Deriving Area From Vertical Slices

From Thin Rectangles to Area

A definite integral can represent more than accumulated change along a line. It can also represent the area of a plane region. To see why, imagine dividing a region into many narrow vertical strips. Each strip is approximately a rectangle: its width is a small change in $x$, and its height is the vertical distance between the boundary curves.

Adding the rectangle areas gives an approximation to the region’s area. Making the strips thinner improves the approximation. In the limit, the sum becomes an integral. This tutorial derives that formula and states the conditions that make the derivation valid.

Describing the Region

Suppose a region extends horizontally from $x=a$ to $x=b$, where $a<b$. Assume its upper boundary is the graph of a continuous function $f$, and its lower boundary is the graph of a continuous function $g$. For the entire interval, suppose $f(x)\geq g(x)$. The region is then

R=\{(x,y):a\leq x\leq b,\;g(x)\leq y\leq f(x)\}.

At a fixed value of $x$, the vertical slice runs from $y=g(x)$ to $y=f(x)$. Its height is therefore $f(x)-g(x)$. The horizontal width of the whole region is $b-a$.

Definition: A region of the form $R=\{(x,y):a\leq x\leq b,\;g(x)\leq y\leq f(x)\}$, with continuous $f$ and $g$ and $f(x)\geq g(x)$ throughout $[a,b]$, has area $$ A=\int_a^b\bigl(f(x)-g(x)\bigr)\,dx. $$ Here $f(x)-g(x)$ is the vertical slice height, and $[a,b]$ is the region’s horizontal extent.

The formula depends on the stated description of the region. In particular, each vertical line with $a\leq x\leq b$ cuts the region in a single interval running between $g(x)$ and $f(x)$. The curves and bounds must describe the whole region being measured.

Deriving the Formula with Rectangle Sums

Partition $[a,b]$ into $n$ subintervals. Write their endpoints as $a=x_0<x_1<\cdots<x_n=b$. The width of the $i$th subinterval is $\Delta x_i=x_i-x_{i-1}$. Choose a sample point $x_i^*$ in that subinterval.

At $x_i^*$, the vertical height of the region is $f(x_i^*)-g(x_i^*)$. A rectangle with this height and width $\Delta x_i$ has area $\bigl(f(x_i^*)-g(x_i^*)\bigr)\Delta x_i$. Adding these areas gives the approximation

S_n=\sum_{i=1}^{n}\bigl(f(x_i^*)-g(x_i^*)\bigr)\Delta x_i.

This is a Riemann sum for the function $h(x)=f(x)-g(x)$. Since $f$ and $g$ are continuous, their difference $h$ is continuous on the closed interval $[a,b]$, and so is Riemann integrable there. As the largest subinterval width tends to zero, the Riemann sums converge to the definite integral of $h$.

Measure each slice.
At horizontal position $x$, the slice height is $h(x)=f(x)-g(x)$.

Approximate each strip.
Over a narrow interval of width $\Delta x_i$, use a rectangle of height $h(x_i^*)$, with area $h(x_i^*)\Delta x_i$.

Add the rectangles.
The total approximation is the Riemann sum $\sum_{i=1}^{n}h(x_i^*)\Delta x_i$.

Pass to the limit.
As the partition becomes finer, the sums converge to $\int_a^b h(x)\,dx$, which is the area.

Why the Approximation Converges

The rectangle picture gives the main idea. We can also justify the limit by bounding the area more precisely. On each subinterval $[x_{i-1},x_i]$, the continuous height function $h=f-g$ has a minimum $m_i$ and a maximum $M_i$. Throughout that strip, its true vertical height lies between $m_i$ and $M_i$. Thus the area of the part of the region in the strip lies between $m_i\Delta x_i$ and $M_i\Delta x_i$.

Summing over all strips bounds the region’s area between the lower sum $L=\sum_{i=1}^{n}m_i\Delta x_i$ and the upper sum $U=\sum_{i=1}^{n}M_i\Delta x_i$. Because $h$ is continuous on a closed, bounded interval, it is uniformly continuous. When the partition widths become small, the difference between the maximum and minimum values of $h$ on each subinterval becomes uniformly small. In particular, if that difference is at most $\varepsilon$ on every subinterval, then

0\leq U-L =\sum_{i=1}^{n}(M_i-m_i)\Delta x_i \leq \varepsilon\sum_{i=1}^{n}\Delta x_i =\varepsilon(b-a).

As the partition is refined, $\varepsilon$ can be made arbitrarily small, so the upper and lower bounds approach one another. The area is therefore the common limit of these sums. That limit is the Riemann integral of $h$, giving $\int_a^b(f(x)-g(x))\,dx$.

Worked Examples

Worked Example: Constant Slice Height

Consider the region between the lines $y=3$ and $y=1$, from $x=0$ to $x=4$. Here the upper boundary is $f(x)=3$, the lower boundary is $g(x)=1$, and every slice has height $3-1=2$.

Divide $[0,4]$ into $n$ equal intervals. Each width is $\Delta x=4/n$. Every rectangle has area $2(4/n)=8/n$, so the sum of the $n$ rectangle areas is

S_n=\sum_{i=1}^{n}\frac{8}{n} =n\frac{8}{n} =8.

The sum is exactly $8$ for every $n$, so its limit is $8$. The integral gives the same result:

A=\int_0^4(3-1)\,dx =\int_0^4 2\,dx =2(4-0) =8.

Check: the region is a rectangle of width $4$ and height $2$, so its geometric area is $4\cdot2=8$, matching the integral.

Worked Example: A Height That Increases Linearly

Suppose the region lies between the specified upper curve $y=x+3$ and lower curve $y=1$ for $0\leq x\leq2$. Its slice height is $h(x)=(x+3)-1=x+2$.

To see the rectangle sum, use $n$ equal intervals, each of width $\Delta x=2/n$, and choose right endpoints $x_i=2i/n$. Then

S_n=\sum_{i=1}^{n}\left(\frac{2i}{n}+2\right)\frac{2}{n} =\frac{4}{n^2}\sum_{i=1}^{n}i+\frac{4}{n}\sum_{i=1}^{n}1.

Using $\sum_{i=1}^{n}i=n(n+1)/2$ and $\sum_{i=1}^{n}1=n$,

S_n=\frac{4}{n^2}\frac{n(n+1)}{2}+\frac{4}{n}n =\frac{2(n+1)}{n}+4 =6+\frac{2}{n}.

Therefore $\lim_{n\to\infty}S_n=6$. Evaluating the area integral gives

A=\int_0^2(x+2)\,dx =\left[\frac{x^2}{2}+2x\right]_0^2 =\left(2+4\right)-0 =6.

Check: the vertical heights at the endpoints are $2$ and $4$. The region is a trapezoid with parallel sides $2$ and $4$, separated by distance $2$, so its area is $\frac{2+4}{2}\cdot2=6$, as calculated.

Worked Example: A Curved Lower Boundary

Consider the region between $y=3$ and $y=x^2$ on $0\leq x\leq1$, with $y=3$ specified as the upper boundary. The slice height is $h(x)=3-x^2$.

Use $n$ equal intervals of width $1/n$ and right endpoints $x_i=i/n$. The rectangle sum is

S_n=\sum_{i=1}^{n}\left(3-\frac{i^2}{n^2}\right)\frac{1}{n} =3-\frac{1}{n^3}\sum_{i=1}^{n}i^2.

Since $\sum_{i=1}^{n}i^2=n(n+1)(2n+1)/6$, this becomes

S_n=3-\frac{n(n+1)(2n+1)}{6n^3} =3-\frac{2n^2+3n+1}{6n^2}.

Taking the limit gives $\lim_{n\to\infty}S_n=3-\frac{2}{6}=\frac{8}{3}$. The integral confirms the result:

A=\int_0^1(3-x^2)\,dx =\left[3x-\frac{x^3}{3}\right]_0^1 =3-\frac{1}{3} =\frac{8}{3}.

Check: differentiating $3x-x^3/3$ gives $3-x^2$, the integrand. Also, the slice height stays between $2$ and $3$ on an interval of width $1$, so an area of $8/3$ is consistent with those bounds.

A Common Mistake: Integrating Only One Boundary

The area is determined by the vertical distance between the two curves, not by the height of the upper curve above the $x$-axis. For the linear example, integrating only $x+3$ gives

\int_0^2(x+3)\,dx =\left[\frac{x^2}{2}+3x\right]_0^2 =2+6 =8.

This is not the desired area: it includes the portion below $y=1$, which is outside the region between the two specified curves. That extra area is $\int_0^2 1\,dx=2$. Subtracting it gives $8-2=6$, or, directly, integrating the slice height gives $\int_0^2((x+3)-1)\,dx=6$.

Reversing the subtraction would also be wrong for geometric area in this setup: $\int_0^2(1-(x+3))\,dx=-6$. The height of a vertical slice is nonnegative, so the formula uses upper boundary minus lower boundary.

When This Formula Applies

The vertical-slice formula is especially direct when the region can be described with one interval of $y$-values at every $x$ between $a$ and $b$, and the boundary functions are continuous there. If the proposed upper and lower curves meet or exchange roles within the interval, the single expression $f(x)-g(x)$ may no longer describe the slice height throughout. The interval must then be treated in pieces or the region described differently.

The central reasoning remains the same: determine the height of a typical vertical slice, multiply by its small width, add the resulting rectangle areas, and take the limit. For a region whose vertical slice height is $f(x)-g(x)$, that limit is the definite integral of the height over the horizontal extent.

Takeaway: Area from vertical slices is the integral of slice height with respect to horizontal position: $$ \text{area}=\int_{\text{left endpoint}}^{\text{right endpoint}} (\text{upper boundary}-\text{lower boundary})\,dx. $$ The formula follows from Riemann sums of thin rectangles and requires that the stated boundaries describe the region over the full interval.

Check Your Understanding

Use the vertical-slice idea to answer these questions.

For a region between $y=f(x)$ and $y=g(x)$ on $[a,b]$, what is the height of the slice at $x$, assuming $f$ is the upper boundary?
Why does a rectangle-sum approximation use the product of slice height and subinterval width?
What role does continuity play in deriving the area integral?
For the region between $y=5$ and $y=2$ on $[1,4]$, write the area integral and evaluate it.
For the region between $y=x+2$ and $y=1$ on $[0,1]$, explain why integrating only $x+2$ does not give the area between the curves.

Deriving Area From Vertical Slices

What You'll Learn

From Thin Rectangles to Area

Describing the Region

Deriving the Formula with Rectangle Sums

Why the Approximation Converges