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

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

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integral from 1 to 4 of-10sqrt(x)ln(x)

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

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

Decimal Notation

- 42.82458 \dots

Solution steps

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

Take the constant out:

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

= - 10 \cdot \int_{1}^{4} x ln (x) d x

Apply Integration By Parts

: [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \int \frac{2}{3} x^{\frac{1}{2}} d x]_{1}^{4}

= - 10 [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \int \frac{2}{3} x^{\frac{1}{2}} d x]_{1}^{4}

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

= - 10 [\frac{2}{3} x^{\frac{3}{2}} ln (x) - \frac{4}{9} x^{\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))

\int_{0}^{4} arctan (\frac{x}{4}) d x

\frac{3 y ^{2} - x ^{2}}{y ^{5}} \frac{d y}{d x} + \frac{x}{2 y ^{4}} = 0

x \to 210 lim (12 x - 350)

\int_{3}^{e^{5}} \frac{1}{x} d x

\frac{\partial}{\partial x} (x^{8} + 6 x y^{5})

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