Question

Evaluate the partial integral. $$ \int_{0}^{x} y e^{x y} d y $$

Short answer

The evaluated integral from 0 to \(x\) of \(y e^{x y} dy\) is \(e^{x^2} - \frac{e^{x^2}}{x}\).

Step-by-step solution

Identify the Function to Integrate

Here, the function to integrate is \(y e^{x y}\). The limits of the integration are from 0 to \(x\).

Perform the Integration

The integral of \(y e^{x y}\) with respect to \(y\) can be computed by parts - popularly known as Integration by Parts method that's often symbolized by the formula: \(\int u v dx = u \int v dx - \int \left( \frac{d(u)}{dx} \int v dx \right) dx \). For our purpose, let u = y and dv = e^{x y} dy. Hence, du = 1 dy and v = \(\frac{1}{x} e^{x y}\). Applying Integration by Parts gives us: \(\int_{0}^{x} y e^{x y} d y = uv \Big|_0^x - \int_{0}^{x} v du\). Substituting u, v, and du into the equation gives us: \(\frac{1}{x} y e^{x y} \Big|_0^x - \int_{0}^{x} \frac{1}{x} e^{x y} dy\).

Evaluate the Integral

Evaluating the integral from 0 to \(x\) gives: \(\left[\frac{1}{x} y e^{x y} - \frac{1}{x^2} e^{x y}\right]_0^x = \frac{1}{x} x e^{x^2} - \frac{1}{x^2} e^{x^2} - 0 = e^{x^2} - \frac{1}{x} e^{x^2}\). Given that \(e^{0} = 1\), we can simplify it further to \(e^{x^2} - \frac{e^{x^2}}{x}\).