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

The integral from 0 to 1 of 4x^3sqrt(3x^4+4) is (14sqrt(7)-16)/9

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(3x^4+4)

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

\frac{14 7 - 16}{9}

Decimal

2.33783 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{7} \frac{u}{12} d u

= 4 \cdot \int_{4}^{7} \frac{u}{12} d u

Take the constant out:

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

= 4 \cdot \frac{1}{12} \cdot \int_{4}^{7} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{7}

= 4 \cdot \frac{1}{12} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{7}

Simplify

4 \cdot \frac{1}{12} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{7} : \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{7}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{7}

Compute the boundaries:

\frac{14 7}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{14 7}{3} - \frac{16}{3})

Simplify

= \frac{14 7 - 16}{9}

\int x (1 - x^{2}) d x

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

\frac{\partial}{\partial x} (x^{2} + y^{2} + 2 x y)

\int 7 e^{- 0.8 x} d x

x \to \infty lim (x \frac{1}{2})

What is the integral from 0 to 1 of 4x^3sqrt(3x^4+4) ?
The integral from 0 to 1 of 4x^3sqrt(3x^4+4) is (14sqrt(7)-16)/9