Area between curves expressed as functions of x

AP Calculus AB· AP Calculus AB CED — Applications of Integration· 14 min read

1. Core Definition of Area Between Curves★★☆☆☆⏱ 3 min

The area between two curves $y = f (x)$ and $y = g (x)$ bounded between vertical lines $x = a$ and $x = b$ is the total positive area of the enclosed region. This topic makes up 7-10% of the total AP Calculus AB exam score, appearing in both multiple-choice and free-response sections. Unlike net area under a single curve (where area below the x-axis counts as negative), area between two curves is always positive by definition. We approximate area with vertical slices of width $D e l t a x$ , then take the limit of the Riemann sum to get a definite integral.

📘 Definition

Area between two curves (functions of x)

The total positive area of a region bounded by two curves $y = f (x)$ , $y = g (x)$ , and vertical bounds $x = a$ and $x = b$ , calculated by integrating the positive difference between the upper and lower function over the interval.

2. Basic Area Formula for Fixed Intervals★★☆☆☆⏱ 4 min

When one function is always above the other on the entire interval $[a, b]$ , the height of each vertical slice is the difference between the y-value of the upper function and the y-value of the lower function. Summing these areas gives the Riemann sum, which becomes the definite integral in the limit:

A = \int_a^b \left[ f_{\text{upper}}(x) - g_{\text{lower}}(x) \right] dx

The x-axis itself is a curve $y = 0$ , so the area between a curve and the x-axis is just a special case of this general formula.

📐 Worked Example

Find the area of the region bounded by $f (x) = x + 4$ , $g (x) = x^{2} + 2 x - 2$ , between $x = 0$ and $x = 2$ .

1
Confirm which function is upper on $[0, 2]$ : Test $x = 1$ , $f (1) = 5$ , $g (1) = 1 + 2 - 2 = 1$ , so $f (x) > g (x)$ across the entire interval.
2
Simplify the integrand (upper minus lower):
$f(x) - g(x) = (x+4) - (x^2 + 2x - 2) = -x^2 - x + 6$
3
Set up the definite integral:
$A = \int_0^2 (-x^2 - x + 6) dx$
4
Find the antiderivative:
$F(x) = -\frac{x^3}{3} - \frac{x^2}{2} + 6x$
5
Evaluate via the Fundamental Theorem of Calculus:
$F(2) - F(0) = \left(-\frac{8}{3} - 2 + 12\right) - 0 = \frac{22}{3} \approx 7.33$

Exam tip:

If an AP question says "set up but do not evaluate the integral," stop after writing the integral expression to avoid wasting time on extra work.

3. Finding Bounds from Intersection Points★★★☆☆⏱ 3 min

Many AP problems do not give explicit interval bounds $[a, b]$ , and instead ask for the area of the region enclosed by two curves. In these cases, the bounds are the x-coordinates of the intersection points of the two curves. After solving $f (x) = g (x)$ , sort the solutions to get the lower bound $a$ (smallest x) and upper bound $b$ (largest x). Always confirm which function is upper on the interval between intersections, do not assume based on leading coefficients or other general properties.

📐 Worked Example

Find the area of the region enclosed by $f (x) = 2 x$ and $g (x) = x^{2} - 4 x$ .

1
Find intersection points by setting functions equal:
$2x = x^2 - 4x \implies 0 = x^2 - 6x = x(x-6)$
2
Solutions are $x = 0$ and $x = 6$ , so our bounds are $a = 0$ , $b = 6$ . Confirm upper function on $(0, 6)$ : Test $x = 1$ , $f (1) = 2$ , $g (1) = - 3$ , so $f (x) > g (x)$ on the entire interval.
3
Set up and simplify the area integral:
$A = \int_0^6 \left[2x - (x^2 - 4x)\right] dx = \int_0^6 (6x - x^2) dx$
4
Evaluate the integral:
$\text{Antiderivative} = 3x^2 - \frac{x^3}{3}, \quad A = \left(3(6)^2 - \frac{6^3}{3}\right) - 0 = 36$

Exam tip:

Always check for extraneous solutions when solving for intersections, especially with square roots or rational functions. Confirm each solution in both original functions before using it as a bound.

4. Regions with Changing Upper/Lower Boundaries★★★★☆⏱ 4 min

When two curves intersect more than once within the interval of interest, the order of the upper and lower functions switches between consecutive intersection points. We cannot use a single integral over the entire interval, because negative area from one subinterval will cancel positive area from another, giving an incorrect final result. Instead, sort all intersection points by x, split the original interval into subintervals where the upper/lower order is constant, calculate area for each subinterval, then add the results.

📐 Worked Example

Find the total area of the region bounded by $f (x) = sin x$ and $g (x) = cos x$ between $x = 0$ and $x = π$ .

1
Find intersection points in $[0, π]$ :
$\sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}$
2
Split into two subintervals: $[0, \frac{π}{4}]$ and $[\frac{π}{4}, π]$ . Confirm upper function for each: On $[0, \frac{π}{4}]$ , $cos x > s in x$ so $g (x)$ is upper. On $[\frac{π}{4}, π]$ , $s in x > cos x$ so $f (x)$ is upper.
3
Split the area integral:
$A = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) dx + \int_{\frac{\pi}{4}}^\pi (\sin x - \cos x) dx$
4
Evaluate each integral: First antiderivative $s in x + cos x$ gives $ \sqrt{2} - 1 $. S eco n d an t i d er i v a t i v e$ -\cos x - \sin x $g i v es$ 1 + \sqrt{2}$.
5
Add the areas:
$A = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2} \approx 2.83$

✓ Quick check

Which of the following is the correct set-up for the total area of the region bounded by $y = x^{3} - 4 x$ and $y = x$ ?

Which option is correct?
- $\int_{- 5}^{5} (x^{3} - 5 x) d x$
- $\int_{- 5}^{0} [(x^{3} - 4 x) - x] d x + \int_{0}^{5} [x - (x^{3} - 4 x)] d x$
- $\int_{- 5}^{5} x^{3} - 5 x d x$
- Both B and C
Reveal answer
3 —
Intersection points are at $x = - 5, 0, 5$ . The upper function changes at $x = 0$ , so the split integral in B is correct. The absolute value form in C automatically accounts for sign changes, so it is also correct. D is the right answer.

Exam tip:

When adding areas of multiple subintervals, always add the results. Do not subtract them, because each integral already gives positive area for its subinterval.

5. Common Pitfalls

Wrong move:

After finding intersection points at $x = - 2$ and $x = 3$ , integrate from $3$ to $- 2$ and leave the negative result as final area.

Why:

Students mix up bound order and forget area must always be positive.

Correct move:

Always order bounds from smaller $x$ to larger $x$ , and take the absolute value of a negative result if you accidentally reverse bounds.

Wrong move:

When curves cross in your interval, write a single integral of $(f (x) - g (x))$ over the entire interval, leading to area cancellation.

Why:

Students forget that switching upper/lower changes the sign of the integrand, leading to incorrect cancellation of positive and negative areas.

Correct move:

Find all intersection points in the interval, sort them by x-value, and split the integral into one subinterval between each consecutive pair of intersections.

Wrong move:

Subtract the larger function from the smaller function, resulting in a negative integrand and negative area.

Why:

Students rush to set up the integral without checking which function is upper on the interval.

Correct move:

After identifying bounds, test one $x$ -value inside the interval to confirm which function gives a larger output, then subtract the smaller output from the larger output.

Wrong move:

When given explicit bounds from $x = a$ to $x = b$ , use intersection points of the curves as your bounds instead of the given $a$ and $b$ .

Why:

Students confuse problems asking for area of a fully enclosed region with problems asking for area between two given vertical lines.

Correct move:

Always read the problem carefully: if it gives explicit bounds $a$ and $b$ , use those values and do not solve for new intersection points.

Wrong move:

When finding area between a curve $f (x)$ and the x-axis, you write $ \int_a^b f(x) dx $in s t e a d o f s pl i tt in g t h e in t e g r a l w h e n$ f(x)$ dips below the x-axis.

Why:

Students forget that the x-axis is the curve $y = 0$ , so it acts as the upper function when $f (x)$ is negative.

Correct move:

Split the integral where $f (x)$ crosses the x-axis, and integrate $(0 - f (x))$ for intervals where $f (x) < 0$ .

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Basic Area (constant order)	$A = \int_{a}^{b} [f_{upper} (x) - g_{lower} (x)] d x$	Applies when $f (x) \geq g (x)$ for all $x \in [a, b]$ ; order bounds $a < b$ for positive area
Find Bounds from Intersections	Solve $f (x) = g (x)$ for $x$	Bounds are x-coordinates of valid intersections; discard solutions outside your domain
Area between curve and x-axis	$A = \int_{a}^{b} ∣ f (x) ∣ d x$	Split integral at x-intercepts; use $(0 - f (x))$ when $f (x) < 0$
Multiple Intersection Points	$A = \sum_{i = 1}^{n - 1} \int_{x_{i}}^{x_{i + 1}} [f_{upper} - g_{lower}] d x$	Sort intersections by x; split into subintervals with constant upper/lower order
Absolute Value Shortcut	$A = \int_{a}^{b} ∣ f (x) - g (x) ∣ d x$	Valid for any number of crossings; automatically handles sign changes
Total Area vs Net Area	Total = $\int ∣ f - g ∣ d x$ , Net = $\int (f - g) d x$	AP almost always asks for total area; net area is not the same as area between curves

What's Next

This topic is the foundational prerequisite for the next core topic in AP Calculus AB Unit 8: finding area between curves expressed as functions of y, which extends the vertical slice logic you learned here to horizontal slices. Mastering the core idea of (upper boundary minus lower boundary, always positive integrand) is critical for that topic, where the core principle becomes (right boundary minus left boundary) instead. This topic also feeds directly into finding volumes of solids of revolution and volumes with known cross-sections, the highest-weight sub-topic in Unit 8. The same slicing logic you use here for area is extended to calculate the area of volume cross-sections, so nailing integral set-up here is essential for scoring well on FRQ questions on the AP exam.