What is the integral from 1 to 2 of ((xe^x)/y) ?

The integral from 1 to 2 of ((xe^x)/y) is xe^xln(2)

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integral from 1 to 2 of ((xe^x)/y)

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

x e^{x} ln (2)

Solution steps

\int_{1}^{2} \frac{x e ^{x}}{y} d y

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= x e^{x} [ln ∣ y ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= x e^{x} ln (2)

\int_{0}^{1} 2 t \cdot 1 + 4 t^{2} d t

\int_{- 1}^{1} (x^{2} - 2 - x^{2}) d x

\int_{0}^{1} (x^{4})^{2} - (x^{7})^{2} d x

\int_{- 2}^{1} (2 x + 3)^{2} d x

\int_{2}^{1} (ln (4 x))^{2} d x

What is the integral from 1 to 2 of ((xe^x)/y) ?
The integral from 1 to 2 of ((xe^x)/y) is xe^xln(2)