Video Transcript
Use the trapezoidal rule to
estimate the integral from zero to two of the cube root of 𝑥 plus one with respect
to 𝑥 using five subintervals. Round your answer to three decimal
places.
The question wants us to
approximate our integral by using the trapezoidal rule. And we recall we can estimate the
integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 as the sum of 𝑛
trapezoids. We set the width of our trapezoid,
Δ𝑥, to be equal to 𝑏 minus 𝑎 divided by 𝑛, where 𝑛 is the number of trapezoids
or the number of subintervals. We then set 𝑥 𝑖 to be equal to 𝑎
plus 𝑖 times Δ𝑥. Then, by using our trapezoidal
rule, we can estimate the area of this integral as Δ𝑥 divided by two multiplied by
𝑓 evaluated at 𝑥 naught plus 𝑓 evaluated at 𝑥 𝑛 plus two times 𝑓 evaluated at
𝑥 one plus 𝑓 evaluated at 𝑥 two. And we add this all the way up to
𝑓 evaluated at 𝑥 𝑛 minus one.
Since we want to use the
trapezoidal rule to estimate the integral from zero to two of the cube root of 𝑥
plus one with respect to 𝑥, we’ll set our function 𝑓 of 𝑥 to be equal to the cube
root of 𝑥 plus one. We see that 𝑎 is the lower limit
of our integral and 𝑏 is the upper limit of our integral. So, we’ll set 𝑎 equal to zero and
𝑏 equal to two. And the question wants us to
approximate this integral by using five subintervals. So, we’ll set 𝑛 equal to five.
We’re now ready to start
calculating some of these values. So, let’s start by calculating the
value of Δ𝑥. We have that Δ𝑥 is equal to 𝑏
minus 𝑎 all divided by 𝑛, which is equal to two minus zero divided by five, which
is equal to two-fifths or 0.4. We see in our estimate we need the
value of Δ𝑥, which we’ve just calculated, but we also need the values of 𝑓
evaluated at each of our values of 𝑥 𝑖. To help us find these values, we’ll
make a table containing the values of 𝑖, 𝑥 𝑖, and 𝑓 evaluated at 𝑥 𝑖.
Since we need to calculate 𝑓
evaluated at 𝑥 zero, 𝑥 one, 𝑥 two all the way up to 𝑥 𝑛, we’re going to need 𝑛
plus one columns for 𝑖. In our case, since we have five
subintervals, we’re going to want the values of 𝑖 from zero to five. We recall that 𝑥 𝑖 is equal to 𝑎
plus 𝑖 times Δ𝑥. We know the value of 𝑎 is zero and
the value of Δ𝑥 is 0.4. So, 𝑥 𝑖 is equal to zero plus 𝑖
times 0.4, which we can simplify to give us 0.4 times 𝑖. And this lets us calculate the
values of 𝑥 𝑖. 𝑥 zero is 0.4 times zero, 𝑥 one
is 0.4 times one, 𝑥 two is 0.4 times two, and this goes up to 𝑥 five, which is 0.4
times five.
We can then calculate each of these
values. We see that 𝑥 zero is zero, 𝑥 one
is 0.4, 𝑥 two is 0.8, 𝑥 three is 1.2, 𝑥 four is 1.6, and 𝑥 five is just two. So, we’ve now found each of our
values of 𝑥 𝑖. To find 𝑓 evaluated at each of
these 𝑥 𝑖, we recall that 𝑓 of 𝑥 is equal to the cube root of 𝑥 plus one. So, we substitute each of our
values of 𝑥 𝑖 into our expression for 𝑓 of 𝑥. For 𝑥 zero, we get the cube root
of zero plus one. For 𝑥 one, we get the cube root of
0.4 plus one, and this goes all the way up to 𝑥 five, where we get the cube root of
two plus one.
We can simplify each of these
expressions by just adding the terms inside of our cube root. This gives us 𝑓 evaluated at 𝑥
zero is the cube root of one, 𝑓 evaluated at 𝑥 one is the cube root of 1.4, 𝑓
evaluated at 𝑥 two is the cube root of 1.8. 𝑓 evaluated at 𝑥 three is the
cube root of 2.2. 𝑓 evaluated at 𝑥 four is the cube
root of 2.6. And 𝑓 evaluated at 𝑥 five is the
cube root of three.
We’re now ready to approximate our
integral by using the trapezoidal rule. First, we need Δ𝑥 divided by two,
in this case, that’s 0.4 divided by two. We multiply this by one lot of 𝑓
evaluated at 𝑥 zero and one lot of 𝑓 evaluated at 𝑥 𝑛. In this case, 𝑛 is five. So, we calculated these earlier to
be the cube root of one and the cube root of three.
We then need to add two times 𝑓
evaluated at the remaining values of 𝑥 𝑖. And we calculated these earlier to
be the cube root of 1.4, the cube root of 1.8, the cube root of 2.2, and the cube
root of 2.6. And we can then calculate this
expression to three decimal places to give us 2.493.
So, by using the trapezoidal rule
with five subintervals, we’ve shown that the integral from zero to two of the cube
root of 𝑥 plus one with respect to 𝑥 is approximately equal to 2.493.
