jsheakatz
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Find an approximation of the area of the region R under the graph of the function f on the interval [0, 2]. Use n = 5 subintervals. Choose the representative points to be the midpoints of the subintervals.
f (x) = x^2 + 5

Respuesta :

LammettHash
Interval: [tex][0,2][/tex]

Partition: [tex]\left[0,\dfrac25\right]\cup\left[\dfrac25,\dfrac45\right]\cup\left[\dfrac45,\dfrac65\right]\cup\left[\dfrac65,\dfrac85\right]\cup\left[\dfrac85,2\right][/tex]

Midpoints: [tex]\left\{\dfrac15,\dfrac35,1,\dfrac75,\dfrac95\right\}[/tex]

Value of [tex]f(x)[/tex] at the midpoints: [tex]\left\{\dfrac{126}{25},\dfrac{134}{25},6,\dfrac{174}{25},\dfrac{206}{25}\right\}[/tex]

So the definite integral is approximated by the sum

[tex]\displaystyle\int_0^2(x^2+5)\,\mathrm dx=\frac25\left(\dfrac{126}{25}+\dfrac{134}{25}+6+\dfrac{174}{25}+\dfrac{206}{25}\right)=\dfrac{316}{25}=12.64[/tex]

Compare to the actual value of the integral,

[tex]\displaystyle\int_0^2(x^2+5)\,\mathrm dx=\dfrac{38}3\approx12.67[/tex]
MrRoyal

The approximated area of the region is an estimate of the area covered by the region

The approximated area of the region is 12.64 square units

The given parameters are:

[tex]\mathbf{Interval = [0,2]}[/tex]

[tex]\mathbf{n = 5}[/tex] --- the sub intervals

[tex]\mathbf{f(x) = x^2 + 5}[/tex] --- the function

Start by calculating the length of each partition

[tex]\mathbf{L = \frac{2 - 0}{5} = \frac 25}[/tex]

So, the partitions are:

[tex]\mathbf{P = [0,\frac 25]\ u\ [\frac 25,\frac 45]\ u\ [\frac 45,\frac 65]\ u\ [\frac 65,\frac 85]\ u\ [\frac 85,2]}[/tex]

Calculate the midpoint of each partition

[tex]\mathbf{x = \{\frac 15, \frac 35, 1 ,\frac 75, \frac 95\}}[/tex]

Calculate f(x) at the midpoints

[tex]\mathbf{f(\frac 15) = (\frac 15)^2 + 5 = 5.04}[/tex]

[tex]\mathbf{f(\frac 35) = (\frac 35)^2 + 5 = 5.36}[/tex]

[tex]\mathbf{f(1) = 1^2 + 5 = 6}[/tex]

[tex]\mathbf{f(\frac 75) = (\frac 75)^2 + 5 = 6.96}[/tex]

[tex]\mathbf{f(\frac 95) = (\frac 95)^2 + 5 = 8.24}[/tex]

Represent the function as a definite integral as follows:

[tex]\mathbf{\int\limits^2_0 {(x^2 + 5)} \, dx = \frac 25(5.04 + 5.36 + 6 + 6.96 + 8.24)}[/tex]

[tex]\mathbf{\int\limits^2_0 {(x^2 + 5)} \, dx = \frac 25(31.6)}[/tex]

Divide

[tex]\mathbf{\int\limits^2_0 {(x^2 + 5)} \, dx = 2 \times 6.32}[/tex]

[tex]\mathbf{\int\limits^2_0 {(x^2 + 5)} \, dx = 12.64}[/tex]

Hence, the approximated area of the region is 12.64 square units

Read more about approximated areas at:

https://brainly.com/question/16654180

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