What is the integral from 0 to 1 of 4x^3sqrt(1+9x^4) ?

The integral from 0 to 1 of 4x^3sqrt(1+9x^4) is (20sqrt(10)-2)/(27)

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(1+9x^4)

\int_{0}^{1} 4 x^{3} 1 + 9 x^{4} d x

\frac{20 10 - 2}{27}

Decimal Notation

2.26835 \dots

Solution steps

\int_{0}^{1} 4 x^{3} 1 + 9 x^{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} 1 + 9 x^{4} d x

Apply u-substitution

: \int_{1}^{10} \frac{u}{36} d u

= 4 \cdot \int_{1}^{10} \frac{u}{36} d u

Take the constant out:

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

= 4 \cdot \frac{1}{36} \cdot \int_{1}^{10} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

= 4 \cdot \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

Simplify

4 \cdot \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10} : \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{10}

Compute the boundaries:

\frac{20 10}{3} - \frac{2}{3}

= \frac{1}{9} (\frac{20 10}{3} - \frac{2}{3})

Simplify

= \frac{20 10 - 2}{27}

\int_{0}^{a} cosh^{2} (\frac{t}{a}) d t

\int_{1}^{b} \frac{1}{x ^{p}} d x

\int_{0}^{\frac{π}{4}} e^{- t} cos (t) d t

\int_{2}^{6} (- x^{2} + 8 x - 12) d x

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

What is the integral from 0 to 1 of 4x^3sqrt(1+9x^4) ?
The integral from 0 to 1 of 4x^3sqrt(1+9x^4) is (20sqrt(10)-2)/(27)