What is the integral from 0 to 2 of (15sqrt(4x+1)) ?

The integral from 0 to 2 of (15sqrt(4x+1)) is 65

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (15sqrt(4x+1))

\int_{0}^{2} (15 4 x + 1) d x

65

Solution steps

\int_{0}^{2} 15 4 x + 1 d x

Take the constant out:

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

= 15 \cdot \int_{0}^{2} 4 x + 1 d x

Apply u-substitution

: \int_{1}^{9} \frac{u}{4} d u

= 15 \cdot \int_{1}^{9} \frac{u}{4} d u

Take the constant out:

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

= 15 \cdot \frac{1}{4} \cdot \int_{1}^{9} u d u

Apply the Power Rule

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

= 15 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Simplify

15 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9} : \frac{15}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

= \frac{15}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Compute the boundaries:

\frac{52}{3}

= \frac{15}{4} \cdot \frac{52}{3}

Simplify

= 65

x \to 2 lim (\frac{x - 3}{x + 4})

\int x cos (7 π) x^{2} d x

f (r) = \frac{3}{r ^{3}}

n = 1 \sum \infty \frac{x ^{n}}{n 3 ^{n}}

\int 3 x^{5} + 4 x d x

What is the integral from 0 to 2 of (15sqrt(4x+1)) ?
The integral from 0 to 2 of (15sqrt(4x+1)) is 65