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

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

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

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

Decimal Notation

- 25.69475 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

= - 6 [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \frac{4}{9} t^{\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))

\int 7 ln (x) d x

\int (cos^{3} (x) sin (x)) d x

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

(x^{2} + y^{2}) d x + (2 x y) d y = 0

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