}{x ^{2} + 2} d x

What is the integral from 0 to 1 of x^6(2+2x^7)^4 ?
The integral from 0 to 1 of x^6(2+2x^7)^4 is 496/35