What is the integral from 0 to 5 of x^3sqrt(1+9x^4) ?

The integral from 0 to 5 of x^3sqrt(1+9x^4) is (5626sqrt(5626)-1)/(54)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 5 of x^3sqrt(1+9x^4)

\int_{0}^{5} x^{3} 1 + 9 x^{4} d x

\frac{5626 5626 - 1}{54}

Decimal

7814.56490 \dots

Solution steps

\int_{0}^{5} x^{3} 1 + 9 x^{4} d x

Apply u-substitution

= \int_{1}^{5626} \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int_{1}^{5626} u d u

Apply the Power Rule

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5626}

Compute the boundaries:

\frac{11252 5626}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{11252 5626}{3} - \frac{2}{3})

Simplify

= \frac{5626 5626 - 1}{54}

What is the integral from 0 to 5 of x^3sqrt(1+9x^4) ?
The integral from 0 to 5 of x^3sqrt(1+9x^4) is (5626sqrt(5626)-1)/(54)