### Video Transcript

Estimate the integral from negative one to one of two times π to the power of negative three π₯ with respect to π₯ using the trapezoidal rule with four subintervals. Round your answer to one decimal place.

The question gives us a definite integral. And it wants us to estimate the value of this definite integral by using the trapezoidal rule with four subintervals. It wants us to round our answer to one decimal place. Letβs start by recalling the trapezoidal rule. This tells us we can estimate the integral from π to π of π of π₯ with respect to π₯ by first splitting the interval from π to π into π equal-width subintervals. These will have width Ξπ₯, which is equal to π minus π divided by π where π is the number of subintervals. Next, we take our sample points π₯ π, which are the endpoints of these intervals. Each π₯ π is equal to π plus π times Ξπ₯.

We then evaluate our function π of π₯ at each of our sample points π₯ π. We can then use this information to create trapezoids on each of our subintervals. Weβll then use these to estimate the integral from π to π of π of π₯ with respect to π₯. We do this by adding the areas of the trapezoids above the π₯-axis and subtracting the areas of the trapezoids below the π₯-axis. This gives us our formula. Our definite integral is approximately equal to Ξπ₯ over two times π of π₯ naught plus π of π₯ one plus two times π of π₯ one plus π of π₯ two, and we add all the way up to π of π₯ π minus one.

From the question, weβre asked to estimate the integral from negative one to one of two π to the power of negative three π₯ by using the trapezoidal rule with four subintervals. So weβll set π equal to negative one, π equal to one, π of π₯ equal to two π to the power of negative three π₯, and π our number of subintervals equal to four. Weβll start by using this information to calculate Ξπ₯. Itβs equal to π minus π over π, which in this case is one minus negative one divided by four, which we can calculate is equal to one-half. We now want to find the values of our sample points π₯ π and π evaluated at these π₯ π. Weβll collect this information in a table.

Remember, since we have four subintervals, weβll have five endpoints. So we need π ranging from zero to four. Letβs start by finding π₯ zero. We can see that itβs equal to π plus zero times Ξπ₯. We have that π is equal to negative one and Ξπ₯ is equal to one-half. So π₯ zero is equal to negative one plus zero times one-half, which we can calculate to give us negative one. We can then do the same to find π₯ one. Itβs equal to negative one plus one times one-half, which we can calculate is equal to negative one-half. We can then do something similar to find the rest of our sample points. We get π₯ two is equal to zero, π₯ three is equal to one-half, and π₯ four is equal to one.

We now want to evaluate π at our sample points. Letβs start with π of π₯ zero. Itβs equal to two times π to the power of negative one times negative three. Weβll just simplify this to give us two π cubed. We can then do something similar to evaluate π for the rest of our sample points. Weβll write our sample points in terms of decimals. This gives us π of π₯ one is two times π to the power of 1.5, π of π₯ two is equal to two, π of π₯ three is two times π to the power of negative 1.5, and π of π₯ four is two times π to the power of negative three. We now have all the information we need to estimate our definite integral by using our trapezoidal rule formula.

Substituting in our values for Ξπ₯ and π evaluated at our sample points, we get the integral from negative one to one of two times π to the power of negative three π₯ with respect to π₯ is approximately equal to. One-half divided by two times two π cubed plus two π to the power of negative three plus two times two π to the power of 1.5 plus two plus two times π to the power of negative 1.5. And we can just calculate this expression. Remember, the question wants us to give our answer to one decimal place. So calculating this to one decimal place, we get 15.8. And this is our final answer.

Therefore, by using the trapezoidal rule with four subintervals, we were able to show the integral from negative one to one of two π to the power of negative three π₯ with respect to π₯ is approximately equal to 15.8.