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

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

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 4x(x^2+1)^3

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

\frac{15}{2}

Decimal

7.5

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{0}^{1} x (x^{2} + 1)^{3} d x

Expand

x (x^{2} + 1)^{3} : x^{7} + 3 x^{5} + 3 x^{3} + x

= 4 \cdot \int_{0}^{1} x^{7} + 3 x^{5} + 3 x^{3} + 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

= 4 (\int_{0}^{1} x^{7} d x + \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} 3 x^{3} d x + \int_{0}^{1} x d x)

\int_{0}^{1} x^{7} d x = \frac{1}{8}

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

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

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

= 4 (\frac{1}{8} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2})

Simplify

4 (\frac{1}{8} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2}) : \frac{15}{2}

= \frac{15}{2}

\int_{0}^{0.2} \frac{1}{1 + x ^{4}} d x

\int_{0}^{1} x sin (π x^{2}) d x

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

\int_{5}^{4} sin (t) d t

\int_{0}^{5} (5 - ∣ x - 5 ∣) d x

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