What is the integral from 36 to x of sqrt(1+9x) ?

The integral from 36 to x of sqrt(1+9x) is (2(1+9x)^{3/2}-3250sqrt(13))/(27)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 36 to x of sqrt(1+9x)

\int_{36}^{x} 1 + 9 x d x

\frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 3250 13}{27}

Solution steps

\int_{36}^{x} 1 + 9 x d x

Apply u-substitution

: \int_{325}^{1 + 9 x} \frac{u}{9} d u

= \int_{325}^{1 + 9 x} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{325}^{1 + 9 x} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{325}^{1 + 9 x}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{325}^{1 + 9 x}

Compute the boundaries:

\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{3250 13}{3}

= \frac{1}{9} (\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{3250 13}{3})

Simplify

= \frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 3250 13}{27}

\frac{t - t}{t ^{\frac{1}{7}}}

\frac{3 x ^{2} + 1}{2 x}

\int cos (x) lo g_{10} (2) 2^{x} d x

\frac{d}{d x} (x sin (y - z))

\int \frac{42}{1 - cos ( 7 x )} d x

What is the integral from 36 to x of sqrt(1+9x) ?
The integral from 36 to x of sqrt(1+9x) is (2(1+9x)^{3/2}-3250sqrt(13))/(27)