What is the integral from 1 to 4 of-3sqrt(t)ln(t) ?

The integral from 1 to 4 of-3sqrt(t)ln(t) is 28/3-32ln(2)

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integral from 1 to 4 of-3sqrt(t)ln(t)

\int_{1}^{4} - 3 t ln (t) d t

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

Decimal

- 12.84737 \dots

Solution steps

\int_{1}^{4} - 3 t ln (t) d t

Take the constant out:

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

= - 3 \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}

= - 3 [\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}}

= - 3 [\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)

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

Simplify

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

\int_{0}^{5} 2 π x (\frac{1}{x + 4}) d x

- 1.621 s - 0.462945095

\frac{\partial}{\partial φ} (ρ sin (φ) cos (θ))

y^{^{''}} - 4 y^{^{'}} + 5 y = 2020 e^{- t}

5 2 x^{3} + 8 x (2.2)

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