Question Video: Finding the Riemann Sum of a Trigonometric Function on a Given Interval by Dividing It into Subintervals and Using the Midpoints of the Subintervals | Nagwa Question Video: Finding the Riemann Sum of a Trigonometric Function on a Given Interval by Dividing It into Subintervals and Using the Midpoints of the Subintervals | Nagwa

# Question Video: Finding the Riemann Sum of a Trigonometric Function on a Given Interval by Dividing It into Subintervals and Using the Midpoints of the Subintervals Mathematics • Higher Education

Estimate β«_(1) ^(9) 5 sin (2β(3π₯)) dπ₯ using the midpoint rule with π = 4, giving your answer to four decimal places.

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### Video Transcript

Estimate the definite integral between one and nine of five sin of two times the square root of three π₯dπ₯ using the midpoint rule with π equals four, giving your answer to four decimal places.

Remember, we can estimate the value of a definite integral by using Riemann sums. We split the region into π sub intervals and create a rectangle in each. The total area of the rectangles gives us an estimate to our integral. In a midpoint Riemann sum, the height of each rectangle is equal to the value of the function at the midpoint of its base. And whilst itβs not entirely necessary, letβs sketch the graph of five sin of two root three π₯ out so we can see whatβs going on.

The graph of π¦ equals five sin of two root three π₯ looks a little something like this. Weβre going to split our region into four sub intervals. So, before we add any rectangles to our diagram, weβll work out the width of each sub interval. Geometrically, this tells us the width of our rectangles. Itβs given by β³π₯ equals π minus π over π, where π and π are the lower and upper limits of the definite integral, and π is the number of sub intervals. In this case, weβre going to let π be equal to one and π be equal to nine. And weβre told that π equals four.

So, the width of each of our rectangles is nine minus one over four, which is two units. This means our first rectangle will be between one and two. The height of this rectangle is the value of the function at the midpoint. So, thatβs at the point where π₯ equals two. In other words, the height of the rectangle will be π of two. Now, of course, this rectangle sits below the π₯-axis, so that will give us a negative value. Which is absolutely fine because eventually weβre going to subtract the area of the rectangles that lie below the π₯-axis from the area of the rectangles that lie above them.

Our next rectangle is going to go from π₯ equals three to π₯ equals five. The height of this rectangle will be the value of the function at the midpoint. Thatβs at the point where π₯ equals four. So, the height of this rectangle will be the value of π of four. Our third rectangle goes from π₯ equals five to π₯ equals seven. And, once again, the height of the rectangle is the value of the function at the midpoint. So, thatβs when π₯ equals six. Its height will be the value of π of six.

Our fourth and final rectangle will go from π₯ equals seven to π₯ equals nine. We notice once again that we have a rectangle that sits below the π₯-axis, so calculating the area will give us a negative value. And thatβs fine because we want to subtract the area from the area of the rectangles that sit above the π₯-axis. This time, the height of this rectangle is the value of the function when π₯ equals eight. So, weβre going to need to work out π of eight.

Now, we said that an estimate for the definite integral is the sum of the areas of the rectangles. Take into account that weβre going to subtract the areas of the rectangle that lie below the π₯-axis. And we know that the area of a rectangle is given by base multiplied by height. And so, the negative area of the first rectangle will be π of two multiplied by two. π of four multiplied by two will give us the area of the second rectangle. π of six multiplied by two will give us the positive area of the third rectangle. And π of eight multiplied by two will give us the negative area of the fourth rectangle.

We can work out π of two, π of four, π of six, and π of eight by substituting two, four, six, and eight into our function, respectively. And so, an estimate for our definite integral is five sin of two root six times two plus five sin of two root 12 times two plus five sin of two root 18 times two plus five sin of two root 24 times two. Thatβs 0.6125207 and so on, which, correct to four decimal places, is 0.6125. And so, using the midpoint rule, weβve estimated the value of our definite integral. Itβs 0.6125.

Now, throughout the video, I mentioned that we would subtract any areas of rectangles that lie below the π₯-axis. And at this step, it may have seen that we were adding these. However, if we work out the value of five sin of two root six and five sin of two root 24, we get a negative value as we would expect. And so, multiplying each of these individually by two maintains a negative. And these end up getting subtracted from the area of the other two rectangles. So, this does all of our work for us. The definite integral is approximately 0.6125.

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