What is the integral from 0 to 6 of 2xsqrt(x^2+1) ?

The integral from 0 to 6 of 2xsqrt(x^2+1) is (2(37sqrt(37)-1))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 6 of 2xsqrt(x^2+1)

\int_{0}^{6} 2 x x^{2} + 1 d x

\frac{2 ( 37 37 - 1 )}{3}

Decimal

149.37480 \dots

Solution steps

\int_{0}^{6} 2 x x^{2} + 1 d x

Take the constant out:

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

= 2 \cdot \int_{0}^{6} x x^{2} + 1 d x

Apply u-substitution

= 2 \cdot \int_{1}^{37} u^{2} d u

Apply the Power Rule

= 2 [\frac{u ^{3}}{3}]_{1}^{37}

Compute the boundaries:

\frac{37 37}{3} - \frac{1}{3}

= 2 (\frac{37 37}{3} - \frac{1}{3})

Simplify

= \frac{2 ( 37 37 - 1 )}{3}

What is the integral from 0 to 6 of 2xsqrt(x^2+1) ?
The integral from 0 to 6 of 2xsqrt(x^2+1) is (2(37sqrt(37)-1))/3