Recipe

Set up an area-between-curves integral

Use when: a region is enclosed by two curves and the boundaries aren't given as explicit integration limits

Goal: a single correctly-oriented integral, in whichever variable avoids splitting the region unnecessarily

Taught in Choosing dx or dy

1. 1

   Sketch both curves. Check whether one curve stays entirely above (or entirely to the right of) the other across the whole region, or whether they swap partway through.

   Example

   f(x)=8-x^2 and g(x)=x^2: sketching shows f opens downward and g opens upward, crossing at two points with f on top between them.

   Why: which curve is "top" (or "right") determines the sign of every subtraction in the integral that follows, so this has to be settled before writing anything down.

2. 2

   Set the two curves equal (in whichever variable they're both naturally functions of) and solve.

   Example

   8-x^2=x^2 \implies 8=2x^2 \implies x^2=4 \implies x=\pm2.

   Why: intersection points are the region's natural bounds whenever none are stated explicitly, since that's exactly where "top minus bottom" would otherwise go wrong.

3. 3

   Slice vertically (dx) when every strip's top and bottom are cleanly functions of x across the whole region. Slice horizontally (dy) instead when a boundary naturally solves for x in terms of y, or vertical slicing would need a split.

   Example

   For f(x)=8-x^2 and g(x)=x^2, both are already clean functions of x with no swap between x=-2 and x=2, so dx is the natural choice. Contrast x=y^2-4 (a parabola) and x=2y-1 (a line): the line stays right of the parabola throughout, so dy avoids splitting the parabola into two branches to solve for y.

   • both curves are naturally functions of x, with no swap in between — use dx, top minus bottom.
   • a boundary is naturally a function of y, or the vertical slicing would need to be split into multiple pieces — use dy, right minus left, instead.

   Why: choosing the orientation that avoids a split, or avoids inverting a curve, is the entire point of this decision; the final area comes out the same either way, just with more or less algebra.

4. 4

   Subtract the lower curve from the upper curve (or the left curve from the right curve), and integrate over the bounds found in Step 2.

   Example

   \text{Area} = \int_{-2}^{2}\big[(8-x^2)-x^2\big]\,dx = \int_{-2}^{2}(8-2x^2)\,dx

   Why: reversing the subtraction produces a negative area, the single most common error once the orientation itself has been chosen correctly.

5. 5

   Antidifferentiate, evaluate at the bounds, and confirm the result is positive.

   Example

   \left[8x-\frac{2x^3}{3}\right]_{-2}^{2} = \left(16-\frac{16}{3}\right)-\left(-16+\frac{16}{3}\right) = 32-\frac{32}{3} = \frac{64}{3}

   \boxed{\text{Area}=\dfrac{64}{3}}, a positive number, as an area must be.

   ⚡ Shortcut

   This region is symmetric about the y-axis, since 8-x^2 and x^2 are both even functions. Integrate from 0 to the positive bound and double the result instead: 2\displaystyle\int_0^2(8-2x^2)\,dx, the same \frac{64}{3} with a shorter arithmetic pass.

   Why: a negative result is a certain sign that the top/bottom or left/right roles from Step 1 got reversed somewhere.