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

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

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integral from 0 to 1 of 2ln(x)

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

- 2

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

- 1

= 2 (- 1)

Simplify

= - 2

\int lo g_{e} (x^{2} + 1 - x) d x

3 - 4 x

\int 2 x^{13} e^{x^{7}} d x

x \to 0 lim (\frac{2 ^{x} - 1}{2})

\frac{d}{d x} (e^{(x + x^{2} + 1)})

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