Difficulty: Intermediate | Prerequisites: Sections 4.3–4.5 (definite integrals, FTC, u-substitution)
To find the area trapped between two curves, you integrate the difference between the "outer" and "inner" function over the region where they overlap. When curves are given as y = f(x), you slice vertically (top minus bottom). When given as x = f(y), you slice horizontally (right minus left). Either way, the integrand is always the larger function minus the smaller one, so the result is always positive.
Area between curves
The total region enclosed between two functions over a given interval, computed by integrating the positive difference between the functions. Think of it as stacking up infinitely thin rectangular slices and adding their areas.
Vertical slicing (integrating with respect to x)
Used when curves are written as y = f(x). Each slice is a thin vertical rectangle of height (top function - bottom function) and width dx.
Horizontal slicing (integrating with respect to y)
Used when curves are written as x = f(y) or when horizontal slices are simpler. Each slice is a thin horizontal rectangle of width (right function - left function) and height dy.
Intersection points
The x- or y-values where two curves meet. These determine the limits of integration and, when the curves cross, where you must split the integral.
Write both curves as y = f(x) and y = g(x).
Determine which function is on top over the interval. If they cross, split at the crossing point.
Area = integral from a to b of (top - bottom) dx.
Write both curves as x = f(y) and x = g(y).
Determine which function is farther right over the interval.
Area = integral from c to d of (right - left) dy.
This approach is often simpler when one curve is given as x = f(y) or when vertical slicing would require splitting the integral.
Find the area bounded by y = 12/x^3, y = 3/x^2, x = 2, and x = 12.
First, check which curve is on top. At x = 2: 12/8 = 1.5 and 3/4 = 0.75, so y = 12/x^3 is above. Check whether they cross: set 12/x^3 = 3/x^2, which gives 12/x = 3, so x = 4.
At x = 4: both equal 3/16. For x > 4, test x = 6: 12/216 < 3/36, so y = 3/x^2 is on top for x in (4, 12).
Area = integral from 2 to 4 of (12/x^3 - 3/x^2) dx + integral from 4 to 12 of (3/x^2 - 12/x^3) dx.
First piece: [-6/x^2 + 3/x] from 2 to 4 = (-6/16 + 3/4) - (-6/4 + 3/2) = (-3/8 + 3/4) - (-3/2 + 3/2) = 3/8.
Second piece: [-3/x + 6/x^2] from 4 to 12 = (-3/12 + 6/144) - (-3/4 + 6/16) = (-1/4 + 1/24) - (-3/4 + 3/8) = -5/24 + 3/8 = -5/24 + 9/24 = 4/24 = 1/6.
Total area = 3/8 + 1/6 = 9/24 + 4/24 = 13/24.
Find the area bounded by x = y^2 + 1 and x = 5.
The parabola x = y^2 + 1 opens to the right with vertex at (1, 0). The line x = 5 is vertical.
Find intersection: y^2 + 1 = 5 gives y^2 = 4, so y = -2 and y = 2.
Horizontal slicing is natural here. For each y in [-2, 2], the right boundary is x = 5 and the left boundary is x = y^2 + 1.
Area = integral from -2 to 2 of (5 - (y^2 + 1)) dy = integral from -2 to 2 of (4 - y^2) dy.
The integrand is even, so this equals 2 times the integral from 0 to 2 of (4 - y^2) dy.
= 2 [4y - y^3/3] from 0 to 2 = 2 (8 - 8/3) = 2 (16/3) = 32/3.
\text{Vertical slicing: } A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx \quad \text{where } f(x) \geq g(x)\text{Horizontal slicing: } A = \int_c^d \bigl[f(y) - g(y)\bigr]\,dy \quad \text{where } f(y) \geq g(y)If the curves cross inside the interval, split at the crossing point and keep (top - bottom) or (right - left) positive in each sub-interval.
Area-between-curves problems show up whenever you need the total difference between two quantities over a period. Economists use the same technique to compute consumer surplus (the area between a demand curve and a price line). Engineers compute the cross-sectional area of channels and ducts this way.
Students often integrate (f - g) without checking which function is on top. If f < g over part of the interval, that piece contributes a negative number, and you get the wrong area.
A common error is forgetting to find intersection points. If the curves cross inside [a, b], you must split the integral there.
Students sometimes try to use vertical slicing when horizontal slicing would be far simpler (as in Problem 7). If a curve is given as x = f(y), slicing horizontally avoids solving for y.
Some students set up the area integral but forget that area is always non-negative. If your answer comes out negative, one of the bounds or the integrand order is wrong.
Area between curves is a staple exam topic. Expect at least one problem, often requiring you to find the intersection points yourself.
Problems where one curve is given as x = f(y) specifically test whether you can set up horizontal slices. These appear frequently on MTH 132 exams.
Sketch the region first. Even a rough sketch prevents most setup errors.
True or false: to find the area between two curves, you always integrate with respect to x. (False. You can integrate with respect to y when horizontal slicing is simpler.)
Fill in the blank: when slicing vertically, the integrand is ______ minus ______ . (top function, bottom function)
True or false: if two curves cross inside your interval, you can ignore the crossing and integrate the whole thing in one piece. (False. You must split the integral at each crossing point.)
Fill in the blank: to find where two curves intersect, set ______ . (f(x) = g(x) and solve for x)
True or false: the area between curves can be negative. (False. Area is always non-negative.)
Q: Find the area between y = x^2 and y = x on [0, 1].
A: On [0, 1], x >= x^2. Area = integral from 0 to 1 of (x - x^2) dx = [x^2/2 - x^3/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
Q: Find the area enclosed by y = x^2 and y = 2x.
A: Intersection: x^2 = 2x gives x = 0 and x = 2. On (0, 2), 2x > x^2. Area = integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2 = 4 - 8/3 = 4/3.
Q: Find the area between x = y^2 and x = 4.
A: Intersection: y^2 = 4 gives y = -2 and y = 2. Horizontal slicing: Area = integral from -2 to 2 of (4 - y^2) dy = 2 integral from 0 to 2 of (4 - y^2) dy = 2[4y - y^3/3] from 0 to 2 = 2(8 - 8/3) = 32/3.
Q: The curves y = 1/x and y = 1/x^2 are bounded by x = 1 and x = 3. Which is on top?
A: At x = 2: 1/2 > 1/4, so y = 1/x is on top. (Check that they do not cross: 1/x = 1/x^2 gives x = 1, which is the left boundary, so y = 1/x stays on top throughout (1, 3).)
Area between curves extends the definite integral work from Section 4.3 by using the integral to measure a geometric quantity rather than a net change. The same setup (top minus bottom, integrated over an interval) reappears in Calc II for volumes of revolution (Sec. 6.2), where you rotate these regions around an axis. If you are comfortable setting up area integrals, the volume problems are a natural extension.
area between curves, area between two functions, vertical slicing, horizontal slicing, top minus bottom, right minus left, enclosed region, intersection points, bounded region, MTH 132, Calculus I, Michigan State University, Section 5.1, recitation 14, consumer surplus, cross-sectional area