Question Video: Estimating the Definite Integral of a Function in a Given Interval by Dividing It into Subintervals Mathematics • Higher Education

Calculate the left endpoint estimate of ∫_(βˆ’2)^(0) π‘₯Β³ + 3π‘₯Β² + 5 dπ‘₯ with 8 subintervals of equal width.

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Video Transcript

Calculate the left endpoint estimate of the integral from negative two to zero of π‘₯ cubed plus three π‘₯ squared plus five with respect to π‘₯ with eight subintervals of equal width.

The question is asking us to find the left endpoint estimate of a definite integral by using eight subintervals of equal width. We can do this by using a left Riemann sum with eight subintervals. Let’s start by recalling how to formulate our left Riemann sum to estimate the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. First, we have our subinterval width Ξ”π‘₯ is equal to 𝑏 minus π‘Ž divided by 𝑛, where 𝑛 is the number of subintervals. Next, we set our sample points to be the left endpoints of our subintervals. These π‘₯ 𝑖 will be equal to π‘Ž plus 𝑖 times Ξ”π‘₯.

Next, we evaluate our function at each of our sample points. We then construct rectangles on each of our subintervals. The width of these rectangles will be Ξ”π‘₯. And the height will be 𝑓 of π‘₯ 𝑖. Finally, we notice Ξ”π‘₯ times 𝑓 of π‘₯ 𝑖 will be the area of our rectangle, which will be positive when it’s above the π‘₯-axis and negative when it’s below the π‘₯-axis. This then allows us to approximate the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ as the sum from 𝑖 equals zero to 𝑛 minus one of Ξ”π‘₯ times 𝑓 of π‘₯ 𝑖.

The question is asking us to estimate the integral from negative two to zero of π‘₯ cubed plus three π‘₯ squared plus five with respect to π‘₯ with eight subintervals of equal width. Since we’re estimating the definite integral from negative two to zero of our function, we’ll set π‘Ž equal to negative two and 𝑏 equal to zero. We’ll set 𝑓 of π‘₯ to be our integrand, π‘₯ cubed plus three π‘₯ squared plus five. And since we want to do this with eight subintervals of equal width, we’ll set 𝑛 equal to eight. We can now use this information to find Ξ”π‘₯. It’s equal to 𝑏 minus π‘Ž divided by 𝑛, which, in this case, is zero minus negative two divided by eight, which we can calculate to be one-quarter. We’ll write this as 0.25.

We now want to find the values of our sample points π‘₯ 𝑖 and 𝑓 evaluated at our sample points. We’ll collect this information in a table. And remember, since we want eight subintervals and we want the left endpoints to be our sample points, we’ll have eight left endpoints. So our values of 𝑖 will range from zero to seven. Let’s start by finding the value of π‘₯ zero. We see that π‘₯ zero will be equal to π‘Ž plus zero times Ξ”π‘₯. And we know that π‘Ž is equal to negative two and Ξ”π‘₯ is equal to 0.25. So we get that π‘₯ zero is equal to negative two plus zero times 0.25, which, of course, we can calculate to give us negative two.

We can then do the same to find π‘₯ one. It’s equal to negative two plus one time 0.25, which we can calculate to give us negative 1.75. We can then do the same to find the rest of our sample points. We get π‘₯ two is negative 1.5, π‘₯ three is negative 1.25, and this goes all the way up to π‘₯ seven, which is negative 0.25. We now need to find 𝑓 evaluated at each of our sample points. Let’s start with 𝑓 of π‘₯ zero. This is 𝑓 evaluated at negative two, which is equal to negative two cubed plus three times negative two squared plus five, which, if we calculate, we see it’s equal to nine.

We can then do the same to find 𝑓 evaluated at π‘₯ one. This is 𝑓 evaluated at negative 1.75. So we substitute this into our integrand. We get negative 1.75 cubed plus three times negative 1.75 squared plus five. And if we calculate this expression, we get 565 divided by 64. We can then do the same to find 𝑓 evaluated at the rest of our sample points. We substitute our values of π‘₯ 𝑖 into our integrand. This allows us to complete our table and means we’ve found 𝑓 evaluated at all of our sample points.

We could now use our Riemann sum to approximate our definite integral. But remember, all of our subintervals have equal width. So we can take our constant factor of Ξ”π‘₯ outside of our sum. This gives us Ξ”π‘₯ times the sum from 𝑖 equals zero to 𝑛 minus one of 𝑓 of π‘₯ 𝑖. So we just have Ξ”π‘₯ multiplied by the sum of 𝑓 evaluated at all of our sample points. And the sum of 𝑓 evaluated at all of our values of π‘₯ 𝑖 is just the sum of the last row in our table. So if we just calculate the total of this row of the table, we get that the sum from 𝑖 equals zero to 𝑛 minus one of 𝑓 of π‘₯ 𝑖 is equal to 58.

We’re now ready to find our approximation of our integral. We have the integral from negative two to zero of π‘₯ cubed plus three π‘₯ squared plus five with respect to π‘₯ is approximately equal to Ξ”π‘₯ times the sum from 𝑖 equals zero to 𝑛 minus one of 𝑓 of π‘₯ 𝑖. And we know that Ξ”π‘₯ is equal to 0.25. And we showed that this sum is equal to 58. So our definite integral is approximately 0.25 times 58, which we can calculate is 14.5. And this is our final answer.

Therefore, by using a left endpoint estimate of eight subintervals of equal width, we were able to show the integral from negative two to zero of π‘₯ cubed plus three π‘₯ squared plus five with respect to π‘₯ is approximately equal to 14.5.

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