What is the integral from 1 to 4 of-5sqrt(t)ln(t) ?

The integral from 1 to 4 of-5sqrt(t)ln(t) is -5(-28/9+32/3 ln(2))

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integral from 1 to 4 of-5sqrt(t)ln(t)

\int_{1}^{4} - 5 t ln (t) d t

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

Decimal Notation

- 21.41229 \dots

Solution steps

\int_{1}^{4} - 5 t ln (t) d t

Take the constant out:

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

= - 5 \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}

= - 5 [\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}}

= - 5 [\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)

= - 5 (- \frac{28}{9} + \frac{32}{3} ln (2))

\frac{\partial}{\partial x} (x^{2} cos (x y^{2}))

t \to + 0 lim (\frac{e ^{t} - 1}{t ^{9}})

f (x) = \frac{3}{x}, (6, \frac{1}{2})

\int 32 sin^{4} (2 x) d x

\frac{d}{d y} ((x^{2} + y^{2}) arctan (\frac{y}{x}))

What is the integral from 1 to 4 of-5sqrt(t)ln(t) ?
The integral from 1 to 4 of-5sqrt(t)ln(t) is -5(-28/9+32/3 ln(2))