What is the integral from 0 to 1 of sqrt(1+12x) ?

The integral from 0 to 1 of sqrt(1+12x) is (13sqrt(13)-1)/(18)

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 sqrt(1+12x)

\int_{0}^{1} 1 + 12 x d x

\frac{13 13 - 1}{18}

Decimal Notation

2.54845 \dots

Solution steps

\int_{0}^{1} 1 + 12 x d x

Apply u-substitution

: \int_{1}^{13} \frac{u}{12} d u

= \int_{1}^{13} \frac{u}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{1}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

= \frac{1}{12} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{2}{3}

= \frac{1}{12} (\frac{26 13}{3} - \frac{2}{3})

Simplify

= \frac{13 13 - 1}{18}

x \to 7 lim (3)

\int (4 x + 2) e^{2 x} d x

x \to 0 lim (ln (6 x + 1))

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x

x^{2} y^{^{'}} + 3 x y = \frac{sin ( 4 x )}{x}

What is the integral from 0 to 1 of sqrt(1+12x) ?
The integral from 0 to 1 of sqrt(1+12x) is (13sqrt(13)-1)/(18)