3) Compute $\int_{C} F \cdot d r$ where $F=y i+(x+y) j$ where $C$ is the curve is the line $(4,31)$ from $(0,2)$ to $(2,3)$, followed by the graph of $y=2 x^{2}-5$ from $(2,3)$ to (4, 31 ).

Step 1 ：We are asked to compute the line integral of the vector field \(F=y i+(x+y) j\) along the curve \(C\). The curve \(C\) is composed of two parts: a line segment from \((0,2)\) to \((2,3)\) and the graph of \(y=2x^2-5\) from \((2,3)\) to \((4,31)\).

Step 2 ：The line integral of a vector field \(F\) along a curve \(C\) is given by \(\int_{C} F \cdot d r\), where \(d r\) is the differential of the position vector \(r\) along the curve.

Step 3 ：We can compute this integral by parameterizing the curve \(C\), computing the differential \(d r\), and then integrating \(F \cdot d r\) over the parameter interval.

Step 4 ：For the line segment from \((0,2)\) to \((2,3)\), we can use the parameterization \(r(t) = (t, 2+t)\) for \(0 \leq t \leq 2\).

Step 5 ：For the graph of \(y=2x^2-5\) from \((2,3)\) to \((4,31)\), we can use the parameterization \(r(t) = (t, 2t^2-5)\) for \(2 \leq t \leq 4\).

Step 6 ：By substituting the parameterizations into the integral, we get two integrals \(I1 = 14\) and \(I2 = 462\).

Step 7 ：Adding these two integrals together, we get the final result \(I = 476\).

Step 8 ：So, the line integral of the vector field \(F=y i+(x+y) j\) along the curve \(C\) is \(\boxed{476}\).