Tasks:
Sometimes we encounter curves where it is more convenient to express its formula by a function
𝑥
=
𝑔
(
𝑦
)
.
Go to Example 9.5 on page 245.
Observe that the curve
𝑦
2
−
𝑥
−
4
=
0
can be written as
𝑥
=
𝑦
2
−
4
, which is a parabola that opens to the right.
This video discusses this idea as well:

Question:

Which of the following integrals represents the same area as
∫
0
2
(
4
−
2
𝑥
)
d
𝑥
?

Question

Tasks:
Sometimes we encounter curves where it is more convenient to express its formula by a function 
𝑥
=
𝑔
(
𝑦
)
.
Go to Example 9.5 on page 245.
Observe that the curve 
𝑦
2
−
𝑥
−
4
=
0
 can be written as 
𝑥
=
𝑦
2
−
4
, which is a parabola that opens to the right.
This video discusses this idea as well:

Question:

Which of the following integrals represents the same area as 
∫
0
2
(
4
−
2
𝑥
)
d
𝑥
?

BlackTom AI · Accepted Answer

We start by restating the problem to ensure clarity: the original area is represented by the integral ∫ from x = 0 to x = 2 of (4 − 2x) dx. The question asks for an equivalent expression of the same area, but written with x as a function of y, i.e., using a dy integral after swapping variables.

Option given: ∫ from y = 0 to y = 4 of (2 − y^2) dy. Now let’s evaluate and compare carefully.

First, determine the relation between x and y. The curve is described by y^2 − x − 4 = 0, which rearranges to x = y^2 − 4. This parabola opens to the right with vertex at (−4, 0). The line corresponding to the top boundary is y = 4 − 2x. To express the area in terms of y, we must find the y-values where the line and the parabola intersect, i.e., solve y = 4 − 2x with x = y^2 − 4. Substituting yields y = 4 − 2(y^2 − 4) = 12 − 2y^2, which leads to 2y^2 + y − 12 = 0. The solutions give the actual y-limits of the region, approximately y ≈ −2.712 and y ≈ 2.212. So the correct dy-integral would have y-limits about −2.712 to 2.212, not 0 to 4.

Second, check the integrand when converting to dy. If x = y^2 − 4, then dx = 2y dy. The original area in dx corresponds to integrating the vertical height (which in dx form is the top minus bottom function in x) with respect to x. When switching to dy, we must express the horizontal width, which is the difference between the rightmost and leftmost x-values for each y, i.e., [x_right(y) − x_left(y)]. Here, x_right from the region is given by x = y^2 − 4, and depending on the region, the left boundary may differ. Merely writing an integrand like (2 − y^2) dy ignores the correct width interpretation and the Jacobian needed for the change of variables.

Third, compare the proposed integrand and limits to the actual geometry. The candidate integral ∫_0^4 (2 − y^2) dy uses a quadratic (2 − y^2) as the height, which does not correspond to the horizontal width of the region expressed as a function of y, nor does it match the correct y-limits derived from the intersection. Moreover, the y-limits 0 to 4 do not align with the intersection points found above (the region’s y-values span roughly from −2.7 to 2.2, not 0 to 4).

In summary, the proposed integral fails on two fronts: (1) the y-limits do not reflect the actual intersection range of the curves, and (2) the integrand does not represent the horizontal width of the region as a function of y (it ignores the proper x(y) structure and the necessary Jacobian-like consideration for swapping variables). Therefore, this integral does not represent the same area as the original ∫_0^2 (4 − 2x) dx.

MAT133 F24-W25 W25 Week 9 Preclass Guided Reading

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