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Using rectangles whose height is given by the value of the function at the midpoint of the​ rectangle's base, estimate the area under the graph using first two and then four rectangles. ​f(x)equals=x squared2 between xequals=1 and xequals=2

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what it means is, check the picture below.

so, the first part shows two rectangles... the rectangle goes from 1.0 to 1.6, so it has a width of 0.6.

and the height comes from the middle point, or 1.3 is f(1.3) = (1.3)² or 1.69.

so you get the area of the first rectangle, 0.6 * 1.69

and then you do the same for the second rectangle, it goes from 1.6 to 2.0, so is of width 0.4, and its height comes from the middle 1.8 point, is f(1.8) = (1.8)²,  or 3.24.

so its area is 0.6 * 3.24

then you add the areas of the rectangles to approximate the area under the parabola.

so, you'd do the same for the other part of the picture with the four rectangles.

the first rectangle of the four for example, has a width of 0.2 and a height of f(1.1) = (1.1)² or 1.21.
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The area under the graph by using the first two and then four rectangles is [tex]2.958[/tex] units square.

For reference use the below-given graph.

Given function is

[tex]f(x)=x^{2}[/tex]  when [tex]x=1[/tex] to [tex]x=2[/tex] .

The first rectangle of the first part graph goes from [tex]1.0[/tex] to [tex]1.6[/tex], so the width will be [tex]0.6[/tex] units. And the height measured from the middle point i.e. [tex]1.3[/tex] is

[tex]f(1.3)=(1.3)^{2}[/tex]

[tex]=1.69[/tex] units.

Then the area of the first rectangle is [tex]0.6\times1.69=1.014[/tex] units square.

Similarly, the second rectangle of the first part graph goes from [tex]1.6[/tex] to [tex]2.0[/tex], so the width will be [tex]0.4[/tex] units. And the height measured from the middle point i.e. [tex]1.8[/tex]  is

[tex]f(1.8)=(1.8)^{2}[/tex]

[tex]=3.24[/tex] units.

So, the area of the second rectangle is [tex]0.6\times3.24=1.944[/tex] units square.

Hence, the final area under the graph will be [tex]1.014+1.944=2.958[/tex] units square.

Further, we can do the same for another part of the graph to find the area under the graph by using four rectangles.

For example,  the first rectangle of the four has a width of [tex]0.6[/tex] units and a height of [tex]f(1.1)=(1.1)^{2}[/tex]

[tex]=1.21[/tex] units.

Therefore, the area under the graph by using the first two and then four rectangles is [tex]2.958[/tex] units square.

Know more about the area under the curve here:

https://brainly.com/question/15122151

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