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

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

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integral from 0 to y of 1/(2y)

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

diverges

Solution steps

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= \frac{1}{2} [ln ∣ y ∣]_{0}^{y}

Compute the boundaries:

diverges

= diverges

\int_{0}^{2} 3 x^{2} + 1 d x

\int_{1}^{3} \frac{ln ( x )}{x ^{2}} d x

\int_{0}^{3} 5 x - 8 d x

\int_{5}^{6} x^{2} + x - 14 d x

\int_{0}^{2} x^{\frac{1}{2}} d x

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