Area Between Curves: Functions of x - AP Calculus AB

1. For $2\le x\le 5$, $f(x)=\ln x+3$ and $g(x)=2$. Which integral gives the area between the curves?
   Answer: $\displaystyle \int_{2}^{5}\big[(\ln x+3)-2\big]dx$ This problem asks for the area between two curves, requiring integration of the difference between upper and lower functions. Since ln(x) is positive for x>1, we have f(x)=ln(x)+3 > 3 on [2,5], while g(x)=2 is constant, so f(x) is above g(x) throughout the interval. The area formula is ∫[upper-lower]dx, giving us ∫₂⁵[(ln x+3)-2]dx. Choice B reverses the subtraction to 2-(ln x+3), which would yield a negative result since ln x+3 > 2 on this interval. Choice D incorrectly starts at x=0, but ln(x) is undefined at x=0, making this integral improper. When finding area between curves, ensure the integrand represents (upper function minus lower function) over the valid domain.
2. For $-2\le x\le 0$, $f(x)=2-x$ is above $g(x)=x^2$; which integral represents the area between them?
   Answer: $\displaystyle \int_{-2}^{0}\big((2-x)-x^2\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 2 - x is the upper curve and g(x) = x² is the lower curve on [-2, 0], so the integrand is (2 - x) - x². Integrating from -2 to 0 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice C, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
3. For $0\le x\le 1$, $f(x)=e^x$ is above $g(x)=1+x$. Which integral gives the area between them?
   Answer: $\displaystyle \int_{0}^{1}\big[e^x-(1+x)\big]dx$ To find the area between curves, we must integrate the upper function minus the lower function over the specified interval. Since f(x) = eˣ is above g(x) = 1 + x on [0, 1], the correct integral is ∫₀¹[eˣ - (1 + x)]dx. This follows the fundamental principle of subtracting the lower function from the upper function to obtain positive area. Choice A incorrectly computes [(1 + x) - eˣ], reversing the subtraction order and yielding a negative result. The strategy for any area problem is to first identify which function is on top, then set up the integral as upper minus lower.
4. On $[0,\pi]$, $f(x)=1+\sin x$ is above $g(x)=\sin^2 x$; which integral represents the area between them?
   Answer: $\displaystyle \int_{0}^{\pi}\big((1+\sin x)-\sin^2 x\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 1 + sin x is the upper curve and g(x) = sin² x is the lower curve on [0, π], so the integrand is (1 + sin x) - sin² x. Integrating from 0 to π ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
5. On $0\le x\le \pi$, $f(x)=2+\sin x$ is above $g(x)=1-\sin x$. Which integral represents the area between them?
   Answer: $\displaystyle \int_{0}^{\pi}\big[(2+\sin x)-(1-\sin x)\big]dx$ This problem involves finding the area between two trigonometric functions by integrating the upper function minus the lower function. Since f(x) = 2 + sin x is above g(x) = 1 - sin x on [0, π], we set up the integral as ∫₀^π[(2 + sin x) - (1 - sin x)]dx. When we distribute the negative sign, this becomes ∫₀^π[2 + sin x - 1 + sin x]dx = ∫₀^π[1 + 2sin x]dx. Choice C incorrectly adds the functions instead of subtracting, which doesn't represent the area between curves. Always remember to subtract the lower function from the upper function to find the enclosed area.
6. For $-1\le x\le 1$, $f(x)=2-x^2$ is above $g(x)=x$. Which integral gives the area between the curves?
   Answer: $\displaystyle \int_{-1}^{1}\big[(2-x^2)-x\big]dx$ Finding the area between curves requires integrating the upper function minus the lower function over the specified interval. Given that f(x) = 2 - x² is above g(x) = x on [-1, 1], we need ∫₋₁¹[(2 - x²) - x]dx. This maintains the proper order of upper minus lower to ensure a positive area result. Choice A incorrectly places x first and then subtracts (2 - x²), which reverses the roles and would give the negative of the actual area. The fundamental principle is to always subtract in the order that preserves the geometric meaning of area as a positive quantity.
7. For $1\le x\le 3$, $f(x)=\frac{1}{x}$ is above $g(x)=\frac{1}{3}$. Which integral gives the area between the curves?
   Answer: $\displaystyle \int_{1}^{3}\left[\frac{1}{x}-\frac{1}{3}\right]dx$ To find the area between curves, we integrate the difference of the upper function minus the lower function. Since f(x) = 1/x is above g(x) = 1/3 on [1, 3], the correct setup is ∫₁³[1/x - 1/3]dx. This follows the standard pattern of subtracting the lower horizontal line from the upper reciprocal function. Choice A incorrectly computes [1/3 - 1/x], which reverses the subtraction and would yield a negative value since 1/x > 1/3 on this interval. The fundamental principle remains consistent: always structure the integral as upper function minus lower function to find enclosed area.
8. On $-2\le x\le 0$, $f(x)=x^2$ is above $g(x)=2x+4$. Which integral represents the area between the curves?
   Answer: $\displaystyle \int_{-2}^{0}\big[x^2-(2x+4)\big]dx$ Finding the area between curves requires careful attention to which function is on top over the given interval. The problem states that f(x) = x² is above g(x) = 2x + 4 on [-2, 0], so we need ∫₋₂⁰[x² - (2x + 4)]dx. This maintains the upper minus lower structure essential for positive area. Choice C reverses this to [(2x + 4) - x²], which would give the negative of the area since it computes lower minus upper. Always verify which function has larger y-values on the given interval before setting up the integral to ensure correct subtraction order.
9. For $-3\le x\le -1$, $f(x)=\frac{1}{x}+5$ lies above $g(x)=4$; which integral gives the area between them?
   Answer: $\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)-4\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 1/x + 5 is the upper curve and g(x) = 4 is the lower curve on [-3, -1], so the integrand is (1/x + 5) - 4. Integrating from -3 to -1 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
10. On $-2\le x\le 0$, $f(x)=x^3+4$ and $g(x)=x+2$. Which integral represents the area between them?
    Answer: $\displaystyle \int_{-2}^{0}\big[(x^3+4)-(x+2)\big]dx$ To find the area between curves, we must integrate the difference between the upper and lower functions. Testing at x=-1: f(-1)=(-1)³+4=3 and g(-1)=-1+2=1, so f(x)=x³+4 is above g(x)=x+2 on [-2,0]. The area formula requires ∫[upper-lower]dx, yielding ∫₋₂⁰[(x³+4)-(x+2)]dx. Choice B incorrectly subtracts in reverse order, (x+2)-(x³+4), which would give a negative area since x+2 < x³+4 on this interval. Choice D adds the functions instead of subtracting, computing total area under both curves rather than the area between them. For area between curves, always verify which function has larger values and integrate (upper minus lower) over the specified bounds.
11. Between $x=-1$ and $x=3$, $f(x)=x+5$ is above $g(x)=2x+1$; which integral represents the area between them?
    Answer: $\displaystyle \int_{-1}^{3}\big((x+5)-(2x+1)\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x + 5 is the upper curve and g(x) = 2x + 1 is the lower curve on [-1, 3], so the integrand is (x + 5) - (2x + 1). Integrating from -1 to 3 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice B, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
12. Between $x=-1$ and $x=2$, $f(x)=x^2+2x+3$ is above $g(x)=x+1$; which integral represents the area?
    Answer: $\displaystyle \int_{-1}^{2}\big((x^2+2x+3)-(x+1)\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x² + 2x + 3 is the upper curve and g(x) = x + 1 is the lower curve on [-1, 2], so the integrand is (x² + 2x + 3) - (x + 1). Integrating from -1 to 2 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice C, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
13. On $\tfrac{\pi}{2}\le x\le \pi$, $f(x)=\cos x+2$ lies above $g(x)=1$. Which integral represents the area between them?
    Answer: $\displaystyle \int_{\pi/2}^{\pi}\big[(\cos x+2)-1\big]dx$ This problem requires finding the area between a trigonometric function and a horizontal line by integrating upper minus lower. Given that f(x) = cos x + 2 lies above g(x) = 1 on [π/2, π], we need ∫_{π/2}^π[(cos x + 2) - 1]dx. This simplifies to ∫_{π/2}^π[cos x + 1]dx, maintaining the correct subtraction order for positive area. Choice A reverses the subtraction to [1 - (cos x + 2)], which would give negative area since it computes lower minus upper. Remember that the area between curves is always computed as the integral of the top function minus the bottom function.
14. Between $x=-2$ and $x=2$, $f(x)=x^2+2$ lies above $g(x)=|x|$; which integral gives the area setup?
    Answer: $\displaystyle \int_{-2}^{2}\big((x^2+2)-|x|\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x² + 2 is the upper curve and g(x) = |x| is the lower curve on [-2, 2], so the integrand is (x² + 2) - |x|. Integrating from -2 to 2 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
15. On $[0,3]$, $f(x)=\sqrt{x}+2$ is above $g(x)=\frac{x}{3}$; which integral represents the area between them?
    Answer: $\displaystyle \int_{0}^{3}\big((\sqrt{x}+2)-\tfrac{x}{3}\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = √x + 2 is the upper curve and g(x) = x/3 is the lower curve on [0, 3], so the integrand is (√x + 2) - x/3. Integrating from 0 to 3 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice B, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
16. On $[2,4]$, $f(x)=x^2$ lies above $g(x)=4x-4$; which integral represents the area between the curves?
    Answer: $\displaystyle \int_{2}^{4}\big(x^2-(4x-4)\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = x² is the upper curve and g(x) = 4x - 4 is the lower curve on [2, 4], so the integrand is x² - (4x - 4). Integrating from 2 to 4 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice A, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
17. On $-\pi\le x\le 0$, $f(x)=\sin x$ lies above $g(x)=-1$. Which integral represents the area between the graphs?
    Answer: $\displaystyle \int_{-\pi}^{0}\big[\sin x-(-1)\big]dx$ Finding the area between curves requires integrating the upper function minus the lower function over the given interval. The problem states that f(x) = sin x lies above g(x) = -1 on [-π, 0], so we need ∫₋π⁰[sin x - (-1)]dx = ∫₋π⁰[sin x + 1]dx. This maintains the proper upper minus lower structure, where subtracting a negative becomes addition. Choice A incorrectly computes [(-1) - sin x], which reverses the roles and would give negative area. The key strategy is to always identify which function has larger y-values, then subtract the lower from the upper to ensure positive area.
18. For $0\le x\le 2$, $f(x)=x^2+1$ and $g(x)=3x+1$ with $g(x)\ge f(x)$. Which integral gives the area between them?
    Answer: $\displaystyle \int_{0}^{2}\big[(3x+1)-(x^2+1)\big]\,dx$ This problem requires finding the area between curves using the fundamental principle of integrating the upper function minus the lower function. Since we're told that g(x) ≥ f(x) on the interval [0, 2], this means g(x) = 3x + 1 is the upper function and f(x) = x² + 1 is the lower function. The area between curves is always calculated as ∫[upper - lower]dx, which gives us ∫₀²[(3x + 1) - (x² + 1)]dx. Choice A incorrectly subtracts in the wrong order, which would give a negative area. The key strategy is to always identify which function is on top before setting up the integral.
19. Between $x=-\pi$ and $x=0$, $f(x)=3-\cos x$ is above $g(x)=1+\sin x$; which integral gives the area?
    Answer: $\displaystyle \int_{-\pi}^{0}\big((3-\cos x)-(1+\sin x)\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 3 - cos x is the upper curve and g(x) = 1 + sin x is the lower curve on [-π, 0], so the integrand is (3 - cos x) - (1 + sin x). Integrating from -π to 0 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice C, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.
20. On $[1,5]$, $f(x)=\frac{5}{x}+1$ lies above $g(x)=1$; which integral gives the area between the curves?
    Answer: $\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)-1\big)\,dx$ This problem involves setting up an integral to find the area between two curves as functions of x. The area is given by the integral of the upper function minus the lower function over the given interval. Here, f(x) = 5/x + 1 is the upper curve and g(x) = 1 is the lower curve on [1, 5], so the integrand is (5/x + 1) - 1. Integrating from 1 to 5 ensures the limits go from left to right, yielding a positive value. A tempting distractor is choice B, which has the lower minus upper, resulting in a negative integral value. Always identify the upper and lower functions and integrate their difference from the left endpoint to the right endpoint for the correct area setup.

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