4.2.13 · D5Calculus II — Integration
Question bank — Area between curves — horizontal and vertical slices
True or false — justify
The naive equals the true area whenever and never cross on
True — if they don't cross, one curve stays on top throughout, so keeps one sign and no cancellation happens. That is exactly the "top minus bottom" setup.
If , the two curves must be identical on
False — it can also mean the positive area where exactly cancels the negative area where . The curves may cross and enclose real, nonzero regions. See Definite Integral as Riemann Sum for why signed pieces cancel.
Choosing horizontal instead of vertical slices can give a smaller numerical area for the same region
False — the region's area is a fixed geometric fact. Slicing direction is only a bookkeeping choice; both must yield the same number (Worked Example 2 got both ways).
For the region between and on , the integrand is non-negative throughout
True — on we have (e.g. at , ), so the top-minus-bottom order never flips inside the interval.
The limits of a -integral are always the same numbers as the limits of the equivalent -integral
False — limits are the -range, limits are the -range; these usually differ. Only by coincidence (a square region) would they match.
Area between two curves is always evaluated over the whole
False in general — this only works if the same function is bigger across the whole interval. If they swap roles you must split at the crossing point and re-order on each piece.
The formula requires the curves to be written as in terms of
True — horizontal slices have thickness and length measured in the -direction, so both boundaries must be expressed as and . This is where Inverse Functions enters.
Spot the error
"Area between and is , so area ." Where's the trap?
The integrand is written bottom-minus-top () because the line is actually on top on . The negative result is a red flag; salvaging it with a final absolute value happens to work here only because they don't cross inside, but the habit is dangerous.
"For and on , area ." Find the flaw
The curves cross at , so inside the parabola is below the axis on and above outside it. The single integral lets negative and positive parts cancel; you must split at and use top-minus-bottom on each piece.
"To slice the region between and horizontally I integrate from to ." What's wrong?
A -integral must use -limits, and the curves meet at and . Feeding -values into a -integral mismatches thickness and limits (a core parent-note mistake).
" can be a top-or-bottom curve for vertical slicing, no problem." Critique this
For a fixed , gives two -values (), so it isn't a single function of . Vertical slices would need it split into and ; horizontal slicing avoids the split entirely.
"I'll take to fix any sign issue." Why does this fail?
Cancellation happens inside the integral before you ever see a final number, so the magnitude you'd take the absolute value of is already too small. You must insert piece by piece, not at the end.
"Both curves pass through the origin, so the lower limit is definitely ." Spot the assumption
Passing through a common point isn't the same as bounding the region there. The limits are the intersection points that actually pinch the region shut — check whether the region truly starts at that shared point, not just that both curves visit it. Limits come from Solving Quadratic & Polynomial Equations.
Why questions
Why do the integration limits come from the intersection points, geometrically?
At an intersection the top and bottom curves meet, so the slice height is zero — the region "pinches shut." Those pinch points are exactly the left/right (or bottom/top) edges of the enclosed area.
Why is the slice height "top minus bottom" and never "bottom minus top"?
Height is a positive vertical distance = (larger ) − (smaller ). Writing it top-minus-bottom guarantees a non-negative integrand so the Riemann sum adds genuine areas, not signed ones.
Why does switching from to never change the answer?
Both are just different ways of tiling the same region with thin rectangles; the total tiled area is a property of the region, not of the tile orientation. The Fundamental Theorem of Calculus evaluates either sum to the same limit.
Why would anyone prefer horizontal slices when vertical ones "always exist"?
When the top or bottom boundary changes partway across, vertical slicing needs multiple integrals; if instead the left/right boundaries are single clean functions of , one horizontal integral does the whole job (Worked Example 3).
Why must we sometimes split the interval even for a single pair of curves?
Because which curve is on top can change at a crossing. Each sub-interval has its own correct top-minus-bottom order, and splitting keeps every integrand non-negative.
Why is testing a single interior point enough to decide which curve is on top?
Between two consecutive intersection points the curves don't cross, so their order can't flip. One test point therefore fixes the order for the whole sub-interval.
Why does the parent note say "the bottom of the rectangle just moved up" when going from to ?
A plain integral is area between and the line . Replacing the baseline with a curve raises the rectangle's bottom edge from the axis to ; nothing else about the slicing idea changes.
Edge cases
What is the area when the two curves are identical, on ?
Zero — every slice has height , so there's no enclosed region. The integral of is .
The curves touch tangentially at one point but don't cross (e.g. and at the origin only) — is there area between them there?
A single touch point has zero width, so it contributes no area; you need the curves to separate over an interval to trap a region. A lone tangent point encloses nothing.
Region unbounded on one side (say between and for ) — does the "intersection = limits" rule still apply?
Not directly; there's no finite crossing on the right, so you take a limit (an improper integral). The area can still be finite even though the interval is infinite.
The two curves cross at three points, giving two enclosed lobes — what does one integral over the full span compute?
It computes the signed difference of the two lobes' areas, which can partially or fully cancel. For total area, split at every crossing and sum over each lobe separately.
A vertical line is one of the boundaries — is it a "curve" ?
No, a vertical line fails the vertical-line test as a function of , but it's a perfectly good constant limit ( or ) for vertical slices, or a boundary handled naturally by horizontal slices.
The region is symmetric about the -axis (top curve , bottom ) — can you exploit that?
Yes; slice height is , so . Symmetry just doubles the area above the axis, saving no conceptual step but simplifying the integrand.
What happens to a horizontal slice at a -value where the left and right curves momentarily coincide?
Its length , so that slice contributes nothing — it marks a top or bottom pinch point of the region and typically sits at a -limit.
Recall One-line summary of every trap here
Order matters (top−bottom, right−left), limits match the thickness variable, and crossings force a split — cancellation and mismatched limits are the two ways almost everyone loses points.
Connections
- Parent: Area between curves
- Definite Integral as Riemann Sum — why signed pieces cancel.
- Fundamental Theorem of Calculus — evaluates either slicing to the same value.
- Inverse Functions — rewriting as for -slices.
- Solving Quadratic & Polynomial Equations — finding the intersection limits.
- Volumes by Slicing & Disks — the same slice-measure-integrate pattern one dimension up.