Video: Use the Trapezoidal Rule to Estimate the Area under a Logarithmic Function

Estimate ∫_(1/2)^(2) 3 ln (4π‘₯) dπ‘₯ using the trapezoidal rule with three subintervals. Round your answer to one decimal place.

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Video Transcript

Estimate the integral from one-half to two of three times the natural logarithm of four π‘₯ with respect to π‘₯ using the trapezoidal rule with three subintervals. Round your answer to one decimal place.

The question wants us to estimate an integral by using the trapezoidal rule. And we recall we can approximate the interval from π‘Ž to 𝑏 of a function 𝑓 of π‘₯ with respect to π‘₯ by using the trapezoidal rule with 𝑛 subintervals. This tells us our integral was approximately equal to 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 all the way up to 𝑓 evaluated at π‘₯ 𝑛 minus one. Where Ξ”π‘₯ is equal to 𝑏 minus π‘Ž over 𝑛, and each of our values π‘₯ 𝑖 is equal to π‘Ž plus 𝑖 times Ξ”π‘₯.

The question wants us to use the trapezoidal rule with three subintervals, so we’ll set our value of 𝑛 equal to three. We also see the lower limit of our integral is a half and the upper limit is two. So we’ll set our value of π‘Ž equal to one-half and our value of 𝑏 equal to two. Finally, we’ll set our function 𝑓 of π‘₯ to be our integrand, three times the natural logarithm of four π‘₯. Let’s start by calculating the value of Ξ”π‘₯. We have that Ξ”π‘₯ is equal to 𝑏 minus π‘Ž divided by 𝑛. We know that 𝑏 is equal to two, π‘Ž is equal to one-half, and 𝑛 is equal to three. So this gives us Ξ”π‘₯ is equal to two minus one-half divided by three, which we can evaluate to give us one-half.

So we’ve found the value of Ξ”π‘₯. To approximate our integral using the trapezoidal rule, we need to evaluate 𝑓 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. Since 𝑛 is equal to three, there’ll be four values of π‘₯ 𝑖. There’ll be π‘₯ zero, π‘₯ one, π‘₯ two, and π‘₯ three. So we’ll need four columns for our value of 𝑖. This is true in general. We’ll always need one more column for our values of 𝑖 than we took subintervals. To find 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 one-half and Ξ”π‘₯ is also equal to one-half, this gives us that π‘₯ 𝑖 is equal to one-half plus 𝑖 times one-half, which simplifies to give us a half plus 𝑖 over two.

So we can substitute our values of 𝑖 into this equation to find π‘₯ 𝑖. This gives us that π‘₯ zero is equal to one-half plus zero over two. And then π‘₯ one is equal to a half plus one over two. And we can do the same to see that π‘₯ two is one-half plus two over two and π‘₯ three is one-half plus three over two. We can then evaluate each of these expressions. We get π‘₯ zero is one-half, π‘₯ one is one, π‘₯ two is three over two, and π‘₯ three is two. Finally, to evaluate 𝑓 for each of these values of π‘₯ 𝑖, we recall that 𝑓 of π‘₯ is equal to three times the natural logarithm of four π‘₯. So we then substitute each of our values of π‘₯ 𝑖 into our function 𝑓.

Substituting π‘₯ zero, which is one-half, into our function 𝑓 gives us three times the natural logarithm of four times one-half. Substituting π‘₯ one, which is equal to one, into our function 𝑓 gives us three times the natural logarithm of four times one. And we can do this all the way up to π‘₯ three which is equal to two. We substitute this into our function 𝑓 to get three times the natural logarithm of four times two. Finally, we can simplify each of these expressions by evaluating the expression inside of our natural logarithm function. Doing this, we get π‘₯ zero is three times the natural logarithm of two. π‘₯ one is three times the natural logarithm of four. π‘₯ two is three times the natural logarithm of six. And π‘₯ three is three times the natural logarithm of eight.

We’ve now calculated all of the values we need to approximate our integral using the trapezoidal rule. We’ve shown that Ξ”π‘₯ is equal to one-half, so Ξ”π‘₯ over two is equal to one-half divided by two. Our number of subintervals 𝑛 was equal to three, so we calculated 𝑓 evaluated at π‘₯ zero and 𝑓 evaluated at π‘₯ three to be three times the natural logarithm of two and three times the natural logarithm of eight. Finally, we need to add two times 𝑓 evaluated at π‘₯ one plus 𝑓 evaluated at π‘₯ two. We calculated these to be three times the natural logarithm of four and three times the natural logarithm of six.

Finally, the question wants us to round our answer to one decimal place. So if we evaluate this expression to one decimal place, we get 6.8. Therefore, by using the trapezoidal rule with three subintervals, we’ve shown that the integral from one-half to two of three times the natural logarithm of four π‘₯ with respect to π‘₯ is approximately equal to 6.8.

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