Unit 8 Applications of Integration: Understanding Area Between Curves

Area between curves

The area of a region in the plane whose boundary is formed by two graphs; typically found using definite integrals that add up infinitesimally thin slices.

Definite integral as area accumulation

A definite integral represents the limit of a sum of areas of many thin rectangles (or slices), giving exact area when boundaries are curved.

Vertical slicing

A method for area where you use rectangles with thickness dx; each rectangle’s height is determined by y-values (top minus bottom).

Top-minus-bottom (dx)

For vertical slices, the slice height is (upper function) − (lower function), so area is ∫(top − bottom) dx over the x-interval.

Area formula with respect to x

If y=f(x) is above y=g(x) on [a,b], then the area is A = ∫_a^b (f(x) − g(x)) dx.

Bounds of integration (a and b)

The left and right x-values that define the region when integrating with dx; they come from intersection points, given vertical lines, or a mix.

Intersection points (for bounds)

Points where two curves cross; found by solving f(x)=g(x), and their x-values often serve as integration limits.

Given vertical lines

Explicit boundaries like x=1 and x=4 that set the integration limits for a dx integral, even if curves intersect elsewhere.

Top function

The curve that gives the larger y-value for a given x on the interval being used (determines the “top” in top-minus-bottom).

Bottom function

The curve that gives the smaller y-value for a given x on the interval being used (determines the “bottom” in top-minus-bottom).

Switching top/bottom

A situation where the curves cross within the interval so the upper curve changes; the area setup must be split into multiple integrals.

Splitting an integral for area

Breaking the area calculation into a sum of integrals over subintervals where the top-minus-bottom (or right-minus-left) relationship stays consistent.

Cancellation (negative area)

When you fail to split at a crossing, parts of the integral become negative and subtract from positive parts, producing the wrong result for area.

Horizontal slicing

A method for area where you use slices with thickness dy; each slice’s length is determined by x-values (right minus left).

Right-minus-left (dy)

For horizontal slices, the slice length is (right boundary) − (left boundary), so area is ∫(right − left) dy over the y-interval.

Area formula with respect to y

If x=R(y) is to the right of x=L(y) on [c,d], then the area is A = ∫_c^d (R(y) − L(y)) dy.

y-bounds (c and d)

The lowest and highest y-values of the region when integrating with respect to y; limits must match the variable dy.

Rewriting boundaries as x in terms of y

To use dy, you often solve equations to express curves as x=… (e.g., y=x^2 becomes x=±√y).

Branch (left/right) selection

Choosing the correct solution when solving for x in terms of y (e.g., using x=√(4−y^2) for the right half of a circle).

Vertical line test (motivation for dy)

If a curve is not a single-valued function y=f(x) over a region (like a circle), using dy with x as a function of y may avoid splitting.

Example: area between y=x and y=x^2

Solve x=x^2 to get bounds 0 and 1; on (0,1), x is above x^2, so A=∫_0^1 (x−x^2) dx = 1/6.

Example: sin x vs cos x on [0,π]

They intersect at x=π/4; top switches, so area is ∫0^{π/4}(cos−sin)dx + ∫{π/4}^{π}(sin−cos)dx = 2√2.

Example (dy): y=x^2 and y=2x

Intersections at x=0,2 give y from 0 to 4; right boundary x=√y, left boundary x=y/2, so A=∫_0^4(√y−y/2)dy=4/3.

Example: right half of circle x^2+y^2=4

Using dy: left boundary x=0, right boundary x=√(4−y^2), y from −2 to 2; area equals semicircle area 2π.

Choosing dx vs dy

Pick dx when vertical slices give one clean top-minus-bottom expression; pick dy when horizontal slices give one clean right-minus-left expression (and avoid extra splitting).