Answer
The angular momentum at $t = 20~s$ is $~~1.5~kg~m^2/s$
Work Step by Step
The area under a torque versus time graph is the change in angular momentum.
From $t = 0$ to $t = 7.0~s$, we can divide the area under the graph into seven sections, including a rectangle (t = 0 to t = 2.0), a triangle (t = 2.0 to t = 4.0), a rectangle (t = 2.0 to t = 7.0), a triangle (t = 7.0 to t = 9.0), a triangle (t = 9.0 to t = 12), a rectangle (t = 12 to t = 18), and a triangle (t = 18 to t = 20).
We can find the area of each section separately:
$A_1 = (4.0~N\cdot m)(2.0~s) = 8.0~kg~m^2/s$
$A_1 = \frac{1}{2}(1.0~N\cdot m)(2.0~s) = 1.0~kg~m^2/s$
$A_3 = (3.0~N\cdot m)(5.0~s) = 15~kg~m^2/s$
$A_4 = \frac{1}{2}(3.0~N\cdot m)(2.0~s) = 3.0~kg~m^2/s$
$A_5 = \frac{1}{2}(-3.0~N\cdot m)(3.0~s) = -4.5~kg~m^2/s$
$A_6 = (-3.0~N\cdot m)(6.0~s) = -18~kg~m^2/s$
$A_7 = \frac{1}{2}(-3.0~N\cdot m)(2.0~s) = -3.0~kg~m^2/s$
We can find the total area:
$A = A_1+A_2+A_3+A_4+A_5+A_6+A_7$
$A = (8.0~kg~m^2/s)+(1.0~kg~m^2/s)+(15~kg~m^2/s)+(3.0~kg~m^2/s)+(-4.5~kg~m^2/s)+(-18~kg~m^2/s)+(-3.0~kg~m^2/s)$
$A = 1.5~kg~m^2/s$
Since the disk is stationary initially, the angular momentum at $t = 20~s$ is $~~1.5~kg~m^2/s$
