Question Video: Using the Trapezoidal Rule with Four Subintervals to Estimate a Definite Integral | Nagwa Question Video: Using the Trapezoidal Rule with Four Subintervals to Estimate a Definite Integral | Nagwa

Question Video: Using the Trapezoidal Rule with Four Subintervals to Estimate a Definite Integral Mathematics

Estimate ∫_(2) ^(6) 2√(3π‘₯) dπ‘₯ using the trapezoidal rule with four subintervals. Round your answer to one decimal place.

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Video Transcript

Estimate the integral from two to six of two times the square root of three π‘₯ with respect to π‘₯ using the trapezoidal rule with four subintervals. Round your answer to one decimal place.

The question wants us to estimate our integral by using the trapezoidal rule. And we recall we can approximate a definite integral by using the trapezoidal rule with 𝑛 subintervals by ... . And we recall we can approximate the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ by using the trapezoidal rule with 𝑛 subintervals by setting Ξ”π‘₯ equal to 𝑏 minus π‘Ž divided by 𝑛. This then gives us the widths of our trapezoids.

To find the heights of our trapezoids, we first need to set π‘₯𝑖 equal to π‘Ž plus 𝑖 times Ξ”π‘₯. Each of these values of π‘₯ 𝑖 will be where on the π‘₯-axis our trapezoids lie. Then we can find the heights of our trapezoids by evaluating 𝑓 at these values of π‘₯ 𝑖. If we then calculate the areas of all of these trapezoids, we’ll get an approximation for our area. This gives us our formula Ξ”π‘₯ over two multiplied by 𝑓 evaluated at π‘₯ zero plus 𝑓 evaluated at π‘₯ 𝑛 plus two times 𝑓 evaluated at π‘₯ one plus 𝑓 evaluated at π‘₯ two. And we add this all the way up to 𝑓 evaluated at π‘₯ 𝑛 minus one.

In our case, we want to use the trapezoidal rule with four subintervals. So we’ll set our value of 𝑛 equal to four. We see that the lower limit of our integral is two and the upper limit is six. So we’ll set π‘Ž equal to two and 𝑏 equal to six. And we’ll set our function 𝑓 of π‘₯ equal to our integrand two times the square root of three π‘₯.

Let’s start by calculating the value of Ξ”π‘₯. We have that Ξ”π‘₯ is equal to 𝑏 minus π‘Ž over 𝑛. We know that 𝑏 is equal to six, π‘Ž is equal to two, and 𝑛 is equal to four. So Ξ”π‘₯ is equal to six minus two divided by four, which we can evaluate to give us one. So we’ve now found the value of Ξ”π‘₯ in our trapezoidal formula.

However, we need to find 𝑓 evaluated at each of our values of π‘₯ 𝑖. To help us calculate these values, we’ll make a table containing our values of 𝑖, π‘₯ 𝑖, and 𝑓 evaluated at π‘₯ 𝑖. In our case, 𝑛 is equal to four. So since we need to calculate 𝑓 evaluated at π‘₯ four, 𝑓 evaluated at π‘₯ three, 𝑓 evaluated at π‘₯ two, 𝑓 evaluated at π‘₯ one, and 𝑓 evaluated at π‘₯ zero, we will need five columns for our values of 𝑖.

It’s true in general we will always need one more column for our values of 𝑖 than the number of subintervals. We now want to calculate the values of π‘₯ 𝑖. We recall that π‘₯ 𝑖 is equal to π‘Ž plus 𝑖 times Ξ”π‘₯. Since π‘Ž is equal to two and Ξ”π‘₯ is equal to one, we get that π‘₯ 𝑖 is equal to two plus 𝑖 times one, which is of course just equal to two plus 𝑖.

So we can find each of our values, π‘₯ 𝑖, by substituting the value of 𝑖 into two plus 𝑖. This gives us π‘₯ zero as two plus zero, π‘₯ one as two plus one, and this goes up to π‘₯ four, which is two plus four. We can then evaluate each of these expressions to get π‘₯ zero is two, π‘₯ one is three, π‘₯ two is four, π‘₯ three is five, and π‘₯ four is six.

Remember that our function 𝑓 of π‘₯ is equal to our integrand two times the square root of three π‘₯. So we just need to substitute our values of π‘₯ 𝑖 into this function. Substituting π‘₯ is equal to two gives us two times the square root of three times two. Substituting π‘₯ is equal to three gives us two times the square root of three times three. And we do this all the way up to substituting π‘₯ is equal to six, which gives us two times the square root of three times six. We can then evaluate the expressions inside of the square root sign to get two root six, two root nine, two root 12, two root 15, and two root 18.

We’ve now calculated all the values we need to estimate our integral by using the trapezoidal rule with four subintervals. Our estimation is Ξ”π‘₯ over two. And we know Ξ”π‘₯ is equal to one. We then multiply this by 𝑓 evaluated at π‘₯ zero plus 𝑓 evaluated at π‘₯ four, which we calculated to be two root six and two root 18. We add to this two times 𝑓 evaluated at the remaining values of π‘₯ 𝑖: two root nine, two root 12, and two root 15. And if we evaluate this expression to one decimal place, we get 27.4.

So by using the trapezoidal rule with four subintervals, we’ve shown that the integral from two to six of two times the square root of three π‘₯ with respect to π‘₯ is approximately equal to 27.4.

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