Find the area under the curve y = 25/x3 from x = 1 to x = t. Evaluate the area under the curve for t = 10, t = 100, and t = 1000. t = 10 t = 100 t = 1000 Find the total area under this curve for x ≥ 1.

One way to find the area under the curve [tex]y = 25/x^3[/tex] from x = 1 to x = t, is to calculate the next integral [tex]\int_{1}^{t}\frac{25}{x^3} dx = \int_{1}^{t} 25x^{-3} dx = 25\frac{x^{-2}}{-2}|_{1}^{t} = (-25/2)x^{-2}|_{1}^{t} = (-25/2)[t^{-2}-1][/tex].

For t = 10 we have [tex](-25/2)[\frac{1}{10^2}-1] = 12.375[/tex].

For t = 100 we have [tex](-25/2)[\frac{1}{100^2}-1] = 12.49875[/tex].

For t = 1000 we have [tex](-25/2)[\frac{1}{1000^2}-1] = 12.49999[/tex].

The total area under this curve for [tex]x\geq1[/tex] is given by

[tex]lim_{t\to\infty}(-25/2)[t^{-2}-1] = (-25/2)(-1) = 25/2 = 12.5[/tex]

astha8579 astha8579 · Answer 2 · 2022-02-25T11:40:38+01:00

Area under a curve can be found by integrating the infinitesimal area on given interval. The area under the given curve for different constraints are:

For x = 1 to x = t: Area under curve and above x axis = [tex]\dfrac{25}{2} - \dfrac{25}{t^2} \: \rm unit^2[/tex]
For t= 10, area = 12.25 sq. units.
For t= 100, area = 12.4975 sq. units.
For t = 1000, area = 12.499975 sq. units.
For x ≥ 1, area = 12.5 unit 12.499975 sq. units..

How to find the area under a curve?

If the region is specified, find the area of an infinitesimal part, then integrate these small areas over the entire region. This will give the area of the desired region.

It must be noted that area is a non-negative quantity and that for some regions, It may be that area comes out to be infinite.

For the given case, referring to the attached figure, we see that function is continuous and thus, at infinitesimally small level , there are infinite rectangles with width [tex]dx[/tex], and height [tex]f(x)[/tex]

Thus, their areas is [tex]f(x)dx[/tex] (bounded between x-axis and the function's curve)

Integrating them on x axis from x = 1 to x = t, we get:

[tex]Area = \int_1^t f(x)dx = \int_1^t\dfrac{25}{x^3}dx = [-\dfrac{25}{2x^2}]^t_1 = \dfrac{25}{2} - \dfrac{25}{t^2} \: \rm unit^2[/tex]

Thus, for t = 10, 100, and 1000, the area comes to be:

For t = 10, A = 12.5 - 0.25 = 12.25 sq. units.
For t = 100, A = 12.5 - 0.0025 = 12.4975 sq. units.
For t = 1000, A = 12.5 - 0.000025 = 12.499975 sq. units.

For area for x ≥ 1, we need t -> infinity, thus,

Area under function (and bounded by x axis) from x ≥ 1 is[tex]lim_{t\rightarrow \infty} (12.5 - 25/t^2) = 12.5 \: \rm unit^2[/tex]

Thus, the area under the given curve for different constraints are:

For x = 1 to x = t: Area under curve and above x axis = [tex]\dfrac{25}{2} - \dfrac{25}{t^2} \: \rm unit^2[/tex]
For t= 10, area = 12.25 sq. units.
For t= 100, area = 12.4975 sq. units.
For t = 1000, area = 12.499975 sq. units.
For x ≥ 1, area = 12.5 unit 12.499975 sq. units..

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