Effective GDC Use #2
Wednesday 19 September 2018
Checking tool
Continuing my discussion on promoting effective and wise use of a calculator (GDC) by our students, I wish to highlight what I consider maybe the first and simplest piece of GDC advice to students: “When possible, use your GDC as a checking tool.” When a Paper 2 question (GDC allowed) requires an exact answer, it usually means that the solution needs to be carried out manually / analytically – and a GDC will not be helpful in obtaining the answer. However, in such a situation a GDC can often be used to quickly and easily check whether the exact answer obtained is correct.
For example, consider the following exam-like (Paper 2) question.
Consider the region R bounded by the graph of \(y = x - \frac{1}{x}\), the x-axis and the line 
.
(a) Find the exact area of the region R.
(b) A solid of revolution is formed by rotating the region R about the x-axis. Find the exact volume of this solid.
Solution:
(a) Region R is shaded in the diagram at right.
area of R\( = \int_1^2 {\left( {x - \frac{1}{x}} \right)dx = \left. {\frac{{{x^2}}}{2} - \ln x} \right]} _1^2\)
\( = \left( {\frac{{{2^2}}}{2} - \ln 2} \right) - \left( {\frac{{{1^2}}}{2} - \ln 1} \right) = \frac{3}{2} - \ln 2\)
(b) volume
\( = {\pi }\int_1^2 {\left( {{{\left( {x - \frac{1}{x}} \right)}^2}} \right)dx = {\pi }} \int_1^2 {\left( {{x^2} + \frac{1}{{{x^2}}} - 2} \right)dx = {\pi }\left. {\left( {\frac{{{x^3}}}{3} - \frac{1}{x} - 2x} \right)} \right]} _1^2\)
\( = {\pi }\left[ {\left( {\frac{{{2^3}}}{3} - \frac{1}{2} - 2 \cdot 2} \right) - \left( {\frac{{{1^3}}}{3} - \frac{1}{1} - 2 \cdot 1} \right)} \right] = {\pi }\left[ { - \frac{{11}}{6} - \left( { - \frac{8}{3}} \right)} \right] = \frac{5}{6}{\pi unit}{{\textrm{s}}^3}\)
With a GDC allowed on this question, it’s certainly a good idea to check the answers for (a) and (b) on a GDC, as shown below.
One of the simplest and most powerful uses of a GDC is to use it to check the correctness, accuracy or reasonableness of an answer.