What is the integral from 0 to 1 of ln(4-2x) ?

The integral from 0 to 1 of ln(4-2x) is 3ln(2)-1

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integral from 0 to 1 of ln(4-2x)

\int_{0}^{1} ln (4 - 2 x) d x

3 ln (2) - 1

Decimal

1.07944 \dots

Solution steps

\int_{0}^{1} ln (4 - 2 x) d x

Apply Integration By Parts

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

\int - \frac{x}{2 - x} d x = 2 ln ∣ 2 - x ∣ + x

= [x ln (4 - 2 x) - (2 ln ∣ 2 - x ∣ + x)]_{0}^{1}

Simplify

= [x ln (4 - 2 x) - 2 ln ∣ 2 - x ∣ - x]_{0}^{1}

Compute the boundaries:

3 ln (2) - 1

= 3 ln (2) - 1

What is the integral from 0 to 1 of ln(4-2x) ?
The integral from 0 to 1 of ln(4-2x) is 3ln(2)-1