What is the integral from 0 to ln(2) of integral from 1 to ln(5) of e^{2x+y} ?

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integral from 0 to ln(2) of integral from 1 to ln(5) of e^{2x+y}

\int_{0}^{l n (2)} \int_{1}^{l n (5)} e^{2 x + y} d y d x

(5 - e) \frac{3}{2}

Decimal

3.42257 \dots

Solution steps

\int_{0}^{l n (2)} \int_{1}^{l n (5)} e^{2 x + y} d y d x

\int_{1}^{l n (5)} e^{2 x + y} d y = e^{2 x} (5 - e)

= \int_{0}^{l n (2)} e^{2 x} (5 - e) d x

\int_{0}^{l n (2)} e^{2 x} (5 - e) d x = (5 - e) \frac{3}{2}

= (5 - e) \frac{3}{2}

\frac{d}{d x} (\frac{1 - x}{( x + 1 ) ^{3}})

sin (2 x) d x + cos (7 y) d y = 0, y (\frac{π}{2}) = \frac{π}{7}

\frac{d}{d x} (\frac{3 x}{1 + 2 x})

\int (cos (x) + \frac{1}{4} x) d x

\int sin^{23} (x) d x

What is the integral from 0 to ln(2) of integral from 1 to ln(5) of e^{2x+y} ?
The integral from 0 to ln(2) of integral from 1 to ln(5) of e^{2x+y} is (5-e) 3/2