Video: Using the Trapezoidal Rule to Estimate the Integration of a Cubic Root Function Over a Given Interval

Use the trapezoidal rule to estimate ∫_(0) ^(2) ³√(π‘₯ + 1) dπ‘₯ using five subintervals. Round your answer to three decimal places.

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Video Transcript

Use the trapezoidal rule to estimate the integral from zero to two of the cube root of π‘₯ plus one with respect to π‘₯ using five subintervals. Round your answer to three decimal places.

The question wants us to approximate our integral by using the trapezoidal rule. And we recall we can estimate the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ as the sum of 𝑛 trapezoids. We set the width of our trapezoid, Ξ”π‘₯, to be equal to 𝑏 minus π‘Ž divided by 𝑛, where 𝑛 is the number of trapezoids or the number of subintervals. We then set π‘₯ 𝑖 to be equal to π‘Ž plus 𝑖 times Ξ”π‘₯. Then, by using our trapezoidal rule, we can estimate the area of this integral as Ξ”π‘₯ divided by two multiplied by 𝑓 evaluated at π‘₯ naught 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.

Since we want to use the trapezoidal rule to estimate the integral from zero to two of the cube root of π‘₯ plus one with respect to π‘₯, we’ll set our function 𝑓 of π‘₯ to be equal to the cube root of π‘₯ plus one. We see that π‘Ž is the lower limit of our integral and 𝑏 is the upper limit of our integral. So, we’ll set π‘Ž equal to zero and 𝑏 equal to two. And the question wants us to approximate this integral by using five subintervals. So, we’ll set 𝑛 equal to five.

We’re now ready to start calculating some of these values. So, let’s start by calculating the value of Ξ”π‘₯. We have that Ξ”π‘₯ is equal to 𝑏 minus π‘Ž all divided by 𝑛, which is equal to two minus zero divided by five, which is equal to two-fifths or 0.4. We see in our estimate we need the value of Ξ”π‘₯, which we’ve just calculated, but we also need the values of 𝑓 evaluated at each of our values of π‘₯ 𝑖. To help us find these values, we’ll make a table containing the values of 𝑖, π‘₯ 𝑖, and 𝑓 evaluated at π‘₯ 𝑖.

Since we need to calculate 𝑓 evaluated at π‘₯ zero, π‘₯ one, π‘₯ two all the way up to π‘₯ 𝑛, we’re going to need 𝑛 plus one columns for 𝑖. In our case, since we have five subintervals, we’re going to want the values of 𝑖 from zero to five. We recall that π‘₯ 𝑖 is equal to π‘Ž plus 𝑖 times Ξ”π‘₯. We know the value of π‘Ž is zero and the value of Ξ”π‘₯ is 0.4. So, π‘₯ 𝑖 is equal to zero plus 𝑖 times 0.4, which we can simplify to give us 0.4 times 𝑖. And this lets us calculate the values of π‘₯ 𝑖. π‘₯ zero is 0.4 times zero, π‘₯ one is 0.4 times one, π‘₯ two is 0.4 times two, and this goes up to π‘₯ five, which is 0.4 times five.

We can then calculate each of these values. We see that π‘₯ zero is zero, π‘₯ one is 0.4, π‘₯ two is 0.8, π‘₯ three is 1.2, π‘₯ four is 1.6, and π‘₯ five is just two. So, we’ve now found each of our values of π‘₯ 𝑖. To find 𝑓 evaluated at each of these π‘₯ 𝑖, we recall that 𝑓 of π‘₯ is equal to the cube root of π‘₯ plus one. So, we substitute each of our values of π‘₯ 𝑖 into our expression for 𝑓 of π‘₯. For π‘₯ zero, we get the cube root of zero plus one. For π‘₯ one, we get the cube root of 0.4 plus one, and this goes all the way up to π‘₯ five, where we get the cube root of two plus one.

We can simplify each of these expressions by just adding the terms inside of our cube root. This gives us 𝑓 evaluated at π‘₯ zero is the cube root of one, 𝑓 evaluated at π‘₯ one is the cube root of 1.4, 𝑓 evaluated at π‘₯ two is the cube root of 1.8. 𝑓 evaluated at π‘₯ three is the cube root of 2.2. 𝑓 evaluated at π‘₯ four is the cube root of 2.6. And 𝑓 evaluated at π‘₯ five is the cube root of three.

We’re now ready to approximate our integral by using the trapezoidal rule. First, we need Ξ”π‘₯ divided by two, in this case, that’s 0.4 divided by two. We multiply this by one lot of 𝑓 evaluated at π‘₯ zero and one lot of 𝑓 evaluated at π‘₯ 𝑛. In this case, 𝑛 is five. So, we calculated these earlier to be the cube root of one and the cube root of three.

We then need to add two times 𝑓 evaluated at the remaining values of π‘₯ 𝑖. And we calculated these earlier to be the cube root of 1.4, the cube root of 1.8, the cube root of 2.2, and the cube root of 2.6. And we can then calculate this expression to three decimal places to give us 2.493.

So, by using the trapezoidal rule with five subintervals, we’ve shown that the integral from zero to two of the cube root of π‘₯ plus one with respect to π‘₯ is approximately equal to 2.493.

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