Tags: area between curves, integral area, bounded region, intersection points, horizontal integration, vertical integration, absolute value integral, Texas A&M calculus
To find the area between two curves, integrate the difference of the top function minus the bottom function over the interval where they overlap. When curves cross each other, split the integral at the intersection points. Sometimes integrating with respect to y (horizontally) is simpler than integrating with respect to x.
Area between curves
The region enclosed between two functions f(x) and g(x) on an interval [a, b]. Computed by integrating the vertical distance between the curves.
Intersection points
The x-values (or y-values) where the bounding curves meet. Found by setting f(x) = g(x) and solving. These determine the limits of integration.
Integrating with respect to y (horizontal slicing)
Rewriting curves as functions of y and integrating along the y-axis. Useful when the region is more naturally described by horizontal slices, or when vertical slicing would require splitting into multiple integrals.
Subdividing into regions
When f and g cross within the interval, the integrand changes sign. Split the interval at each crossing point and integrate each sub-interval separately, always subtracting the lower curve from the upper.
The area under a single curve y = f(x) from a to b is approximated by summing rectangles of height f(x) and width dx.
This gives the familiar definite integral from first principles.
When a region has two curved boundaries, the height of each rectangle becomes f(x) − g(x), where f is the upper curve and g is the lower.
If f(x) ≥ g(x) for all x in [a, b]:
Area = ∫ from a to b of [f(x) − g(x)] dx
If the curves cross (f is not always above g), use the absolute value form:
Area = ∫ from a to b of |f(x) − g(x)| dx
In practice, you handle the absolute value by subdividing into regions at each intersection point, then integrating each piece with the correct "top minus bottom" order.
Set the two curve equations equal and solve for x (or y).
These solutions become your limits of integration.
Always sketch the region or check a test point in each sub-interval to confirm which curve is on top.
Some regions are awkward with vertical slices but straightforward with horizontal ones.
Rewrite the boundary curves as x = h(y) and x = k(y).
The area becomes:
A = ∫ from c to d of [h(y) − k(y)] dy
where h(y) is the rightmost curve and k(y) is the leftmost, and c, d are the y-coordinates of the corners.
For each region, identify the "corners" (intersection points and boundary values).
Decide whether integrating with respect to x or y gives a simpler single integral.
Integrating with respect to y avoids splitting when a region has a boundary that is multi-valued as a function of x.
Area (f on top, vertical slices):
A = ∫ from a to b of [f(x) − g(x)] dx, where f(x) ≥ g(x)
Area (curves cross, vertical slices):
A = ∫ from a to b of |f(x) − g(x)| dx (split at crossings in practice)
Area (horizontal slices):
A = ∫ from c to d of [h(y) − k(y)] dy, where h(y) ≥ k(y)
⚠️ The most common error is forgetting to check which curve is on top in each sub-interval. A negative area result is the giveaway.
⚠️ Always find intersection points first. They set your limits and tell you where the "top" and "bottom" swap.
⚠️ When given a region bounded by y = x − 1 and y² = 2x + 6 (a parabola opening sideways), horizontal integration is far cleaner. Recognise sideways parabolas as a cue to integrate with respect to y.
⚠️ If an integral has an absolute value inside, the exam expects you to split it into pieces and remove the absolute value with the correct sign in each piece.
Q: Find the area of the region bounded by y = x² and y = −2x − x². What are the first steps?
A: Set x² = −2x − x², which gives 2x² + 2x = 0, so x(x + 1) = 0, meaning x = 0 and x = −1. These are the limits. Then determine which curve is above the other on [−1, 0] and integrate (top − bottom) dx.
Q: For the region bounded by y = sin x, y = cos x, x = 0, x = π, why do you need to split the integral?
A: Because sin x and cos x cross at x = π/4. On [0, π/4], cos x ≥ sin x, and on [π/4, π], sin x ≥ cos x. You integrate each piece separately with the correct order.
Q: When is it better to integrate with respect to y instead of x?
A: When the boundary curves are more naturally expressed as functions of y (e.g. sideways parabolas), or when integrating with respect to x would force you to split into multiple integrals but a single integral in y covers the whole region.
Q: Evaluate ∫ from 0 to 2 of |x² − x³| dx. What is the geometric interpretation?
A: Find where x² = x³, i.e. x²(1 − x) = 0, so x = 0 and x = 1. On [0, 1], x² ≥ x³; on [1, 2], x³ ≥ x². Split into two integrals: ∫₀¹ (x² − x³) dx + ∫₁² (x³ − x²) dx. The result equals the total area between the curves y = x² and y = x³ from 0 to 2.
Q: Set up the integral for the area bounded by y = x² + 1, y = 3 − x², x = −2, x = 2.
A: Find where x² + 1 = 3 − x², giving x² = 1, so x = ±1. On [−2, −1] and [1, 2], x² + 1 > 3 − x². On [−1, 1], 3 − x² > x² + 1. Set up three integrals (or two by symmetry) with the correct top-minus-bottom in each.
area between curves, area under a curve, bounded region, integration limits, intersection of curves, horizontal vs vertical integration, integrating with respect to y, absolute value integral, splitting integrals, region enclosed by curves, top minus bottom, right minus left, calculus area problems, MATH 151, MATH 152, Texas A&M