What is the integral from 1 to 1.5 of x^2ln(x) ?

The integral from 1 to 1.5 of x^2ln(x) is 0.19225…

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integral from 1 to 1.5 of x^2ln(x)

\int_{1}^{1.5} x^{2} ln (x) d x

0.19225 \dots

Solution steps

\int_{1}^{1.5} x^{2} ln (x) d x

Apply Integration By Parts

: [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \int \frac{x ^{2}}{3} d x)]_{1}^{1.5}

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \int \frac{x ^{2}}{3} d x)]_{1}^{1.5}

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

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \frac{x ^{3}}{9})]_{1}^{1.5}

Simplify

[\frac{1}{3} (x^{3} ln (x) - 3 \cdot \frac{x ^{3}}{9})]_{1}^{1.5} : [0.33333 \dots (x^{3} ln (x) - 0.33333 \dots x^{3})]_{1}^{1.5}

= [0.33333 \dots (x^{3} ln (x) - 0.33333 \dots x^{3})]_{1}^{1.5}

Compute the boundaries:

0.19225 \dots

= 0.19225 \dots

f (x) = \frac{sin ( x )}{e ^{x}}

\frac{x}{2}, 3 x^{2} - 1, 0, 0.5

- 9 x + 5

\frac{s ^{2} - 4}{s + 1}

\int \frac{1}{cos ( θ )} d θ

What is the integral from 1 to 1.5 of x^2ln(x) ?
The integral from 1 to 1.5 of x^2ln(x) is 0.19225…