Area between curves expressed as functions of x — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Deriving and applying the definite integral formula for area between two curves as functions of x, finding intersection bounds, handling regions with changing boundary order, and setting up/evaluating integrals for bounded regions.

You should already know: Definite integration of functions of x, finding intersection points of two functions, Fundamental Theorem of Calculus Part 2.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Area between curves expressed as functions of x?

The area between two curves expressed as functions of x is the total positive area of any bounded region enclosed by two curves $y=f(x)$ and $y=g(x)$, between two vertical bounds at $x=a$ and $x=b$. Per the College Board AP Calculus AB CED, this topic (and its related applications) makes up approximately 7-10% of the total exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Unlike the net area under a single curve (where area below the x-axis counts as negative), the area between two curves is always positive by definition. We use vertical slices of width $\Delta x$ to approximate the area, then take the limit as the number of slices goes to infinity to get a definite integral. Common synonyms include "area between two curves with respect to x" and "vertical slice area between curves." On the AP exam, you will be asked to both set up integrals for area (common in MCQ) and compute exact or approximate area (common in FRQ), often in real-world contexts.

2. The Basic Area Formula for Constant Bounds

When you are given a fixed interval $[a,b]$ and two continuous functions where one function is always above the other on the entire interval, we can derive the area formula directly from the definition of the definite integral. To find the height of each vertical slice of width $\Delta x$, we subtract the y-value of the lower (bottom) function from the y-value of the upper (top) function. This gives a positive height for every slice, so the total area of all slices is the sum of $(f_{\text{upper}}(x_i)-g_{\text{lower}}(x_i))\Delta x$. Taking the limit as $n\to \infty$ (number of slices) gives the definite integral formula: $$A={\int }_a^b\left[f_{\text{upper}}(x)-g_{\text{lower}}(x)\right]dx$$ The key requirement here is that the order of upper and lower does not change on $[a,b]$, so we can use a single integral for the entire region. 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 formula.

Worked Example

Find the area of the region bounded by $f(x)=x+4$, $g(x)=x^2+2x-2$, 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. Do not waste time integrating, as you will not get extra points and will lose time for other questions.

3. Finding Bounds from Intersection Points

Many AP problems do not give explicit interval bounds $[a,b]$. Instead, you are asked to find the area of the region enclosed by two curves, which means the bounds are the x-coordinates of the points where the two curves intersect. The process is straightforward: first solve $f(x)=g(x)$ for all x-values, the smallest solution is your lower bound $a$, and the largest solution is your upper bound $b$. For two curves that intersect exactly twice, you only have one interval, so you can use the basic formula from the previous section after confirming which function is upper. Always confirm the order of upper and lower after finding your bounds: never assume one function is always upper just because it has a higher leading coefficient or other general property.

Worked Example

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

1. Find intersection points by setting the functions equal: $2x=x^2-4x⟹0=x^2-6x=x(x-6)$. The solutions are $x=0$ and $x=6$, so our bounds are $a=0$, $b=6$.
2. Confirm upper function on $(0,6)$: Test $x=1$, $f(1)=2$, $g(1)=1-4=-3$, so $f(x)>g(x)$ on the entire interval.
3. Set up 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: Antiderivative is $3x^2-\frac{x^3}{3}$. $A=\left(3(6{)}^2-\frac{6^3}{3}\right)-0=108-72=36$.

Exam tip: Always check for extraneous solutions when solving for intersection points, especially with square roots or rational functions. Plug each solution back into both original functions to confirm it is a valid intersection before using it as a bound.

4. Regions with Changing Upper/Lower Boundaries

When two curves intersect more than twice within your interval of interest, the order of the upper and lower functions switches between consecutive intersection points. Since we always need a positive integrand to get positive area, we cannot use a single integral over the entire interval. Instead, we sort all intersection points in increasing order of x, split the original interval into subintervals where the upper/lower order is constant, apply the basic formula to each subinterval, and add the results. Even if the final integral from the start to end bound comes out positive, it will be wrong if it does not account for sign changes, because the negative area from one subinterval will cancel the positive area from another.

Worked Example

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

1. Find intersection points in $[0,\pi ]$: Set $\sin x=\cos x⟹\tan x=1$, so the only solution is $x=\frac{\pi }{4}$. We have two subintervals: $[0,\frac{\pi }{4}]$ and $[\frac{\pi }{4},\pi ]$.
2. Confirm upper function on each subinterval: Test $x=0$: $\cos 0=1>\sin 0=0$, so $g(x)$ is upper on $[0,\frac{\pi }{4}]$. Test $x=\frac{\pi }{2}$: $\sin \frac{\pi }{2}=1>\cos \frac{\pi }{2}=0$, so $f(x)$ is upper on $[\frac{\pi }{4},\pi ]$.
3. Split the 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 integral antiderivative is $\sin x+\cos x$, so it evaluates to $\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)-(0+1)=\sqrt{2}-1$. Second integral antiderivative is $-\cos x-\sin x$, which evaluates to $(1-0)-\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)=1+\sqrt{2}$.
5. Add the areas: $A=(\sqrt{2}-1)+(1+\sqrt{2})=2\sqrt{2}\approx 2.83$.

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

5. Common Pitfalls (and how to avoid them)

• Wrong move: After finding intersection points at $x=-2$ and $x=3$, you integrate from $3$ to $-2$ instead of $-2$ to $3$, and leave the resulting negative value as your final area. Why: Students mix up the order of bounds when solving for intersections, and forget that area must be positive even if the definite integral comes out negative. Correct move: Always order your bounds from smaller $x$ to larger $x$, and if you end up with a negative area, take its absolute value as the final answer.
• Wrong move: When curves cross in your interval, you write a single integral of $(f(x)-g(x))$ over the entire interval, leading to cancellation of positive and negative areas. Why: Students remember the basic formula for constant bounds and forget that switching upper/lower changes the sign of the integrand. Correct move: After finding all intersection points in your interval, sort them in increasing order and split the integral into one subinterval between each consecutive pair of intersection points.
• Wrong move: You 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 the bounds, always 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 bounds from $x=a$ to $x=b$, you use the intersection points of the curves as your bounds instead of the given $a$ and $b$. Why: Students confuse finding intersection points for enclosed regions with given bounds for a region between two vertical lines. Correct move: Always read the problem carefully: if it says "between $x=a$ and $x=b$", use $a$ and $b$ as your bounds, do not solve for new intersection points.
• Wrong move: When finding area between a curve $f(x)$ and the x-axis, you write ${\int }_a^{b}f(x)dx$ instead of splitting the integral when $f(x)$ 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 below. Correct move: Split the integral where $f(x)$ crosses the x-axis, and integrate $(0-f(x))$ for intervals where $f(x)<0$.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is the correct set-up for the total area of the region bounded by $y=x^3-4x$ and $y=x$? A) ${\int }_{-\sqrt{5}}^{\sqrt{5}}(x^3-5x)dx$ B) ${\int }_{-\sqrt{5}}^0\left[(x^3-4x)-x\right]dx+{\int }_0^{\sqrt{5}}\left[x-(x^3-4x)\right]dx$ C) ${\int }_{-\sqrt{5}}^{\sqrt{5}}\left|x^3-5x\right|dx$ D) Both B and C

Worked Solution: First, find intersection points by setting $x^3-4x=x$, which simplifies to $x(x^2-5)=0$, giving roots at $x=-\sqrt{5},0,\sqrt{5}$. On the interval $(-\sqrt{5},0)$, $x^3-4x>x$, so the integrand is $(x^3-4x-x)$, and on $(0,\sqrt{5})$, $x>x^3-4x$, so the integrand becomes $x-(x^3-4x)$. This matches option B. The absolute value of $(f(x)-g(x))$ automatically accounts for the sign change, so the integral of $|x^3-5x|$ over the entire interval is also equal to total area, matching option C. Both B and C are correct. Correct answer: D.

Question 2 (Free Response)

Let $R$ be the region bounded by $y=e^{x/2}$, $y=1$, and $x=2$, where the curves intersect at $x=0$. (a) Find the area of region $R$. (b) Suppose the line $x=k$, where $0<k<2$, divides $R$ into two regions of equal area. Write, but do not solve, an equation involving one or more integrals that can be solved to find $k$. (c) Explain why your equation from part (b) has exactly one solution for $k$ between 0 and 2.

Worked Solution: (a) Upper function is $e^{x/2}$, lower function is $1$, bounds $[0,2]$. Set up: $$A={\int }_0^2\left(e^{x/2}-1\right)dx$$ Antiderivative: $2e^{x/2}-x$. Evaluate: $$A=(2e^1-2)-(2e^0-0)=2e-4\approx 1.437$$

(b) The area of the region left of $x=k$ equals half the total area, so the equation is: $${\int }_0^k\left(e^{x/2}-1\right)dx=e-2$$ (An equivalent correct answer is ${\int }_0^k\left(e^{x/2}-1\right)dx={\int }_k^2\left(e^{x/2}-1\right)dx$.)

(c) The integrand $e^{x/2}-1$ is positive for all $x>0$, so the function $F(k)={\int }_0^k\left(e^{x/2}-1\right)dx$ is strictly increasing on $(0,2)$. Since $F(0)=0<e-2<F(2)=2e-4$, by the Intermediate Value Theorem there is exactly one solution for $k$.

Question 3 (Application / Real-World Style)

A city is planning a new community park shaped as the bounded region between two rivers modeled by $y=0.5x^2-4x+10$ and $y=-0.5x^2+4x+2$, where $x$ and $y$ are measured in hundreds of meters. Find the total area of the park in hectares (note: 1 square unit in the model = $(100\text{m}{)}^2=1$ hectare).

Worked Solution:

1. Find intersection points by setting the functions equal: $$0.5x^2-4x+10=-0.5x^2+4x+2⟹x^2-8x+8=0$$ Solutions are $x=4-2\sqrt{2}$ and $x=4+2\sqrt{2}$, the lower and upper bounds.
2. Confirm upper function: $y=-0.5x^2+4x+2$ is upper, so the integrand is: $$(-0.5x^2+4x+2)-(0.5x^2-4x+10)=-x^2+8x-8$$
3. Substitute $u=x-4$ to simplify the integral, resulting in ${\int }_{-2\sqrt{2}}^{2\sqrt{2}}(-u^2+8)du=2{\left[-\frac{u^3}{3}+8u\right]}_0^{2\sqrt{2}}=\frac{64\sqrt{2}}{3}\approx 30.17$.

The total area of the park is approximately 30.2 hectares, meaning the planned park covers just over 300,000 square meters of land between the two rivers.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for the next core topic in 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 for AP Calculus AB; the same slicing logic you use here for area is extended to calculate the area of volume cross-sections. Without nailing the process of setting up integrals for area with respect to x, you will struggle to correctly set up volume integrals, which make up a large share of the FRQ section on the AP exam.

• Area between curves expressed as functions of y
• Volumes of solids of revolution: disk method
• Volumes of solids of revolution: washer method