# What is the area of the region in the first quadrant bounded on the left by the y-axis, below by the line y = x/4, above left by the curve y=1 + sqrt(x), and above right by the curve y=2\/sqrt(x)?

Elias, would you please explain what would be y=2\/sqrt(x)?

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To find the area of the region in the first quadrant bounded by the given curves, you need to integrate the difference between the upper and lower curves with respect to x from the starting point to the ending point.

The starting point is where the lower curve intersects the y-axis, which is at the origin (0,0).

The ending point is where the upper curve intersects each other. To find this point, set the upper curves equal to each other and solve for x: (1 + \sqrt{x} = \frac{2}{\sqrt{x}}).

After solving for x, you integrate the difference between the upper and lower curves from 0 to the obtained x value.

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

Evaluate this integral to find the area of the region.

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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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