What is the integral from 1 to 4 of 7sqrt(t)ln(t) ?

The integral from 1 to 4 of 7sqrt(t)ln(t) is 7(-28/9+32/3 ln(2))

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integral from 1 to 4 of 7sqrt(t)ln(t)

\int_{1}^{4} 7 t ln (t) d t

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

Decimal

29.97721 \dots

Solution steps

\int_{1}^{4} 7 t ln (t) d t

Take the constant out:

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

= 7 \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}

= 7 [\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}}

= 7 [\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)

= 7 (- \frac{28}{9} + \frac{32}{3} ln (2))

y = (7 ln (x))^{2}

\frac{\partial}{\partial x} (x^{e})

\int_{- \infty}^{0} z e^{9 z} d z

\int \frac{x}{( x + t ) ^{2}}

\frac{d}{d x} (4 arcsin (\frac{2 x - 4}{3 x + 2}))

What is the integral from 1 to 4 of 7sqrt(t)ln(t) ?
The integral from 1 to 4 of 7sqrt(t)ln(t) is 7(-28/9+32/3 ln(2))