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

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

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

\frac{2 ( 4 2 + 4 )}{15}

Decimal Notation

1.28758 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{( u - 1 ) u}{2} d u

= 4 \cdot \int_{1}^{2} \frac{( u - 1 ) u}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{2} (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{2} u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{2} (\int_{1}^{2} u^{\frac{3}{2}} d u - \int_{1}^{2} u d u)

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

\int_{1}^{2} u d u = \frac{4 2 - 2}{3}

= 4 \cdot \frac{1}{2} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3})

Simplify

4 \cdot \frac{1}{2} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}) : \frac{2 ( 4 2 + 4 )}{15}

= \frac{2 ( 4 2 + 4 )}{15}

\int_{1}^{2} \frac{x ^{2}}{2} + \frac{1}{2 x ^{2}} d x

\int_{- 1}^{1} \frac{5 x}{( x + 5 ) ^{2}} d x

\int_{0}^{r} (b r^{2} + cr) \cdot 2 π r d r

\int_{0}^{2} (- 8 y^{2} + \frac{256}{3} + 8 y) d y

\int_{5}^{10} \frac{2.6}{t} d t

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