\( \int _ { 0 } ^ { 1 } 2 x d x , F ( x ) = x ^ { 2 } + 5 \)

To evaluate the integral \( \int_{0}^{1} 2x \, dx \), we first find the antiderivative of \( 2x \), which is \( x^2 \). Then we apply the limits of integration: \[ \int_{0}^{1} 2x \, dx = \left[ x^2 \right]_{0}^{1} = 1^2 - 0^2 = 1. \] Now, to understand the function \( F(x) = x^2 + 5 \), it’s essentially a quadratic function shifted up by 5 units. At \( x = 0 \), \( F(0) = 5 \) and at \( x = 1 \), \( F(1) = 6 \). This means that in the context of our integral, the area under the curve \( 2x \) from 0 to 1 adds a nice linear visual alongside the quadratic nature of \( F(x) \).