11. Evaluate the definite integral. \[ \int_{1}^{6} \frac{e^{1 / x}}{x^{2}} d x \] \( \square \)

To evaluate the definite integral \[ \int_{1}^{6} \frac{e^{1 / x}}{x^{2}} \, dx, \] we can use the substitution \( u = \frac{1}{x} \), which implies that \( du = -\frac{1}{x^2} \, dx \) or \( dx = -\frac{1}{u^2} \, du \). When \( x = 1 \), \( u = 1 \), and when \( x = 6 \), \( u = \frac{1}{6} \). Changing the limits and the variables gives us: \[ \int_{1}^{\frac{1}{6}} e^u (-du) = \int_{\frac{1}{6}}^{1} e^u \, du. \] Now, integrating \( e^u \) from \( \frac{1}{6} \) to \( 1 \): \[ \int e^u \, du = e^u + C. \] Therefore, \[ \int_{\frac{1}{6}}^{1} e^u \, du = \left[ e^u \right]_{\frac{1}{6}}^{1} = e^1 - e^{\frac{1}{6}} = e - e^{\frac{1}{6}}. \] Thus, the value of the definite integral is \[ \int_{1}^{6} \frac{e^{1/x}}{x^2} \, dx = e - e^{\frac{1}{6}}. \]