Work Done by a Variable Force

Work done by a variable force: $W=\int_{y_i}^{y_f}F\,dy$, with $y_i=0$, $y_f=1\ m$ $W=\int_0^1(20+10y)\,dy=\left[20y+\frac{10y^2}{2}\right]_0^1$ $W=20+5=25\ J$

Work done = area under the force–distance graph.
From the graph, the force is 2 N from d = 3 m to d = 7 m and then falls uniformly to zero at d = 12 m.
Area of the rectangle (3 m to 7 m) = $2\times(7 - 3) = 8$ J
Area of the triangle (7 m to 12 m) = $\frac{1}{2}\times(12 - 7)\times2 = 5$ J
Total work = 8 + 5 = 13 J

Work done by a variable force: $W = \int F\,dx$ = area under the F-x graph.
From $x$ = 0 to 3 m the force is constant at 3 N: area of the rectangle $= 3 \times 3 = 9$ J
From $x$ = 3 m to 6 m the force falls linearly from 3 N to 0: area of the triangle $= \dfrac{1}{2} \times 3 \times 3 = 4.5$ J
Total work $W = (3 \times 3) + \dfrac{1}{2} \times (3)(3)$
$W = 9 + 4.5 = 13.5$ J