What is the integral from 1 to 16 of sqrt(t)ln(t) ?

The integral from 1 to 16 of sqrt(t)ln(t) is -28+512/3 ln(2)

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integral from 1 to 16 of sqrt(t)ln(t)

\int_{1}^{16} t ln (t) d t

- 28 + \frac{512}{3} ln (2)

Decimal Notation

90.29711 \dots

Solution steps

\int_{1}^{16} 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}^{16}

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \int \frac{2}{3} t^{\frac{1}{2}} d t]_{1}^{16}

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

= [\frac{2}{3} t^{\frac{3}{2}} ln (t) - \frac{4}{9} t^{\frac{3}{2}}]_{1}^{16}

Compute the boundaries:

- 28 + \frac{512}{3} ln (2)

= - 28 + \frac{512}{3} ln (2)

x tan (x)

θ \to 0 lim (\frac{1 - cos ( θ )}{θ})

y = x^{4}, y = 8 x

\int_{1}^{6} x^{2} d x

x \to 0 lim (\frac{1 - x}{2 + x})

What is the integral from 1 to 16 of sqrt(t)ln(t) ?
The integral from 1 to 16 of sqrt(t)ln(t) is -28+512/3 ln(2)