What is the integral from 1 to 4 of-9sqrt(x)ln(x) ?

The integral from 1 to 4 of-9sqrt(x)ln(x) is 28-96ln(2)

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integral from 1 to 4 of-9sqrt(x)ln(x)

\int_{1}^{4} - 9 x ln (x) d x

28 - 96 ln (2)

Decimal Notation

- 38.54212 \dots

Solution steps

\int_{1}^{4} - 9 x ln (x) d x

Take the constant out:

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

= - 9 \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}

= - 9 [\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}}

= - 9 [\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)

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

Simplify

= 28 - 96 ln (2)

x \to 0 lim (\frac{2 e ^{- x} - x ^{2} + 2 x - 2}{4 x ^{3}})

y^{^{'}} + x \cdot y = x \cdot y^{3}

n = 0 \sum \infty 5 (- \frac{4}{3})^{n}

\int (x^{3} - \frac{1}{x ^{2}} - x^{\frac{2}{3}}) d x

\int 36 - (0.6 x^{2} + 2 x + 11) d x

What is the integral from 1 to 4 of-9sqrt(x)ln(x) ?
The integral from 1 to 4 of-9sqrt(x)ln(x) is 28-96ln(2)