Area Between Curves: Functions of x - AP Calculus BC

Let $f(x)=\sin x$ and $g(x)=0$ on $[0,\pi]$ , where $f(x)\ge g(x)$ . Which integral gives the area?
Answer: $\displaystyle \int_{0}^{\pi}\big(f(x)-g(x)\big)\,dx$ 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.
For $f(x)=e^x$ and $g(x)=1+x$ on $[0,1]$ , with $f$ above $g$ , which integral gives area?
Answer: $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ This problem involves finding the area between two curves, f(x) = $e^x$ and g(x) = 1 + x, over the interval [0, 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) on (0, 1), since $e^x$ > 1 + x for x > 0, based on the Taylor series expansion. Therefore, the integral setup is ∫ from 0 to 1 of (f(x) - g(x)) dx, providing the positive area. A tempting distractor is integrating g(x) - f(x), but this results in a negative value. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.
On $[-1,1]$ , $f(x)=2-x^2$ and $g(x)=x$ with $f(x)\ge g(x)$ ; which integral gives the area between them?
Answer: $\displaystyle \int_{-1}^{1}\big(f(x)-g(x)\big)\,dx$ This is an area between curves problem where we must integrate the difference of the upper and lower functions. Given that f(x) ≥ g(x) on [-1,1], f(x) = 2 - x² is the upper curve and g(x) = x is the lower curve. The area is calculated as ∫[f(x) - g(x)]dx from -1 to 1, which matches choice A. Choice B reverses the subtraction order, which would produce a negative result—area must always be positive. When setting up area integrals, verify which function is greater throughout the interval, then integrate (upper - lower).
Let $f(x)=x^2+1$ and $g(x)=3x+1$ . On $[0,3]$ , which integral gives the area between the curves?
Answer: $\displaystyle \int_{0}^{3}\big(g(x)-f(x)\big)\,dx$ 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.
On $[-2,2]$ , $f(x)=4-x^2$ and $g(x)=0$ with $f(x)\ge g(x)$ ; which integral is the area between the curves?
Answer: $\displaystyle \int_{-2}^{2}\big(f(x)-g(x)\big)\,dx$ 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.
Let $f(x)=4-x^2$ and $g(x)=0$ . On $[-2,2]$ , which integral gives the area between the curves?
Answer: $\displaystyle \int_{-2}^{2}\big(f(x)-g(x)\big)\,dx$ To find the area between curves, we integrate (upper - lower) over the given interval. With f(x)=4-x² and g(x)=0 on [-2,2], we see that f(x) is a downward parabola with vertex at (0,4). At the endpoints, f(-2)=f(2)=0, and f(x)>0 for x in (-2,2), so f(x) is above g(x)=0 throughout the open interval. The area is ∫[f(x)-g(x)]dx = ∫[4-x²]dx from -2 to 2. Choice A would give the negative of the area since it uses g(x)-f(x) where g is below f. When one function is the x-axis (y=0), the area calculation simplifies to just integrating the other function.
For $1\le x\le 4$ , $f(x)=\sqrt{x}$ and $g(x)=\frac{x}{2}$ with $f(x)\ge g(x)$ ; which integral represents the area?
Answer: $\displaystyle \int_{1}^{4}\big(f(x)-g(x)\big)\,dx$ To find the area between curves, we need to integrate the difference between the upper and lower functions over the specified interval. The problem states that f(x) ≥ g(x) on [1,4], making f(x) the upper function and g(x) the lower function. The area formula is ∫[upper - lower]dx = ∫[f(x) - g(x)]dx from 1 to 4. Choice A incorrectly uses g(x) - f(x), which would yield a negative value since g(x) < f(x) on this interval. Remember that the area between curves is always positive, so we must subtract the lower function from the upper function.
Let $f(x)=e^x$ and $g(x)=1+x$ . On $[0,1]$ , which integral gives the area between the graphs?
Answer: $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ 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.
For $0\le x\le \pi$ , $f(x)=\sin x+2$ and $g(x)=2$ with $f(x)\ge g(x)$ ; which integral is the area?
Answer: $\displaystyle \int_{0}^{\pi}\big(f(x)-g(x)\big)\,dx$ Finding the area between curves requires integrating the difference between the upper and lower functions. Since f(x) = sin x + 2 and g(x) = 2, we have f(x) - g(x) = sin x, which is non-negative on [0,π], confirming f(x) ≥ g(x). The area is ∫[f(x) - g(x)]dx from 0 to π, corresponding to choice D. Choice A incorrectly subtracts in the reverse order, yielding -∫sin x dx, which would be negative. Always ensure you're subtracting the lower function from the upper function to get positive area.
On $1\le x\le 4$ , the region is between $y=\sqrt{x}$ (upper) and $y=\frac{x}{4}$ (lower). Choose the area integral.
Answer: $\displaystyle \int_{1}^{4}\big(\sqrt{x}-\tfrac{x}{4}\big)\,dx$ This problem asks for the area between two curves where y = √x is the upper function and y = x/4 is the lower function on [1, 4]. The area integral follows the standard form ∫[upper - lower]dx, giving us ∫₁⁴[√x - x/4]dx. This matches choice D perfectly, as it correctly subtracts the lower function from the upper function. Choice A reverses this subtraction (x/4 - √x), which would yield a negative value since it computes (lower - upper). Always identify which function is on top before writing the integral to ensure correct subtraction order.
On $[0, \pi]$ , $f(x) = \sin x + 2$ is above $g(x) = 2$ ; which integral gives the area between curves?
Answer: $\displaystyle \int_{0}^{\pi}\big(f(x)-g(x)\big)\,dx$ 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.
On $[-1,1]$ , $f(x)=2-x^2$ lies above $g(x)=x$ ; which integral equals the enclosed area?
Answer: $\displaystyle \int_{-1}^{1}\big(f(x)-g(x)\big)\,dx$ 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.
On $-1\le x\le 3$ , $y=2x+5$ is above $y=x^2+1$ . Which integral represents the area between the curves?
Answer: $\displaystyle \int_{-1}^{3}\big((2x+5)-(x^2+1)\big)\,dx$ 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.
For $0\le x\le 2$ , the region is bounded by $y=4-x^2$ above and $y=x$ below. Which integral gives its area?
Answer: $\displaystyle \int_{0}^{2}\big((4-x^2)-x\big)\,dx$ This problem requires finding the area between curves using integration. Since we're told that y = 4 - x² is above and y = x is below on the interval [0, 2], we apply the fundamental principle: area = ∫[upper function - lower function]dx. This gives us ∫₀²[(4 - x²) - x]dx, which simplifies to ∫₀²[(4 - x²) - x]dx. Choice A incorrectly subtracts in the wrong order (lower - upper), which would give a negative area. The key strategy is to always verify which function is on top by checking a test point or sketching the curves before setting up the integral.
On $[1,4]$ , $f(x)=\frac{4}{x}$ and $g(x)=1$ with $f(x)\ge g(x)$ ; which integral gives the area?
Answer: $\displaystyle \int_{1}^{4}\big(f(x)-g(x)\big)\,dx$ 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.
Let $f(x)=\cos x$ and $g(x)=\cos^2 x$ on $\left[0,\frac{\pi}{2}\right]$ , where $f(x)\ge g(x)$ . Which setup is correct?
Answer: $\displaystyle \int_{0}^{\pi/2}\big(f(x)-g(x)\big)\,dx$ 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.
Let $f(x)=\ln(x+1)+1$ and $g(x)=1$ . On $[0,2]$ , which integral represents the area between them?
Answer: $\displaystyle \int_{0}^{2}\big(f(x)-g(x)\big)\,dx$ Finding the area between curves involves integrating (upper function - lower function). Here we have f(x)=ln(x+1)+1 and g(x)=1 on [0,2]. Since ln(x+1) is positive for x>0 and equals 0 at x=0, f(x)=ln(x+1)+1 starts at f(0)=1 and increases, making f(x)≥g(x)=1 throughout [0,2]. The area is ∫[f(x)-g(x)]dx = ∫[ln(x+1)+1-1]dx = ∫ln(x+1)dx from 0 to 2. Choice A incorrectly uses g(x)-f(x), which would yield a negative result. The strategy is to recognize that when one function is a horizontal line, you only need to check if the other function is above or below it.
For $f(x)=e^x$ and $g(x)=1+x$ on $[0,1]$ with $f(x)$ above $g(x)$ , which integral is the area?
Answer: $\displaystyle \int_{0}^{1}\big(f(x)-g(x)\big)\,dx$ 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.
On $[0,3]$ , $f(x)=x+1$ and $g(x)=1+\frac{x^2}{3}$ with $f(x)\ge g(x)$ ; which integral gives the area?
Answer: $\displaystyle \int_{0}^{3}\big(f(x)-g(x)\big)\,dx$ This problem requires finding the area between two curves by integrating the difference of the upper and lower functions. We're told f(x) ≥ g(x) on [0,3], so f(x) = x + 1 is the upper curve and g(x) = 1 + x²/3 is the lower curve. The area equals ∫[f(x) - g(x)]dx from 0 to 3, which is choice A. Choice B reverses the subtraction, giving ∫[g(x) - f(x)]dx, which would be negative since g(x) < f(x). For area between curves, always integrate (upper function - lower function) over the given bounds.
For $f(x)=\cos x$ and $g(x)=0$ on $[0,\pi/2]$ , with $f$ above $g$ , which integral gives area?
Answer: $\displaystyle \int_{0}^{\pi/2}\big(f(x)-g(x)\big)\,dx$ This problem involves finding the area between two curves, f(x) = cos x and g(x) = 0, over the interval [0, π/2] 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, π/2), since cos x > 0 in that region. Therefore, the integral setup is ∫ from 0 to π/2 of (f(x) - g(x)) dx, yielding the positive area. A tempting distractor is integrating g(x) - f(x), but this gives a negative value. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.

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