### Video Transcript

Estimate the integral from two to
six of two times the square root of three π₯ with respect to π₯ using the
trapezoidal rule with four subintervals. Round your answer to one decimal
place.

The question wants us to estimate
our integral by using the trapezoidal rule. And we recall we can approximate a
definite integral by using the trapezoidal rule with π subintervals by ... . And we recall we can approximate
the definite integral from π to π of π of π₯ with respect to π₯ by using the
trapezoidal rule with π subintervals by setting Ξπ₯ equal to π minus π divided by
π. This then gives us the widths of
our trapezoids.

To find the heights of our
trapezoids, we first need to set π₯π equal to π plus π times Ξπ₯. Each of these values of π₯ π will
be where on the π₯-axis our trapezoids lie. Then we can find the heights of our
trapezoids by evaluating π at these values of π₯ π. If we then calculate the areas of
all of these trapezoids, weβll get an approximation for our area. This gives us our formula Ξπ₯ over
two multiplied by π evaluated at π₯ zero plus π evaluated at π₯ π plus two times
π evaluated at π₯ one plus π evaluated at π₯ two. And we add this all the way up to
π evaluated at π₯ π minus one.

In our case, we want to use the
trapezoidal rule with four subintervals. So weβll set our value of π equal
to four. We see that the lower limit of our
integral is two and the upper limit is six. So weβll set π equal to two and π
equal to six. And weβll set our function π of π₯
equal to our integrand two times the square root of three π₯.

Letβs start by calculating the
value of Ξπ₯. We have that Ξπ₯ is equal to π
minus π over π. We know that π is equal to six, π
is equal to two, and π is equal to four. So Ξπ₯ is equal to six minus two
divided by four, which we can evaluate to give us one. So weβve now found the value of Ξπ₯
in our trapezoidal formula.

However, we need to find π
evaluated at each of our values of π₯ π. To help us calculate these values,
weβll make a table containing our values of π, π₯ π, and π evaluated at π₯
π. In our case, π is equal to
four. So since we need to calculate π
evaluated at π₯ four, π evaluated at π₯ three, π evaluated at π₯ two, π evaluated
at π₯ one, and π evaluated at π₯ zero, we will need five columns for our values of
π.

Itβs true in general we will always
need one more column for our values of π than the number of subintervals. We now want to calculate the values
of π₯ π. We recall that π₯ π is equal to π
plus π times Ξπ₯. Since π is equal to two and Ξπ₯ is
equal to one, we get that π₯ π is equal to two plus π times one, which is of
course just equal to two plus π.

So we can find each of our values,
π₯ π, by substituting the value of π into two plus π. This gives us π₯ zero as two plus
zero, π₯ one as two plus one, and this goes up to π₯ four, which is two plus
four. We can then evaluate each of these
expressions to get π₯ zero is two, π₯ one is three, π₯ two is four, π₯ three is
five, and π₯ four is six.

Remember that our function π of π₯
is equal to our integrand two times the square root of three π₯. So we just need to substitute our
values of π₯ π into this function. Substituting π₯ is equal to two
gives us two times the square root of three times two. Substituting π₯ is equal to three
gives us two times the square root of three times three. And we do this all the way up to
substituting π₯ is equal to six, which gives us two times the square root of three
times six. We can then evaluate the
expressions inside of the square root sign to get two root six, two root nine, two
root 12, two root 15, and two root 18.

Weβve now calculated all the values
we need to estimate our integral by using the trapezoidal rule with four
subintervals. Our estimation is Ξπ₯ over two. And we know Ξπ₯ is equal to
one. We then multiply this by π
evaluated at π₯ zero plus π evaluated at π₯ four, which we calculated to be two
root six and two root 18. We add to this two times π
evaluated at the remaining values of π₯ π: two root nine, two root 12, and two root
15. And if we evaluate this expression
to one decimal place, we get 27.4.

So by using the trapezoidal rule
with four subintervals, weβve shown that the integral from two to six of two times
the square root of three π₯ with respect to π₯ is approximately equal to 27.4.