We remind that the formula for the work is: A=∫x1x2F(s)ds, where x1=0, x2=8.5.
F′(x)=16.3arctan(x−8.5). We receive that: F=16.3∫arctan(x−8.5)dx. Make the change t=x−8.5. We get: F=16.3∫arctan(t)dt=16.3tarctan(t)−16.3∫1+t2tdt=16.3tarctan(t)−16.3∫211+t2d(1+t2)=16.3tarctan(t)−216.3ln(1+t2)+C
We substitute F(8.5)=0 and receive: C=−(138.55arctan(8.5)−8.15ln(73.25)).
Calculate integrals: ∫tarctan(t)dt=2t2arctan(t)−∫2(1+t2)t2dt=2t2arctan(t)−∫(21−2(1+t2)1)dt=2t2+1arctan(t)−2t+C1
∫ln(1+t2)dt=tln(1+t2)−∫(1+t2)2t2dt=tln(1+t2)−∫(2−(1+t2)2)dt=tln(1+t2)−2t+2arctan(t)+C2
We point out that after the change t=x−8.5 the boundary points become: t1=−8.5 and t2=0.
Thus, we receive:
A=16.3∫−8.50(tarctan(t)−21ln(1+t2))dt=16.3(2t2+1arctan(t)−2t)∣−8.50−216.3(tln(1+t2)−2t+2arctan(t))∣−8.50≈615.95
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