Questions: Evaluate the integral ∫ from -∞ to -7 of (2/(2x-3)^2) dx or state that the integral diverges. Express your answer in simplest form.

Solution Steps

We start with the improper integral we want to evaluate: \[ \int_{-\infty}^{-7} \frac{2}{(2x-3)^{2}} \, dx \]

To evaluate the integral, we first find the antiderivative of the integrand \(\frac{2}{(2x-3)^{2}}\). The antiderivative is given by: \[ -\frac{2}{4x - 6} \]

Next, we evaluate the definite integral from a limit \(a\) approaching \(-\infty\) to \(-7\): \[ \lim_{a \to -\infty} \left( -\frac{2}{4a - 6} - \left(-\frac{2}{4(-7) - 6}\right) \right) \]

Calculating the value at the upper limit: \[ -\frac{2}{4(-7) - 6} = -\frac{2}{-28 - 6} = -\frac{2}{-34} = \frac{1}{17} \] Thus, the expression simplifies to: \[ \lim_{a \to -\infty} \left( -\frac{2}{4a - 6} + \frac{1}{17} \right) \]

As \(a\) approaches \(-\infty\), the term \(-\frac{2}{4a - 6}\) approaches \(0\). Therefore, we have: \[ 0 + \frac{1}{17} = \frac{1}{17} \]

The integral converges to: \[ \int_{-\infty}^{-7} \frac{2}{(2x-3)^{2}} \, dx = \frac{1}{17} \]

Final Answer