# How do you find the area between y=x, #y=1/x^2#, the xaxis and x=3?

First we note that:

#1/x^2 > 0# for any#x#

so the intercept between the curves and the

Then we analyze the inequality:

This means that in the interval

A picture can clarify that based on these considerations the area we seek is:

Performing the integral we obtain:

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To find the area between the curves (y = x), (y = \frac{1}{x^2}), the x-axis, and (x = 3):

- Identify the points of intersection by setting the two functions equal to each other and solving for (x).
- Determine the limits of integration based on the points of intersection and the boundaries given ((x = 0) and (x = 3)).
- Integrate the function that lies above the other with respect to (x) within the determined limits.
- Subtract the integral of the function below from the integral of the function above to find the area between the curves.

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To find the area between the curves (y = x), (y = \frac{1}{x^2}), and the x-axis from (x = 0) to (x = 3), you need to calculate the definite integral of the absolute difference between the two functions over the given interval.

First, identify the points of intersection of the curves. Setting (y = x) equal to (y = \frac{1}{x^2}), we have:

[x = \frac{1}{x^2}]

Solving for (x) gives:

[x^3 = 1]

[x = 1]

So, the curves intersect at (x = 1).

Next, determine which function is above the other on the interval [0, 1] and which one is above on the interval [1, 3].

For (x) between 0 and 1, (y = x) is above (y = \frac{1}{x^2}).

For (x) between 1 and 3, (y = \frac{1}{x^2}) is above (y = x).

Now, set up the definite integral:

[Area = \int_{0}^{1} (x - \frac{1}{x^2}) , dx + \int_{1}^{3} (\frac{1}{x^2} - x) , dx]

Integrate each part separately:

[Area = \left[\frac{x^2}{2} + \frac{1}{x}\right]*{0}^{1} + \left[-\frac{1}{x} - \frac{x^2}{2}\right]*{1}^{3}]

[Area = \left(\frac{1}{2} + 1\right) - \left(0 + \frac{1}{0}\right) + \left(-\frac{1}{3} - \frac{9}{2}\right) - \left(-1 - \frac{1}{2}\right)]

[Area = \frac{3}{2} - \frac{1}{3} - \frac{9}{2} + 1 + \frac{1}{2}]

[Area = \frac{1}{2} - \frac{8}{2} + 1 + \frac{1}{2}]

[Area = \frac{2}{2}]

[Area = 1]

So, the area between the curves (y = x), (y = \frac{1}{x^2}), the x-axis, and (x = 3) is 1 square unit.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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