Express the definite integral
between three and nine of three 𝑥 to the sixth power with respect to 𝑥 as the
limit of Riemann sums.
Remember, if 𝑓 is integrable on
the closed interval 𝑎 to 𝑏, then the definite integral between 𝑎 and 𝑏 of 𝑓 of
𝑥 with respect to 𝑥 can be expressed as the limit of Riemann sums as shown. Let’s compare everything in our
theorem to our integral. Our function is a polynomial. And we know that all polynomials
are continuous over their domain, which means that they are therefore integrable
over their domain. So the function 𝑓 of 𝑥 equals
three 𝑥 to the sixth power is continuous and therefore integrable over the closed
interval defined by the lower limit three and the upper limit nine.
So we’re going to let 𝑎 be equal
to three and 𝑏 be equal to nine. We’ll move on and define Δ𝑥. It’s 𝑏 minus 𝑎 over 𝑛. Well, we said 𝑏 is nine and 𝑎 is
three. And this is all over 𝑛. That gives us that Δ𝑥 is equal to
six over 𝑛. And we can now define 𝑥𝑖. It’s 𝑎 plus 𝑖 lots of Δ𝑥. Well, 𝑎 is three. And we need 𝑖 lots of Δ𝑥, which
we worked out to be six over 𝑛. Let’s write that as three plus six
𝑖 over 𝑛.
In our limit, we’re going to need
to work out 𝑓 of 𝑥𝑖. So it follows that we can find this
by substituting the expression for 𝑥𝑖 into our function. That gives us three times three
plus six 𝑖 over 𝑛 to the sixth power. And we can now substitute
everything we have into our definition for the integral. When we do, we see that the
definite integral between six and nine of three 𝑥 to the sixth power with respect
to 𝑥 is equal to the limit as 𝑛 approaches ∞ of the sum of three times
three plus six 𝑖 over 𝑛 to the sixth power times six over 𝑛 evaluated between 𝑖
equals one and 𝑛.
Since multiplication is
commutative, we can rewrite three times six over 𝑛 as 18 over 𝑛. And we have our definite integral
expressed as the limit of Riemann sums. It’s the limit as 𝑛 approaches
∞ of the sum of 18 over 𝑛 times three plus six 𝑖 over 𝑛 to the sixth power
evaluated from 𝑖 equals one to 𝑛.