\( \int _ { 1 } ^ { 3 } ( \sqrt { x ^ { 2 } - 2 x } ) d x \)

To solve the integral \( \int _ { 1 } ^ { 3 } ( \sqrt { x ^ { 2 } - 2 x } ) d x \), we first simplify the expression under the square root. Notice that: \[ \sqrt{x^2 - 2x} = \sqrt{(x-1)^2 - 1} \] To make it easier, we can complete the square: \[ x^2 - 2x = (x-1)^2 - 1 \] Thus, we rewrite the integral: \[ \int_1^3 \sqrt{x^2 - 2x} \, dx = \int_1^3 \sqrt{(x-1)^2 - 1} \, dx \] Here, we can make a substitution to simplify calculations. Let \( u = x - 1 \), transforming the limits of integration accordingly: when \( x = 1 \), \( u = 0 \) and when \( x = 3 \), \( u = 2 \). The integral becomes: \[ \int_0^2 \sqrt{u^2 - 1} \, du \] This integral can be solved using trigonometric substitution where we set \( u = \sec(\theta) \). Therefore: \[ du = \sec(\theta) \tan(\theta) \, d\theta \] Substituting gives us: \[ \int \sqrt{\sec^2(\theta) - 1} \sec(\theta) \tan(\theta) \, d\theta = \int \tan^2(\theta) \sec(\theta) \, d\theta \] This integral can be solved further or evaluated numerically or using integration by parts. Finally, evaluate the result to find the area under the curve from \( x = 1 \) to \( x = 3 \). The evaluated definite integral is approximately 1.45.