What is the integral from 1 to 4 of 8sqrt(t)ln(t) ?

The integral from 1 to 4 of 8sqrt(t)ln(t) is 8(-28/9+32/3 ln(2))

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integral from 1 to 4 of 8sqrt(t)ln(t)

\int_{1}^{4} 8 t ln (t) d t

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

Decimal

34.25967 \dots

Solution steps

\int_{1}^{4} 8 t ln (t) d t

Take the constant out:

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

= 8 \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}

= 8 [\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}}

= 8 [\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)

= 8 (- \frac{28}{9} + \frac{32}{3} ln (2))

y^{^{'}} = - 0.02 (y - 20)

n = 1 \sum \infty 2 - \frac{5}{2 ^{n}}

\frac{4 x - 1}{2 x + 3}

\int_{1}^{b} \frac{1}{( x + 8 ) ^{2}} d x

\int \frac{e ^{x}}{3 + 16 e ^{2 x}} d x

What is the integral from 1 to 4 of 8sqrt(t)ln(t) ?
The integral from 1 to 4 of 8sqrt(t)ln(t) is 8(-28/9+32/3 ln(2))