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 $\int (\text{top} - \text{bottom}) \, dx$ 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 = \textcolor{blue}{\textbackslash}int_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 $\int (\text{right} - \text{left}) \, dy$ 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 = \textcolor{blue}{\textbackslash}int_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 $A=\frac{1}{6}$.

Example: sin x vs cos x on [0,π]

They intersect at $x=\frac{\pi}{4}$; top switches, so area is $\textcolor{blue}{\int_0^{\frac{\pi}{4}}(\cos - \sin) \, dx} + \textcolor{blue}{\int_{\frac{\pi}{4}}^{\pi}(\sin - \cos) \, dx} = 2\textcolor{blue}{\sqrt{2}}$.

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

Intersections at $x=0,2$ give $y$ from $0$ to $4$; right boundary x=\textcolor{blue}{\textbackslash}sqrt{y}, left boundary $x=\frac{y}{2}$, so A=\textcolor{blue}{\textbackslash}int_0^4(\textcolor{blue}{\textbackslash}sqrt{y}−\frac{y}{2})dy=\frac{4}{3}.

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

Using $dy$: left boundary $x=0$, right boundary x=\textcolor{blue}{\textbackslash}sqrt{4−y^2}, $y$ from $-2$ to $2$; area equals semicircle area 2\textcolor{blue}{\textbackslash}pi.

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).