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

The integral from 0 to 2 of 2x(x^2+4)^2 is 448/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{448}{3}

Decimal

149.33333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2} x (x^{2} + 4)^{2} d x

Expand

x (x^{2} + 4)^{2} : x^{5} + 8 x^{3} + 16 x

= 2 \cdot \int_{0}^{2} x^{5} + 8 x^{3} + 16 x 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

= 2 (\int_{0}^{2} x^{5} d x + \int_{0}^{2} 8 x^{3} d x + \int_{0}^{2} 16 x d x)

\int_{0}^{2} x^{5} d x = \frac{32}{3}

\int_{0}^{2} 8 x^{3} d x = 32

\int_{0}^{2} 16 x d x = 32

= 2 (\frac{32}{3} + 32 + 32)

Simplify

2 (\frac{32}{3} + 32 + 32) : \frac{448}{3}

= \frac{448}{3}

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

\int_{1}^{10} \frac{5}{x} d x

\int_{0}^{2} (x^{2} - 10 x + 16) d x

\int_{0}^{\infty} 2 e^{- 2 y} d y

\int_{2}^{x^{2}} \frac{1}{t ^{3}} d t

What is the integral from 0 to 2 of 2x(x^2+4)^2 ?
The integral from 0 to 2 of 2x(x^2+4)^2 is 448/3