# Question Video: Using the Trapezoidal Rule to Estimate Integrals Mathematics • Higher Education

Estimate β«_(1) ^(2) (π^(π₯))/(π₯) dπ₯, using the trapezoidal rule with four subintervals. Approximate your answer to two decimal places.

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### Video Transcript

Estimate the definite integral between the limits of one and two of π to the power of π₯ over π₯ dπ₯, using the trapezoidal rule with four subintervals. Approximate your answer to two decimal places.

Remember, the trapezoidal rule says that we can find an estimate for the definite integral of some function π of π₯ between the limits of π and π by performing the calculation Ξπ₯ over two times π of π₯ nought plus π of π₯ π plus two times π of π₯ one plus π of π₯ two all the way through to π of π₯ π minus one. Where Ξπ₯ is π minus π over π, where π is the number of subintervals. And π₯ π is π plus π lots of Ξπ₯. Weβll begin then just simply by working out Ξπ₯. Contextually, Ξπ₯ is the width of each of our subinterval. Here weβre working with four subintervals. So π is equal to four. π is equal to one. And π is equal to two. Ξπ₯ is therefore two minus one over four, which is a quarter or 0.25. Thatβs the perpendicular height of each trapezoid.

The values for π of π₯ nought and π of π₯ one and so on require a little more work. But we can make this as simple as possible by including a table. We recall that there will always be one more π of π₯ value than the number of subintervals. So here, thatβs going to be four plus one, which is five π of π₯ values. The π₯-values themselves run from π to π. Thatβs here from one to two. And the ones in between are found by repeatedly adding Ξπ₯, thatβs 0.25, to π, which is one. So these values are 1.25, 1.5, and 1.75. And that gives us our four strips of width 0.25 units. Weβre then going to substitute each π₯-value into our function.

Here, weβre going to need to make a decision on accuracy. Whilst the question tells us to use an accuracy of two decimal places, thatβs only for our answer. A good rule of thumb is to use at least five decimal places. We begin with π of one. Thatβs π to the power of one over one, which is 2.71828, correct to five decimal places. We have π of 1.25, which is π to the power of 1.25 divided by 1.25. Thatβs, correct to five decimal places, 2.79227. We repeat this process for 1.5. π of 1.5 is 2.98779. π of 1.75 is 3.28834. And π of two is 3.69453 rounded to five decimal places. All thatβs left is to substitute what we know into our formula for the trapezoidal rule. Itβs Ξπ₯ over two. Thatβs 0.25 over two times π of one. Thatβs 2.71828 plus π of two. Thatβs 3.69453 plus two lots of everything else essentially. Thatβs 2.79227, 2.98779, and 3.28834. That gives us 3.0687, which, correct to two decimal places, is 3.07.

Itβs useful to remember that we can check whether this answer is likely to be sensible by using the integration function on our calculator. And when we do, we get 3.06 correct to two decimal places. Thatβs really close to the answer we got, suggesting weβve probably performed our calculations correctly. And so an approximation to the integral evaluated between one and two of π to the power of π₯ over π₯ dπ₯ is 3.07.