Use the trapezoidal rule to
estimate the definite integral between zero and two of 𝑥 cubed with respect to 𝑥
using four subintervals.
Remember, the trapezoidal rule says
that we can find an approximation to the definite integral between the limits of 𝑎
and 𝑏 of 𝑓 of 𝑥 by using the calculation Δ𝑥 over two times 𝑓 of 𝑥 nought plus
𝑓 of 𝑥 𝑛 plus two lots of 𝑓 of 𝑥 one plus 𝑓 of 𝑥 two all the way through to
𝑓 of 𝑥 𝑛 minus one. Where Δ𝑥 is equal to 𝑏 minus 𝑎
over 𝑛. And 𝑥 subscript 𝑖 is equal to 𝑎
plus 𝑖 times Δ𝑥.
Let’s break this down and just
begin by working out the value of Δ𝑥. Contextually, Δ𝑥 is the width of
each of our subintervals. In this case, we’re working with
four subintervals. So we could say that 𝑛 is equal to
four. 𝑎 is the lower limit of our
integral. So 𝑎 is equal to zero, where 𝑏 is
the upper limit. And that’s equal to two. Δ𝑥 is therefore two minus zero
over four, which is one-half or 0.5. The values for 𝑓 of 𝑥 nought and
𝑓 of 𝑥 one and so on require a little more work. But we can make this as simple as
possible by adding a table.
It’s useful to remember that there
will always be 𝑛 plus one columns required in the table. So here, that’s four plus one,
which is five. We have five columns in our
table. The 𝑥-values run from 𝑎 to
𝑏. That’s zero to two. And the ones in between are found
by repeatedly adding Δ𝑥, that’s 0.5, to 𝑎, which is zero. That’s 0.5, one, and 1.5. And that gives us our four strips
of width 0.5 units. We’re then simply going to
substitute each 𝑥-value into our function. We begin with 𝑓 of zero. That’s zero cubed, which is
zero. Next, we have 𝑓 of 0.5. That’s 0.5 cubed, which is
0.125. 𝑓 of one is one cubed, which is
still one. And we obtain the final two values
in a similar manner. 𝑓 of 1.5 is 3.375. And 𝑓 of two is eight. And that’s the tricky bit done
with. All that’s left is to substitute
what we know into our formula for the trapezoidal rule.
It’s Δ𝑥 over two, which is 0.5
over two, times the first 𝑓 of 𝑥 value plus the last 𝑓 of 𝑥 value. That’s zero plus eight plus two
lots of everything else. That’s two times 0.125 plus one
plus 3.375. And that gives us a value of 17
over four. So using four subintervals, the
trapezoidal rule gives us the estimate to the definite integral of 𝑥 cubed between
zero and two to be 17 over four. Now where possible, this can be
checked in a number of ways. You could evaluate a Riemann or
midpoint sum or here simply perform the integration.
When we integrate 𝑥 cubed, we get
𝑥 to the fourth power divided by four. Evaluating this between the limits
of zero and two gives us two to the fourth power divided by four minus zero to the
fourth power divided by four, which is 16 over four. And that’s really close to the
answer we got, suggesting we’ve probably performed our calculations correctly.