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

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))

x \to - 3 lim (- \frac{5}{( x + 3 ) ^{2}})

\frac{d}{d x} (\frac{8 - x e ^{x}}{x + e ^{x}})

4 y^{^{'}} = \frac{4 x y + 8 x ^{2} + 2 y ^{2}}{x ^{2}}

y^{^{'}} + csc (y) = 0

\int \frac{e ^{3 x} + 1}{e ^{x} + 1} d 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))