Final answer:
To find the integral from infinity to 6 of 1/(x-4)² dx, evaluate the antiderivative at the limits, taking into account the limit as one boundary approaches infinity. The integral results in a value of -1/2.
Step-by-step explanation:
The student has asked to find the integral of the function 1/(x-4)² from infinity to 6. To solve this integral, we would usually find the indefinite integral first and then evaluate the definite integral by applying the boundaries of integration. However, since the lower limit of the integration is infinity, this is an improper integral. To proceed, we recognize that the antiderivative of 1/(x-4)² is -1/(x-4). We then evaluate the definite integral by taking the limit as the lower boundary approaches infinity:
\[ \int_{\infty}^{6} \frac{1}{(x-4)^2} dx = \lim_{a \to \infty} \int_{a}^{6} \frac{1}{(x-4)^2} dx \]
This results in:
\[ = \lim_{a \to \infty} \left[-\frac{1}{(x-4)}\right]_{a}^{6} \]
\[ = \lim_{a \to \infty} \left(-\frac{1}{(6-4)} + \frac{1}{(a-4)}\right) \]
\[ = \lim_{a \to \infty} \left(-\frac{1}{2} + \frac{1}{(a-4)}\right) \]
As a approaches infinity, 1/(a-4) approaches 0, leaving us with:
\[ = -\frac{1}{2} \]
Therefore, the value of the integral from infinity to 6 of 1/(x-4)² dx is -1/2.