What is the integral from 1 to 2 of w^2ln(w) ?

The integral from 1 to 2 of w^2ln(w) is -7/9+8/3 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 2 of w^2ln(w)

\int_{1}^{2} w^{2} ln (w) d w

- \frac{7}{9} + \frac{8}{3} ln (2)

Decimal Notation

1.07061 \dots

Solution steps

\int_{1}^{2} w^{2} ln (w) d w

Apply Integration By Parts

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

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

\int \frac{w ^{2}}{3} d w = \frac{w ^{3}}{9}

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

Compute the boundaries:

- \frac{7}{9} + \frac{8}{3} ln (2)

= - \frac{7}{9} + \frac{8}{3} ln (2)

\frac{d}{d x} (2 sin (π x - y))

\frac{2}{x}, - 3 x + 7, - 2 x + 4

\frac{d}{d x} (ln (ln (10 x)))

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

x \to - 2 lim (tan (\frac{π x}{4}))

What is the integral from 1 to 2 of w^2ln(w) ?
The integral from 1 to 2 of w^2ln(w) is -7/9+8/3 ln(2)