What is the integral from 0 to 4 of 3x^2ln(2x) ?

The integral from 0 to 4 of 3x^2ln(2x) is 3(64ln(2)-64/9)

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integral from 0 to 4 of 3x^2ln(2x)

\int_{0}^{4} 3 x^{2} ln (2 x) d x

3 (64 ln (2) - \frac{64}{9})

Decimal Notation

111.75092 \dots

Solution steps

\int_{0}^{4} 3 x^{2} ln (2 x) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{4} x^{2} ln (2 x) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} ln (2 x) - \int \frac{x ^{2}}{3} d x]_{0}^{4}

= 3 [\frac{1}{3} x^{3} ln (2 x) - \int \frac{x ^{2}}{3} d x]_{0}^{4}

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

= 3 [\frac{1}{3} x^{3} ln (2 x) - \frac{x ^{3}}{9}]_{0}^{4}

Compute the boundaries:

64 ln (2) - \frac{64}{9}

= 3 (64 ln (2) - \frac{64}{9})

\int_{0}^{x} y e^{- y} d y

\int_{0}^{1} t t^{2} + 1 d t

\int_{- \infty}^{1} e^{3 x} d x

\int_{0}^{1} 2 t 1 + t^{2} d t

\int_{- 1}^{1} m x^{2} d x

What is the integral from 0 to 4 of 3x^2ln(2x) ?
The integral from 0 to 4 of 3x^2ln(2x) is 3(64ln(2)-64/9)