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

The integral from 1/2 to 1 of ln(x) is -1/2+1/2 ln(2)

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integral from 1/2 to 1 of ln(x)

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

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

Decimal

- 0.15342 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

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

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

\int_{0}^{\frac{π}{2}} \frac{1}{2} sin (2 θ) d θ

\int_{1}^{4} x^{2} - 1 d x

\int_{1}^{4} \frac{1}{x ^{2} + x - 6} d x

\int_{0}^{1} (2 t - 1)^{3} d t

\int_{- 1}^{1} (e^{- 3 x}) d x

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