Use finite approximations to estimate the area under the graph of the function using a. a lower

Use finite approximations to estimate the area under the...

Key Concepts

Riemann Sum

A Riemann sum is a method for approximating the area under a curve by dividing the domain into small subintervals, calculating the function value at specific points within these subintervals, and then summing the areas of the resulting rectangles. This method is fundamental in the definition of a definite integral.

Partitioning of the Interval

Partitioning involves dividing the overall interval of integration into smaller subintervals of equal or varied widths. This division is necessary for constructing Riemann sums and allows for the approximation of the area under the curve by summing up the areas computed over each subinterval.

Lower Sum

A lower sum approximation is obtained by using the minimum function value in each subinterval to determine the height of the rectangle. This method typically underestimates the area under the curve because it chooses the lowest value on each interval, ensuring that the combined rectangle areas are less than or equal to the actual area.

Upper Sum

An upper sum approximation involves using the maximum function value in each subinterval for the height of the rectangle. This approach generally overestimates the area under the curve since the highest function value on each interval is used, making the total area of the rectangles greater than or equal to the actual area.

Transcript

00:01 In problem number four, we're asked to approximate the area of the graph f of x equals not x squared.

00:06 I'm apologies.

00:07 I'm thinking quadratic, four minus x squared.

00:10 Now, i've already got that drawn out four times over here because we've got four parts.

00:14 So four minus x squared is an upside down parabola, the y intercept of four, and then it intersects the x axis at native 2 and 2.

00:23 And we want to approximate the area from negative 2 to 2, so that's pretty convenient for us.

00:28 So part a, we want to approximate the area using a lower sum with two rectangles of equal width.

00:38 So two rectangles, we divide this up into two equal portions, and that's going to give us a portion from negative 2 to 0 and 0 .2.

00:46 And if you recall, you can calculate your delta x by doing max minus men over number of rectangles.

00:57 So here we've got 2 minus negative 2 over 2 rectangles.

01:05 So we end up with 4 over 2 equals 2.

01:08 Our delta x is 2.

01:09 So negative 2 plus 2 is 0 and 0 plus 2 is 2.

01:14 And that's our delta x.

01:16 So for a lower sum, we want to use the minimum value of f of x on each sub interval for the height of our rectangle.

01:24 So from negative 2 to 0, our lower 1.

01:30 Value of f of x for that is going to be zero so our height of the rectangle here has a value of zero and similarly from zero to two our lowest value of f of x is zero so our height has zero again so our approximation here is there's zero area for part a because our or rectangles both have a height of zero so we'd end up with equals delta x which in this case is times zero plus zero, which is zero, so area still equals zero.

02:08 Part b, we want to use four rectangles in a lower sum.

02:16 So now our delta x is going to be one.

02:18 We can find that using this formula up here.

02:22 We get negative two, native one, zero, one, and two for our division points, if you might call it that.

02:29 And we're still using lower sum, so we want to use the lowest point switch here on each of these intervals use the lowest point of f of x so from negative 2 to 1 that's still going to be 0 from negative 1 to 0 it's going to be f of negative 1 so we get this from 0 to 1 it's going to be like this and that should be even it's just my poor rd abilities and then from 1 to 2 it's going to be 0 again so now our area is equal to our delta x and this is just width times length with the area of a rectangle.

03:10 So we got, then we just add them up.

03:13 So our delta x is 1 times 0 plus negative 1 squared is 1, 4 minus 1 is 3 plus 1 plus 1 plus 1.

03:25 Plus 1 is going to be 3 again plus 0.

03:27 So we get area equals 2 6...

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