Evaluate double integral of&nbsp;∫01∫0xexydxdy

Question

Evaluate double integral of&amp;nbsp;∫01∫0xexydxdy

nick1337 · Accepted Answer

To evaluate the given double integral, let's start by integrating with respect to x and then with respect to y.
The double integral is:

\int_{0}^{1} \int_{0}^{x} e^{x / y} d x d y

Integrating with respect to x first, we treat y as a constant:

\int_{0}^{1} [\int_{0}^{x} e^{x / y} d x] d y

To evaluate the inner integral, we use the substitution u = x/y, which implies du = dx/y:

\int_{0}^{1} [\int_{0}^{x} e^{x / y} d x] d y = \int_{0}^{1} [\int_{0}^{x} y e^{u} d u] d y

Now we can evaluate the inner integral:

\int_{0}^{1} [{y e^{u} |}_{0}^{x}] d y = \int_{0}^{1} (y e^{x} - y e^{0}) d y

Simplifying, we have:

\int_{0}^{1} (y e^{x} - y) d y

Integrating with respect to y, we treat x as a constant:

{(\frac{1}{2} y^{2} e^{x} - \frac{1}{2} y^{2}) |}_{0}^{1}

Evaluating at the limits, we get:

(\frac{1}{2} (1^{2} e^{x} - 1^{2})) - (\frac{1}{2} (0^{2} e^{x} - 0^{2}))

Simplifying further, we have:

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

Therefore, the value of the given double integral is

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

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Evaluate double integral of ∫01∫0xexydxdy