What is the integral from 1 to 4 of-6sqrt(x)ln(x) ?

The integral from 1 to 4 of-6sqrt(x)ln(x) is -6(-28/9+32/3 ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of-6sqrt(x)ln(x)

\int_{1}^{4} - 6 x ln (x) d x

- 6 (- \frac{28}{9} + \frac{32}{3} ln (2))

Decimal Notation

- 25.69475 \dots

Solution steps

\int_{1}^{4} - 6 x ln (x) d x

Take the constant out:

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

= - 6 \cdot \int_{1}^{4} x ln (x) d x

Apply Integration By Parts

: [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \int \frac{2}{3} x^{\frac{1}{2}} d x]_{1}^{4}

= - 6 [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \int \frac{2}{3} x^{\frac{1}{2}} d x]_{1}^{4}

\int \frac{2}{3} x^{\frac{1}{2}} d x = \frac{4}{9} x^{\frac{3}{2}}

= - 6 [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \frac{4}{9} x^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

- \frac{28}{9} + \frac{32}{3} ln (2)

= - 6 (- \frac{28}{9} + \frac{32}{3} ln (2))

x \to 4 - lim (3)

\int \frac{16}{25 + 9 r ^{2}} d r

\frac{d}{d x} ((4 x - x^{2})^{3} (1 - x + 2 x^{2})^{2})

\int c x^{2} e^{- x} d x

\frac{d}{d x} (\frac{x ^{3}}{4})

What is the integral from 1 to 4 of-6sqrt(x)ln(x) ?
The integral from 1 to 4 of-6sqrt(x)ln(x) is -6(-28/9+32/3 ln(2))