# How do you find the area between #y^2=-4(x-1)# and #y^2=-2(x-2)#?

Please see below.

Here is a picture of the region whose area we seek.

When we first learn to find areas by integration, we take representative rectangles vertically.

The rectangles have base

That approach is quite challenging in this problem.

Here it is very valuable to learn to reflect our thinking

We will take representative rectangles horiontally.

The rectangles have height

Notice the duality

The phrase "from the smallest

The phrase "from the smallest

Here is a picture of the region with a small rectangle indicated:

The

Solving for

The

Solving for

# = int_-2^2 (1-y^2/4) dy = 8/3#

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To find the area between the curves (y^2 = -4(x - 1)) and (y^2 = -2(x - 2)), we first need to determine the points of intersection. Setting the expressions equal to each other gives us:

[-4(x - 1) = -2(x - 2)]

Solving this equation yields the x-coordinate of the point of intersection. After finding the x-coordinate, substitute it into one of the equations to find the corresponding y-coordinate.

Next, we integrate the difference of the top curve and the bottom curve with respect to x over the interval where they intersect. This gives us the area between the curves.

If the curves intersect at multiple points, repeat the process for each interval and sum up the areas.

Finally, ensure to take absolute values when necessary to avoid negative area values.

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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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