What is the integral from e to 2e of yln(z)ln(2) ?

The integral from e to 2e of yln(z)ln(2) is ln(2)y(2eln(2e)-2e)

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integral from e to 2e of yln(z)ln(2)

\int_{e}^{2 e} y ln (z) ln (2) d z

ln (2) y (2 e ln (2 e) - 2 e)

Solution steps

\int_{e}^{2 e} y ln (z) ln (2) d z

Take the constant out:

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

= ln (2) y \cdot \int_{e}^{2 e} ln (z) d z

Apply Integration By Parts

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

= ln (2) y [z ln (z) - \int 1 d z]_{e}^{2 e}

\int 1 d z = z

= ln (2) y [z ln (z) - z]_{e}^{2 e}

Compute the boundaries:

2 e ln (2 e) - 2 e

= ln (2) y (2 e ln (2 e) - 2 e)

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

\int_{0}^{3} 2 e^{- 2 x} d x

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

\int_{x}^{y^{3}} e^{t^{2}} d t

\int_{0}^{16} 4 x d x

What is the integral from e to 2e of yln(z)ln(2) ?
The integral from e to 2e of yln(z)ln(2) is ln(2)y(2eln(2e)-2e)