Answer to Question #334942 in Calculus for Quân Jason

Question

When a particle is located a distance x meter from the origin, a force F( x )(newton) acts on it. Given F'(x) = 16.3.arctan(x - 8.5), F(8.5)=0. How much work is done in moving it from x=0 to x=8.5.

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Answer 1

We remind that the formula for the work is: $A=\int_{x_1}^{x_2}F(s)ds$, where $x_1=0$, $x_2=8.5$.

$F'(x)=16.3\,\arctan(x-8.5)$. We receive that: $F=16.3\int\arctan(x-8.5)dx$. Make the change $t=x-8.5$. We get: $F=16.3\int\arctan(t)dt=16.3\,t\,\arctan(t)-16.3\int\,\frac{t}{1+t^2}dt=16.3\,t\,\arctan(t)-16.3\int\,\frac12\frac{d(1+t^2)}{1+t^2}=16.3\,t\,\arctan(t)-\frac{16.3}{2}ln(1+t^2)+C$

We substitute $F(8.5)=0$ and receive: $C=-(138.55\,\arctan(8.5)-8.15\,ln(73.25))$.

Calculate integrals: $\int\,t\,\arctan(t)dt=\frac{t^2}{2}\arctan(t)-\int\frac{t^2}{2(1+t^2)}dt=\frac{t^2}{2}\arctan(t)-\int\left(\frac12-\frac{1}{2(1+t^2)}\right)dt=\frac{t^2+1}{2}\arctan(t)-\frac{t}{2}+C_1$

$\int\,ln(1+t^2)dt=t\,ln(1+t^2)-\int\frac{2t^2}{(1+t^2)}dt=t\,ln(1+t^2)-\int\left(2-\frac{2}{(1+t^2)}\right)dt=t\,ln(1+t^2)-2t+2\,\arctan(t)+C_2$

We point out that after the change $t=x-8.5$ the boundary points become: $t_1=-8.5$ and $t_2=0.$

Thus, we receive:

$A=16.3\int_{-8.5}^{0}\left(t\,\arctan(t)-\frac12ln(1+t^2)\right)dt=16.3\left(\frac{t^2+1}{2}\arctan(t)-\frac{t}{2}\right)|_{-8.5}^0-\frac{16.3}{2}\left(t\,ln(1+t^2)-2t+2\,\arctan(t)\right)|_{-8.5}^0\approx615.95$