What is the integral from 0 to 1 of (ln(2)+y) ?

The integral from 0 to 1 of (ln(2)+y) is ln(2)+1/2

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integral from 0 to 1 of (ln(2)+y)

\int_{0}^{1} (ln (2) + y) d y

ln (2) + \frac{1}{2}

Decimal Notation

1.19314 \dots

Solution steps

\int_{0}^{1} ln (2) + y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} ln (2) d y + \int_{0}^{1} y d y

\int_{0}^{1} ln (2) d y = ln (2)

\int_{0}^{1} y d y = \frac{1}{2}

= ln (2) + \frac{1}{2}

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What is the integral from 0 to 1 of (ln(2)+y) ?
The integral from 0 to 1 of (ln(2)+y) is ln(2)+1/2