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

Express β«_(0)^(2π) 3 sin 5π₯ dπ₯ as the limit of Riemann sums.

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### Video Transcript

Express the definite integral from zero to two π of three times the sin of five π₯ with respect to π₯ as the limit of Riemann sums.

Weβre given the definite integral of a trigonometric function. And instead of evaluating this definite integral, we instead need to express this as the limit of a Riemann sum. So to answer this question, we need to recall how we would represent a definite integral as the limit of Riemann sums. We recall if π is an integrable function on a closed interval from π to π, then we know the definite integral from π to π of π of π₯ with respect to π₯ will be equal to the limit as π approaches infinity of its Riemann sum.

For the right Riemann sum, this will be the limit as π approaches infinity of the sum from π equals one to π of π of π₯ π times β³π₯, where β³π₯ is our interval width, π minus π all divided by π, and π₯ π will be our sample points, π plus π times β³π₯. We want to use this on the definite integral given to us in the question. First, our function π of π₯ will be our integrand three times the sin of five π₯. Next, our values of π and π will be the upper and lower limits of integration, in this case zero and two π.

So to use this, we first need to check that our function π is integrable on our closed interval of integration. In our case, we need to check that our function three sin of five π₯ is integrable on the closed interval from zero to two π. And in this case, the easiest way to do this is to notice that three sin of five π₯ is a trigonometric function. And trigonometric functions are continuous across their entire domain. In fact, the domain of this function is all real values of π₯. And a function being continuous on an interval means that it must be integrable on this interval. So in this case, our function π is integrable on any interval. So in particular, it will be integrable on the closed interval from zero to two π.

This justifies our use of this limit result. But before we can do that, we need to find expressions for β³π₯ and π₯ π. Letβs start with β³π₯. We know β³π₯ is our interval width π minus π all divided by π. We know π is two π and π is zero. So we get β³π₯ is two π minus zero all divided by π. And of course, we can simplify this to give us two π over π. Now that we found β³π₯, weβre ready to find our sample points π₯ π. We know that our sample points π₯ π will be equal to π plus π times β³π₯. We know π is zero and β³π₯ is two π over π. So we get π₯ π is equal to zero plus π times two π over π. And weβll write this as two π π divided by π.

The last thing we need to do to express this limit is find an expression for π evaluated at π₯ π. And to find this, we just need to substitute our expression for π₯ π into our expression for π of π₯. So we substitute two π π over π into three sin of five π₯. This gives us three sin of five times two π π over π. And we can simplify this since five multiplied by two π will give us 10π. So weβve shown that π evaluated at π₯ π is three sin of 10π times π divided by π.

Weβre now ready to evaluate our definite integral as the limit of a right Riemann sum. We can just write in our expression for π evaluated at each of our sample points π₯ π and our expression for β³π₯. This gives us the limit as π approaches infinity of the sum from π equals one to π of three sin of 10π times π divided by π multiplied by two π over π. And weβll just simplify this expression slightly. Instead of multiplying by two π by π at the end of our expression, weβll do this at the start of our expression. And we can also notice three multiplied by two can simplify to give us six.

And doing this gives us our final answer, the limit as π approaches infinity of the sum from π equals one to π of six π by π multiplied by the sin of 10π times π divided by π. And itβs also worth pointing out since we know our integrand is integrable, if we evaluated our definite integral by using techniques such as what we know about integrating trigonometric functions and if we were to evaluate this limit by using what we know about limit results, we would get exactly the same answer.