Approximating net area The following functions are negative on the given interval. a. Sketch the

Approximating net area The following functions are negative...

Key Concepts

Sketching a Function

Sketching a function involves drawing a rough graph that captures the overall shape and key features of the function over a certain interval, such as intercepts, turning points, and behavior at the endpoints. This visual representation helps in understanding the behavior of the function and the corresponding area approximations.

Net Area

The concept of net area refers to the total area between a function's graph and the x-axis, taking into account the sign of the function. When parts of the function are below the x-axis, these areas are considered negative. Approximating the net area means summing these signed areas, which is especially important when the function involves regions where the graph falls below the axis.

Subinterval Partitioning

Partitioning an interval means dividing the overall interval into a number of equal or unequal subintervals. This division is fundamental to approximating areas via Riemann sums because it establishes the basis for calculating the width (?x) of each rectangle used in the approximation, directly impacting the accuracy of the estimated area.

Riemann Sums

A Riemann sum is a method used to approximate the definite integral of a function, which represents the area under the curve. It involves partitioning the interval of interest into smaller subintervals, then using sample points from these subintervals to evaluate the function. The product of each function value with the width of its respective subinterval forms an approximation for the area, which are then summed across all subintervals.

Left, Right, and Midpoint Approximations

These terms refer to the different methods for choosing the representative point within each subinterval when calculating a Riemann sum. With left endpoint approximation, the left-most point of each subinterval is used; with right endpoint approximation, the right-most point is chosen; and with midpoint approximation, the center of each subinterval is used. Each method can yield a different error bound in approximating the true area.

Transcript

00:01 So in this question, we are first given a function, f of x, and asked to first sketch it, which we'll do on the left, and then asked to approximate the net area, which is between the graph and the x -axis, inside the interval which we're given, which is between 3 and 7, using left, right, and mid -point ream and sums.

00:27 And we're using n equals 4 but we'll come back to that later we'll start off by drawing the graph so the best way of doing this is to plot a couple of points so we only know the values between three and seven so we can ignore everything else at three we've got the value of f of x at x equals 3 is minus 4 minus 3 cubed which is 27 so we've got minus 31 so it's sort of roughly there and then if we go for the most extreme value, we can see at 7, 7 cubed is 343.

01:05 So minus 4, minus 343 is minus 347, which will be somewhere like that down there.

01:13 And we know that the graph is a negative x cubed graph, so it will have this sort of shape.

01:23 And so at this point we can give it a rough sketch of something like that it's only a sketch so it doesn't need to be perfectly accurate i might draw that slightly oh i might just draw that slightly better there that's a bit better um so that's our sketch that's the first part done now we can look over to our ream and sums so we had n equals 4 given to us in the question h is b minus a over n.

01:56 So b is this value at the right hand end of the interval, which is 7.

02:03 And a is our value on the left hand, which is 3, divided by m, which is 4, which is 4 over 4, which is 1.

02:11 So our h is 1.

02:14 If we now look at our riemann sums, we'll go for green.

02:21 So for our left riemann sum, we can see that our...

02:25 X values that we're going to use in the formula are x i's and then if we just plug into the formula we've got all of the values we need we know a that was given to us up there we know i takes these values here and we know that n is four so we've got i between zero and three and uh that just tells us that our values are four oh sorry three, four, five, and six.

03:04 And hopefully some of you've already noticed the mistake i made.

03:08 So with the graph here, i started at four, but really i should have started at three.

03:15 So i'll come back and redraw that like that.

03:19 It's a bit better.

03:21 And you could also break this up.

03:24 So we know we then equals four that our four intervals are between three and four, four and five, five, five and five, and six and six and seven so you could look at your x's at on the left -hand side of all those points and you can see the left -hand side of all these is three four five and six um we then plug those into our well we then plug those into our function um which here i've called f of x but it's actually f of x so if we plug those into that f of x function at the top we have uh our values of f f of x i are minus four minus three cubed which we already said was minus 31 you can obviously use a calculator for these so that would be minus four minus four cubed which is minus 68 then it's minus four minus five cubed which is minus 129 and minus four minus six cubed which is minus 220...

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