What is the integral from 1 to 4 of 10sqrt(t)ln(t) ?

The integral from 1 to 4 of 10sqrt(t)ln(t) is 10(-28/9+32/3 ln(2))

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integral from 1 to 4 of 10sqrt(t)ln(t)

\int_{1}^{4} 10 t ln (t) d t

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

Decimal Notation

42.82458 \dots

Solution steps

\int_{1}^{4} 10 t ln (t) d t

Take the constant out:

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

= 10 \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}

= 10 [\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}}

= 10 [\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)

= 10 (- \frac{28}{9} + \frac{32}{3} ln (2))

\frac{\partial}{\partial x} (6 y^{2} + 478 x - 10)

\frac{d}{d x} (- \frac{2 x}{( 4 + x ^{2} ) ^{2}})

\int \frac{x ^{2} + x - 3}{x} d x

\frac{d}{d x} (2 (x - 2))

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

What is the integral from 1 to 4 of 10sqrt(t)ln(t) ?
The integral from 1 to 4 of 10sqrt(t)ln(t) is 10(-28/9+32/3 ln(2))