Question Video: Estimating a Definite Integral Using the Trapezoidal Rule with Five Subintervals

Estimate ∫_(0)^(1) 2𝑒^(π‘₯Β² + 3π‘₯) dπ‘₯ using the trapezoidal rule with five subintervals. Round your answer to one decimal place.

05:19

Video Transcript

Estimate the integral from zero to one of two times 𝑒 to the power of π‘₯ squared plus three π‘₯ with respect to π‘₯ using the trapezoidal rule with five subintervals. Round your answer to one decimal place.

The question is asking us to estimate this integral by using the trapezoidal rule. And we recall we can estimate the definite integral from π‘Ž to 𝑏 of a function 𝑓 of π‘₯ with respect to π‘₯ by using the trapezoidal rule with 𝑛 subintervals. As 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.

Well, Ξ”π‘₯ is equal to 𝑏 minus π‘Ž over 𝑛. And each of our π‘₯ 𝑖 are equal to π‘Ž plus 𝑖 times Ξ”π‘₯. We get this approximation since the definite integral is an area. And we can approximate this area by using 𝑛 trapezoids of width Ξ”π‘₯. Since we want to use the trapezoidal rule with five subintervals, we’ll set our value of 𝑛 equal to five. We see the lower limit of our integral is zero and the upper limit is one. So we’ll set our value of π‘Ž equal to zero and our value of 𝑏 equal to one. Finally, we’ll set our function 𝑓 of π‘₯ equal to our integrand two times 𝑒 to the power of π‘₯ squared plus three π‘₯.

The first thing we want to do is calculate 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 one, π‘Ž is equal to zero, and 𝑛 is equal to five. So Ξ”π‘₯ is equal to one minus zero over five, which is equal to one-fifth or 0.2. So we’ve found the value of Ξ”π‘₯. We’ve found the first part of our trapezoidal formula.

We now need to calculate 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 𝑛 is equal to five, we need to find the value of π‘₯ five. We also need to find the value of π‘₯ four, π‘₯ three, π‘₯ two, π‘₯ one, and π‘₯ zero. So we’ll need six columns for our values of 𝑖. In fact, we’ll always need one more column than the value of 𝑛 for our values of 𝑖.

To find our 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 zero and Ξ”π‘₯ is equal to 0.2, this gives us that π‘₯ 𝑖 is equal to zero plus 𝑖 times 0.2, which simplifies to give us 0.2 times 𝑖. We can then substitute the values of 𝑖 into this equation to find the values for π‘₯ 𝑖.

Substituting 𝑖 is equal to zero gives us that π‘₯ zero is equal to 0.2 times zero, which is zero. Substituting 𝑖 is equal to one gives us that π‘₯ one is equal to 0.2 times one, which is 0.2. We can do the same to find the rest of our values of π‘₯ 𝑖. We have π‘₯ two is equal to 0.4, π‘₯ three is equal to 0.6, π‘₯ four is equal to 0.8, and π‘₯ five is equal to one.

Now we need to find our function 𝑓 evaluated at each of these values of π‘₯ 𝑖. We recall that 𝑓 of π‘₯ is equal to our integrand two times 𝑒 to the power of π‘₯ squared plus three π‘₯. Substituting π‘₯ 𝑖 is equal to zero, we get two times 𝑒 to the power of zero squared plus three times zero. And our exponent of zero squared plus three times zero simplifies to give us zero. And 𝑒 to the power of zero is just equal to one. So 𝑓 evaluated at π‘₯ zero is just equal to two.

We can do the same for π‘₯ one. We substitute 0.2 into our function 𝑓 of π‘₯, to give us two times 𝑒 to the power of 0.2 squared plus three times 0.2. And again, we can evaluate the expression in our exponent. This time, we get an exponent of 0.64. So we have 𝑓 evaluated at π‘₯ one is two times 𝑒 to the power of 0.64.

We can do the same to find 𝑓 evaluated at the rest of our values of π‘₯ 𝑖. We get 𝑓 evaluated at π‘₯ two is two times 𝑒 to the power of 1.36. 𝑓 evaluated at π‘₯ three is two times 𝑒 to the power of 2.16. 𝑓 evaluated at π‘₯ four is two times 𝑒 to the power of 3.04. And 𝑓 evaluated at π‘₯ five is two times 𝑒 to the fourth power.

We’ve now calculated all of the values we need to estimate our integral by using the trapezoidal rule with five subintervals. We’ve shown that Ξ”π‘₯ is equal to 0.2. So Ξ”π‘₯ over two is equal to 0.2 divided by two. Since 𝑛 is equal to five, we’ve shown that 𝑓 evaluated at π‘₯ zero and 𝑓 evaluated at π‘₯ five are equal to two and two times 𝑒 to the fourth power, respectively. Then we need to add 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‘₯ four. And we calculated 𝑓 evaluated at π‘₯ one, two, three, and four in our table.

So we just need to calculate this expression. And remember, the question wants us to give our answer to one decimal place. So we need to calculate the value of this expression to one decimal place. And we get 25.3. Therefore, by using the trapezoidal rule with five subintervals, we’ve shown that the integral from zero to one of two times 𝑒 to the power of π‘₯ squared plus three π‘₯ with respect to π‘₯ is approximately equal to 25.3.

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