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

Answer 1

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 #dx# (a small change in #x#) and heights equal to the greater #y# (the one on upper curve) minus the lesser #y# value (the one on the lower curve). We then integrate from the smallest #x# value to the greatest #x# value.

That approach is quite challenging in this problem.
Here it is very valuable to learn to reflect our thinking #90^@#.

We will take representative rectangles horiontally.
The rectangles have height #dy# (a small change in #y#) and bases equal to the greater #x# (the one on rightmost curve) minus the lesser #x# value (the one on the leftmost curve). We then integrate from the smallest #y# value to the greatest #y# value.

Notice the duality

#{:("vertical ", iff ," horizontal"), (dx, iff, dy), ("upper", iff, "rightmost"), ("lower", iff, "leftmost"), (x, iff, y):}#

The phrase "from the smallest #x# value to the greatest #x# value." indicates that we integrate left to right. (In the direction of increasing #x# values.)

The phrase "from the smallest #y# value to the greatest #y# value." indicates that we integrate bottom to top. (In the direction of increasing #y# values.)

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

The #x# on the right (the greater #x# value) lies on the graph of #y^2=-2(x-2)#.
Solving for #x#, we get #x_"right" = -y^2/2+2#

The #x# on the left (the lesser #x# value) lies on the graph of #y^2=-4(x-1)#.
Solving for #x#, we get #x_"left" = -y^2/4+1#

#y# varies from #-2# to #2#, so the area of the region is

#int_-2^2 ((-y^2/2+2)-(-y^2/4+1)) dy#

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

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Answer 2

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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Answer from HIX Tutor

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.

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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