Question Video: Expressing a Definite Integral as the Limit of a Riemann Sum | Nagwa Question Video: Expressing a Definite Integral as the Limit of a Riemann Sum | Nagwa

# Question Video: Expressing a Definite Integral as the Limit of a Riemann Sum Mathematics • Higher Education

Express β«_(3)^(9) 3π₯βΆ dπ₯ as the limit of Riemann sums.

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### Video Transcript

Express the definite integral between three and nine of three π₯ to the sixth power with respect to π₯ as the limit of Riemann sums.

Remember, if π is integrable on the closed interval π to π, then the definite integral between π and π of π of π₯ with respect to π₯ can be expressed as the limit of Riemann sums as shown. Letβ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 π of π₯ equals three π₯ 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 weβre going to let π be equal to three and π be equal to nine. Weβll move on and define Ξπ₯. Itβs π minus π over π. Well, we said π is nine and π is three. And this is all over π. That gives us that Ξπ₯ is equal to six over π. And we can now define π₯π. Itβs π plus π lots of Ξπ₯. Well, π is three. And we need π lots of Ξπ₯, which we worked out to be six over π. Letβs write that as three plus six π over π.

In our limit, weβre going to need to work out π of π₯π. So it follows that we can find this by substituting the expression for π₯π into our function. That gives us three times three plus six π over π 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 π₯ to the sixth power with respect to π₯ is equal to the limit as π approaches β of the sum of three times three plus six π over π to the sixth power times six over π evaluated between π equals one and π.

Since multiplication is commutative, we can rewrite three times six over π as 18 over π. And we have our definite integral expressed as the limit of Riemann sums. Itβs the limit as π approaches β of the sum of 18 over π times three plus six π over π to the sixth power evaluated from π equals one to π.

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