Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width). (Round your answers to three decimal places.) y = 1 − x2.

Respuesta :

Answer:

The answer to this question can be defined as follows:

The Lower sum ="0.659"

The Upper sum ="0.859"

Step-by-step explanation:

In the given equation there is some mistype error, so the correct equation and its solution can be defined as follows:

Equation:

[tex]y= \sqrt{1-x^2}[/tex]

calculating the Δx:

[tex]=\frac{(1 - 0)}{5}\\\\=\frac{1}{5}[/tex]

calculating the Upper sum value:

[tex]=\bigtriangleup x \times (f(0) + f(\frac{1}{5}) + f(\frac{2}{5}) + f(\frac{3}{5}) + f(\frac{4}{5})) \\\\= \frac{1}{5} \times (1 + \sqrt{(\frac{24}{25})} + \sqrt{\frac{21}{25}} + \frac{4}{5} + \frac{3}{5})\\\\= 0.859[/tex]

calculating the Lower sum value:

Answer:

The upper sum value is 0.8592

The lower sum value is 0.6592.

Step-by-step explanation:

Given information:

The equation [tex]y=\sqrt{1-x^2}[/tex]

And [tex]n=5[/tex]

now, first find derivative of the above equation:

As, [tex]\Delta x=(1/5)\\\Delta x=0.2[/tex]

Now calculate the upper sum value as :

[tex]U=0.2[\sqrt{1-0}+\sqrt{1-0.2^2}+\sqrt{1-0.4^2}+\sqrt{1-0.6^2}]+ \sqrt{1-0.8^2}\\U=0.2\times 4.296\\U=0.8592\\[/tex]

Hence, the upper sum value is 0.8592

Now ,calculate the lower sum value as:

[tex]L=0.2[\sqrt{1-0.2^2}+\sqrt{1-0.4^2}+\sqrt{1-0.6^2}+\sqrt{1-0.8^2}+\sqrt{1-1^2}]\\L= 0.2\times 3.296\\L=0.6592\\[/tex]

Hence, the lower sum value is 0.6592.

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