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

The integral from 0 to ln(2) of e^{x-y} is 1/2 e^x

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integral from 0 to ln(2) of e^{x-y}

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

\frac{1}{2} e^{x}

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

= - [e^{u}]_{x}^{x - l n (2)}

Compute the boundaries:

- \frac{1}{2} e^{x}

= - (- \frac{1}{2} e^{x})

Simplify

= \frac{1}{2} e^{x}

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