AP Calculus BC Quiz: Area Between Curves Functions Of X

What this quiz covers

This quiz focuses on Area Between Curves Functions Of X, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1}\big(g(x)-f(x)\big)\,dx$
3. $\displaystyle \int_{1}^{0}\big(f(x)-g(x)\big)\,dx$
4. $\displaystyle \int_{-1}^{0}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{2}\big(f(x)-g(x)\big)\,dx$

Explanation: To find the area between curves, we must integrate the difference (upper - lower) over the interval. Comparing f(x)=eˣ and g(x)=1+x on [0,1], at x=0: f(0)=1 and g(0)=1 (they start together), but at x=0.5: f(0.5)≈1.649 and g(0.5)=1.5, so f(x) is above g(x). Since eˣ grows faster than the linear function 1+x, f remains above g throughout [0,1]. The area is ∫[f(x)-g(x)]dx = ∫[eˣ-(1+x)]dx from 0 to 1. Choice B would give the negative of the area since it reverses the subtraction. Always verify which function is on top before setting up your integral.

Question 2

1. $\displaystyle \int_{0}^{\sqrt{3}}\big(f(x)-g(x)\big)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{\sqrt{3}}\big(g(x)-f(x)\big)\,dx$
3. $\displaystyle \int_{\sqrt{3}}^{0}\big(f(x)-g(x)\big)\,dx$
4. $\displaystyle \int_{0}^{3}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{\sqrt{3}}\big(f(x)+g(x)\big)\,dx$

Explanation: This problem involves finding the area between two curves, f(x) = 3x and g(x) = x³, over the interval [0, √3] using integration, a key skill in AP Calculus BC. To compute the area, we integrate the difference between the upper curve and the lower curve over the given interval. Here, f(x) lies above g(x) on (0, √3), since 3x > x³ when x² < 3. Therefore, the integral setup is ∫ from 0 to √3 of (f(x) - g(x)) dx, ensuring a positive area. A tempting distractor is integrating g(x) - f(x), but this gives a negative result instead of the area. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.

Question 3

1. $\displaystyle \int_{0}^{1}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{1}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{1}\big(f(x)+g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)^2\,dx$

Explanation: This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. On [0,1], with f(x) = ln(x+1) ≥ g(x) = x/2, the area integral uses upper minus lower as f(x) - g(x). This structure holds as the logarithm starts equal but stays above due to its concavity. Integrating from 0 to 1 sums the positive differences effectively. A tempting distractor is choice A, using g(x) - f(x), but it results in negative area, not the desired positive value. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_a^b (upper - lower) dx, first confirming which is upper via graphing or comparison.

Question 4

1. $\displaystyle \int_{-1}^{2}\big(f(x)-g(x)\big)\,dx$
2. $\displaystyle \int_{2}^{-1}\big(g(x)-f(x)\big)\,dx$
3. $\displaystyle \int_{-1}^{2}\big(g(x)-f(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{2}\big(g(x)-f(x)\big)\,dx$
5. $\displaystyle \int_{-1}^{1}\big(g(x)-f(x)\big)\,dx$

Explanation: To find the area between curves, we must integrate the difference between the upper and lower functions. The problem states g(x) ≥ f(x) on [-1,2], making g(x) = x + 5 the upper curve and f(x) = x³ + 5 the lower curve. The area is ∫[g(x) - f(x)]dx = ∫[(x + 5) - (x³ + 5)]dx = ∫[x - x³]dx from -1 to 2, which corresponds to choice C. Choice A incorrectly uses f(x) - g(x), reversing the subtraction order and yielding a negative result. Always integrate (upper function - lower function) to ensure positive area.

Question 5

1. $\displaystyle \int_{1}^{4}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{4}^{1}\big(g(x)-f(x)\big)\,dx$
3. $\displaystyle \int_{1}^{4}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{4}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{1}^{2}\big(f(x)-g(x)\big)\,dx$

Explanation: This problem asks for the area between two curves, requiring integration of the difference between upper and lower functions. Given f(x) = 4/x and g(x) = 1 with f(x) ≥ g(x) on [1,4], f(x) is the upper curve. The area is ∫[f(x) - g(x)]dx = ∫[4/x - 1]dx from 1 to 4, matching choice C. Choice A reverses the order to g(x) - f(x) = 1 - 4/x, which is negative on [1,4] and would give negative area. Remember: area between curves equals ∫[upper - lower]dx over the interval.

Question 6

1. $\displaystyle \int_{-2}^{2}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{2}^{-2}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{2}\big(f(x)-g(x)\big)\,dx$
4. $\displaystyle \int_{-2}^{2}\big(f(x)-g(x)\big)\,dx$ (correct answer)
5. $\displaystyle \int_{-2}^{0}\big(f(x)-g(x)\big)\,dx$

Explanation: This is an area between curves problem where we need to integrate the difference between the upper and lower functions. Since f(x) = 4 - x² and g(x) = 0 with f(x) ≥ g(x) on [-2,2], f(x) is the upper curve. The area is ∫[f(x) - g(x)]dx = ∫[4 - x²]dx from -2 to 2, corresponding to choice D. Choice A reverses the subtraction to g(x) - f(x) = -(4 - x²), which would yield a negative value. When finding area between curves, always integrate (upper - lower) to ensure a positive result.

Question 7

1. $\displaystyle \int_{0}^{1}\big((1+x)-e^x\big)\,dx$
2. $\displaystyle \int_{1}^{0}\big(e^x-(1+x)\big)\,dx$
3. $\displaystyle \int_{0}^{1}\big(e^x+(1+x)\big)\,dx$
4. $\displaystyle \int_{0}^{1}\big(e^x-(1+x)\big)\,dx$ (correct answer)
5. $\displaystyle \int_{0}^{1}\big(e^x-(1+x)\big)\,dy$

Explanation: This area between curves problem requires careful identification of which function is on top. Given that y = eˣ is the upper curve and y = 1 + x is the lower curve on [0, 1], we compute the area as ∫₀¹[eˣ - (1 + x)]dx. This follows the standard formula for area between curves: integrate the difference (upper - lower) over the given interval. Choice A reverses the subtraction, computing (1 + x) - eˣ, which would give a negative result. To avoid this error, always verify the relative positions of the curves before setting up the integral.

Question 8

1. $\displaystyle \int_{-\pi/2}^{\pi/2}\big(\sin x-3\cos x\big)\,dx$
2. $\displaystyle \int_{\pi/2}^{-\pi/2}\big(3\cos x-\sin x\big)\,dx$
3. $\displaystyle \int_{-\pi/2}^{\pi/2}\big(3\cos x+\sin x\big)\,dx$
4. $\displaystyle \int_{-\pi/2}^{\pi/2}\big(3\cos x-\sin x\big)\,dx$ (correct answer)
5. $\displaystyle \int_{-\pi/2}^{\pi/2}\big(3\cos x-\sin x\big)\,dy$

Explanation: To find the area between trigonometric curves, we apply the same principle: integrate (upper - lower). Since y = 3cos x is above y = sin x on [-π/2, π/2], the area integral is ∫₋π/₂^π/₂[3cos x - sin x]dx. This matches choice D, which correctly orders the subtraction to ensure a positive result. Choice A incorrectly computes (sin x - 3cos x), reversing the subtraction and yielding a negative integrand. When working with trig functions, it's especially helpful to sketch or evaluate at key points to confirm which function is on top.

Question 9

1. $\displaystyle \int_{-1}^{1}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{1}^{-1}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{-1}^{1}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{-1}^{0}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{-1}^{1}\big(f(x)+g(x)\big)\,dx$

Explanation: This problem involves finding the area between two curves, f(x) = 2 - x² and g(x) = x, over the interval [-1, 1] using integration, a key skill in AP Calculus BC. To compute the area, we integrate the difference between the upper curve and the lower curve over the given interval. Here, f(x) lies above g(x) throughout [-1, 1], as f - g = 2 - x² - x remains positive, verified by its minimum value. Therefore, the integral setup is ∫ from -1 to 1 of (f(x) - g(x)) dx, providing the enclosed area. A tempting distractor is integrating g(x) - f(x), but this would give a negative value instead of the positive area. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.

Question 10

1. $\displaystyle \int_{0}^{\pi}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{\pi}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{\pi}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{\pi/2}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{\pi}\big(f(x)+g(x)\big)\,dx$

Explanation: This problem involves finding the area between two curves, $f(x) = \sin x + 2$ and $g(x) = 2$, over the interval $[0, \pi]$ using integration, a key skill in AP Calculus BC. To compute the area, we integrate the difference between the upper curve and the lower curve over the given interval. Here, $f(x)$ lies above $g(x)$ on $(0, \pi)$, since $\sin x > 0$ in that interval, making $f(x) > 2$. Therefore, the integral setup is $\int_0^\pi (f(x) - g(x)) \, dx$, which simplifies to $\int_0^\pi \sin x \, dx$ and gives a positive area. A tempting distractor is integrating $g(x) - f(x)$, but this results in a negative value, not the area. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.

Question 11

1. $\displaystyle \int_{0}^{2}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{2}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{2}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{2}\big(f(x)+g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{2}\big(g(x)-f(x)\big)^2\,dx$

Explanation: This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. For f(x) = x + 2 and g(x) = x² on [0,2], with f(x) ≥ g(x), the integral is upper minus lower as f(x) - g(x). This is suitable because the line stays above the parabola in the interval, with positive difference. The setup from 0 to 2 covers the region up to their intersection at x=2. A tempting distractor is choice A, integrating g(x) - f(x), but this gives negative area, missing the positive requirement. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_a^b (upper - lower) dx, first confirming which is upper via graphing or comparison.

Question 12

1. $\displaystyle \int_{0}^{1}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{1}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{1}\big(f(x)+g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)^2\,dx$

Explanation: This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. On [0,1], with f(x) = e^x above g(x) = 1 + x, the area uses upper minus lower as f(x) - g(x). This structure applies since e^x grows faster than the linear function, keeping the difference positive. Integrating from 0 to 1 accumulates the area from their touch at x=0 onward. A tempting distractor is choice A, using g(x) - f(x), but this integrates a negative function, not yielding positive area. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_a^b (upper - lower) dx, first confirming which is upper via graphing or comparison.

Question 13

1. $\displaystyle \int_{0}^{1}\big(\sqrt{x}-x\big)\,dx$ (correct answer)
2. $\displaystyle \int_{1}^{0}\big(\sqrt{x}-x\big)\,dx$
3. $\displaystyle \int_{0}^{1}\big(x-\sqrt{x}\big)\,dx$
4. $\displaystyle \int_{0}^{1}\big(x+\sqrt{x}\big)\,dx$
5. $\displaystyle \int_{0}^{1}\big(x-\sqrt{x}\big)^2\,dx$

Explanation: This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. For f(x) = √x and g(x) = x on [0,1], with f(x) ≥ g(x), the area is found by integrating upper minus lower, which is √x - x. This applies because √x lies above x in the interval, making the difference positive and representative of the height at each x. The integral from 0 to 1 sums these heights, giving the total area without intersections requiring splits. A tempting distractor is choice C, integrating x - √x, but this gives a negative result since x < √x, not the positive area. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_a^b (upper - lower) dx, first confirming which is upper via graphing or comparison.

Question 14

1. $\displaystyle \int_{0}^{\pi}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{\pi}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{\pi}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{\pi}\big(f(x)+g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{\pi}\big(g(x)-f(x)\big)^2\,dx$

Explanation: This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. For f(x) = sin x and g(x) = 0 on [0, π], where f(x) ≥ g(x), the integral subtracts lower from upper as f(x) - g(x). This works because sin x is nonnegative over the interval, ensuring the difference measures positive heights. The limits from 0 to π cover the full hump above the x-axis without crossings. A tempting distractor is choice A, integrating g(x) - f(x), but this produces a negative area, which doesn't make sense for geometric area. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_a^b (upper - lower) dx, first confirming which is upper via graphing or comparison.

Question 15

1. $\displaystyle \int_{0}^{\pi/2}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{\pi/2}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{\pi/2}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{\pi/2}\big(f(x)+g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{\pi/2}\big(g(x)-f(x)\big)^2\,dx$

Explanation: This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. On [0, π/2], with f(x) = cos x ≥ g(x) = cos² x, the integral uses upper minus lower as f(x) - g(x). This applies because cos x exceeds its square in the interval, with positive difference. The limits from 0 to π/2 cover the region without internal crossings. A tempting distractor is choice A, integrating g(x) - f(x), but this negatives the area, failing geometrically. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_a^b (upper - lower) dx, first confirming which is upper via graphing or comparison.

Question 16

1. $\displaystyle \int_{0}^{\pi/2}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{\pi/2}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{\pi/2}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{\pi}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$

Explanation: Finding the area between curves requires integrating (upper - lower) over the interval. For f(x)=cos(x)+1 and g(x)=1 on [0,π/2], since cos(x) ranges from 1 to 0 on this interval, f(x)=cos(x)+1 ranges from 2 to 1. This means f(x)≥g(x)=1 throughout [0,π/2], with f(x) strictly above g(x) except at the endpoint x=π/2. The area is ∫[f(x)-g(x)]dx = ∫[cos(x)+1-1]dx = ∫cos(x)dx from 0 to π/2. Choice A incorrectly reverses the subtraction, giving a negative result. Remember that adding a constant to a function shifts it vertically, helping determine which function is on top.

Question 17

1. $\displaystyle \int_{0}^{1}\big(g(x)-f(x)\big)\,dx$
2. $\displaystyle \int_{1}^{0}\big(f(x)-g(x)\big)\,dx$
3. $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{2}\big(f(x)-g(x)\big)\,dx$
5. $\displaystyle \int_{-1}^{1}\big(f(x)-g(x)\big)\,dx$

Explanation: To find the area between curves, we must integrate the difference (upper function - lower function) over the given interval. Comparing f(x)=2-x and g(x)=x² on [0,1], let's test at x=0.5: f(0.5)=1.5 and g(0.5)=0.25, so f(x) is above g(x). We can verify this holds throughout [0,1] since both functions are continuous and they intersect at x=1 where f(1)=g(1)=1. The area is therefore ∫[f(x)-g(x)]dx from 0 to 1. Choice A incorrectly reverses the subtraction order, yielding a negative value. The essential strategy is to identify the upper function first, then integrate (upper - lower) to ensure a positive area.

Question 18

1. $\displaystyle \int_{0}^{2}\big(\tfrac{1}{x+1}-(\tfrac{1}{x+1}+3)\big)\,dx$
2. $\displaystyle \int_{2}^{0}\big((\tfrac{1}{x+1}+3)-\tfrac{1}{x+1}\big)\,dx$
3. $\displaystyle \int_{0}^{2}\big((\tfrac{1}{x+1}+3)+\tfrac{1}{x+1}\big)\,dx$
4. $\displaystyle \int_{0}^{2}\big((\tfrac{1}{x+1}+3)-\tfrac{1}{x+1}\big)\,dy$
5. $\displaystyle \int_{0}^{2}\big((\tfrac{1}{x+1}+3)-\tfrac{1}{x+1}\big)\,dx$ (correct answer)

Explanation: This area problem involves rational functions where y = 1/(x+1) + 3 is above y = 1/(x+1) on [0, 2]. The area integral is ∫₀²[(1/(x+1) + 3) - 1/(x+1)]dx, which simplifies to ∫₀²[3]dx. Choice E correctly shows this setup, maintaining the proper (upper - lower) order. Choice D has the wrong differential (dy instead of dx), making it incorrect despite having the right integrand. When curves differ by a constant, the area calculation becomes particularly simple, but proper integral notation remains essential.

Question 19

1. $\displaystyle \int_{0}^{3}\big(f(x)-g(x)\big)\,dx$
2. $\displaystyle \int_{0}^{3}\big(g(x)-f(x)\big)\,dx$ (correct answer)
3. $\displaystyle \int_{3}^{0}\big(g(x)-f(x)\big)\,dx$
4. $\displaystyle \int_{0}^{1}\big(g(x)-f(x)\big)\,dx$
5. $\displaystyle \int_{1}^{3}\big(f(x)-g(x)\big)\,dx$

Explanation: This problem requires finding the area between curves, which involves integrating the difference between the upper and lower functions. To determine which function is on top, we can test a point in the interval $[0,3]$: at $x=1$, $f(1)=2$ and $g(1)=4$, so g(x) is above f(x). Since g(x) is the upper function throughout $[0,3]$, the area is given by $\int [g(x)-f(x)] \, dx$ from 0 to 3. Choice A incorrectly subtracts in the wrong order (f-g), which would give a negative result since f is below g. The key strategy is to always check which function is on top by evaluating at a test point, then integrate (upper - lower) over the given interval.

Question 20

1. $\displaystyle \int_{-1}^{3}\big((x^2+1)-(2x+5)\big)\,dx$
2. $\displaystyle \int_{3}^{-1}\big((2x+5)-(x^2+1)\big)\,dx$
3. $\displaystyle \int_{-1}^{3}\big((2x+5)-(x^2+1)\big)\,dx$ (correct answer)
4. $\displaystyle \int_{-1}^{3}\big((2x+5)+(x^2+1)\big)\,dx$
5. $\displaystyle \int_{-1}^{3}\big((2x+5)-(x^2+1)\big)\,dy$

Explanation: This is an area between curves problem where we need to integrate the difference of the functions. Given that y = 2x + 5 is above y = x² + 1 on the interval [-1, 3], we set up the integral as ∫[upper - lower]dx. This becomes ∫₋₁³[(2x + 5) - (x² + 1)]dx, which matches choice C exactly. Choice A reverses the subtraction order, yielding a negative result since it computes (lower - upper). When setting up area integrals, always ensure the upper function comes first in the subtraction to guarantee a positive area value.