How much work is done as the spring stretches from $20$ to $40\ \mathrm{cm}$?

$A_{total}=12\ \mathrm{J}$

The area under the curve will be equal to the work done. In this case, we have some choices. You may recognize this shape as a trapezoid (it might help to momentarily rotate your head 90 degrees) to see this.

$$A=\frac{1}{2}(b_1+b_2)h$$

$$A=\frac{1}{2}(40\ \mathrm{N+80\ N)(0.20\ m)}$$

$$A=12\ \mathrm{N\cdot m}$$

$$\mathrm{or}$$

$$A=12\ \mathrm{J}$$

An alternative choice is to recognize this shape as a triangle sitting on top of a rectangle. The total area is simply the area of the rectangle plus the area of the triangle.

$$A_{total}=A_{rectangle}+A_{triangle}$$

$$A_{total}=bh_1+\frac{1}{2}(bh_2)$$

$$A_{total}=(0.20\ \mathrm{m)(40\ N-0\ N)+\frac{1}{2}(0.20\ m)(80\ N-40\ N)}$$

$$A_{total}=8\ \mathrm{N\cdot m+4\ N\cdot m}$$

$$A_{total}=12\ \mathrm{J}$$