Science:Math Exam Resources/Courses/MATH102/December 2015/Question 19 (b)/Solution 1

It is impossible for the resident to make any errors at time $t=0$, so $E(0)=0$.

Next, recall that the fundamental theorem of calculus states that

                                                      $f(v)-f(u)=\int _{u}^{v}f'(x)dx$

where $f'(x)$ is the derivative of $f(x)$.

Now, calculate $E(s)$ for any $s>0$ as follows. Let $s>0$. Then in the above equation, set $f=E$, $u=0$, $v=s$, and $x=t$. By the fundamental theorem of calculus

                                                      $E(s)-E(0)=\int _{0}^{s}E'(t)dt$

As $E(0)=0$, it follows that

$E(s)=\int _{0}^{s}E'(t)dt=\int _{0}^{s}4-t+{\frac {1}{16}}t^{2}$

To calculate the above definite integral, we calculate the following indefinite integral (below, recall that 48 is 16 times 3).

$\int 4-t+{\frac {1}{16}}t^{2}dt=\int 4dt-\int tdt+\int {\frac {1}{16}}t^{2}dt=4t-{\frac {1}{2}}t^{2}+{\frac {1}{48}}t^{3}$

Now compute

$\int _{0}^{s}4-t+{\frac {1}{16}}t^{2}=(4s-{\frac {1}{2}}s^{2}+{\frac {1}{48}}s^{3})-(4*0-{\frac {1}{2}}0^{2}+{\frac {1}{48}}0^{3})$

Hence

$E(s)=4s-{\frac {1}{2}}s^{2}+{\frac {1}{48}}s^{3}$.

By simply changing notation (write t instead of s) we can re-write our solution as

$\color {blue}E(t)=4t-{\frac {1}{2}}t^{2}+{\frac {1}{48}}t^{3}$

which is the answer to the question.