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

The integral from 1 to 2 of t^2sqrt(t^3+1) is (54-4sqrt(2))/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of t^2sqrt(t^3+1)

\int_{1}^{2} t^{2} t^{3} + 1 d t

\frac{54 - 4 2}{9}

Decimal

5.37146 \dots

Solution steps

\int_{1}^{2} t^{2} t^{3} + 1 d t

Apply u-substitution

= \int_{2}^{9} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{2}^{9} u d u

Apply the Power Rule

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

Compute the boundaries:

18 - \frac{4 2}{3}

= \frac{1}{3} (18 - \frac{4 2}{3})

Simplify

= \frac{54 - 4 2}{9}

What is the integral from 1 to 2 of t^2sqrt(t^3+1) ?
The integral from 1 to 2 of t^2sqrt(t^3+1) is (54-4sqrt(2))/9