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AP Calculus AB Quiz

AP Calculus AB Quiz: Area Between Curves Functions Of X

Practice Area Between Curves Functions Of X in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

On [0,π][0,\pi][0,π], f(x)=1+sin⁡xf(x)=1+\sin xf(x)=1+sinx is above g(x)=sin⁡2xg(x)=\sin^2 xg(x)=sin2x; which integral represents the area between them?

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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 AB.

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.

All questions

Question 1

On [0,π][0,\pi][0,π], f(x)=1+sin⁡xf(x)=1+\sin xf(x)=1+sinx is above g(x)=sin⁡2xg(x)=\sin^2 xg(x)=sin2x; which integral represents the area between them?

  1. ∫0π(sin⁡2x−(1+sin⁡x)) dx\displaystyle \int_{0}^{\pi}\big(\sin^2 x-(1+\sin x)\big)\,dx∫0π​(sin2x−(1+sinx))dx
  2. ∫π0((1+sin⁡x)−sin⁡2x) dx\displaystyle \int_{\pi}^{0}\big((1+\sin x)-\sin^2 x\big)\,dx∫π0​((1+sinx)−sin2x)dx
  3. ∫0π((1+sin⁡x)+sin⁡2x) dx\displaystyle \int_{0}^{\pi}\big((1+\sin x)+\sin^2 x\big)\,dx∫0π​((1+sinx)+sin2x)dx
  4. ∫0π((1+sin⁡x)−sin⁡2x) dx\displaystyle \int_{0}^{\pi}\big((1+\sin x)-\sin^2 x\big)\,dx∫0π​((1+sinx)−sin2x)dx (correct answer)
  5. ∫π0(sin⁡2x−(1+sin⁡x)) dx\displaystyle \int_{\pi}^{0}\big(\sin^2 x-(1+\sin x)\big)\,dx∫π0​(sin2x−(1+sinx))dx

Explanation: 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.

Question 2

Between x=0x=0x=0 and x=2x=2x=2, f(x)=3xf(x)=3^xf(x)=3x is above g(x)=2xg(x)=2^xg(x)=2x; which integral represents the area between them?

  1. ∫02(2x−3x) dx\displaystyle \int_{0}^{2}\big(2^x-3^x\big)\,dx∫02​(2x−3x)dx
  2. ∫20(3x−2x) dx\displaystyle \int_{2}^{0}\big(3^x-2^x\big)\,dx∫20​(3x−2x)dx
  3. ∫02(3x+2x) dx\displaystyle \int_{0}^{2}\big(3^x+2^x\big)\,dx∫02​(3x+2x)dx
  4. ∫20(2x−3x) dx\displaystyle \int_{2}^{0}\big(2^x-3^x\big)\,dx∫20​(2x−3x)dx
  5. ∫02(3x−2x) dx\displaystyle \int_{0}^{2}\big(3^x-2^x\big)\,dx∫02​(3x−2x)dx (correct answer)

Explanation: 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^x is the upper curve and g(x) = 2^x is the lower curve on [0, 2], so the integrand is 3^x - 2^x. Integrating from 0 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.

Question 3

For 1≤x≤41\le x\le 41≤x≤4, f(x)=x2+12f(x)=\frac{x^2+1}{2}f(x)=2x2+1​ is above g(x)=xg(x)=xg(x)=x; which integral represents the area?

  1. ∫14(x2+12−x) dx\displaystyle \int_{1}^{4}\big(\tfrac{x^2+1}{2}-x\big)\,dx∫14​(2x2+1​−x)dx (correct answer)
  2. ∫14(x−x2+12) dx\displaystyle \int_{1}^{4}\big(x-\tfrac{x^2+1}{2}\big)\,dx∫14​(x−2x2+1​)dx
  3. ∫41(x2+12−x) dx\displaystyle \int_{4}^{1}\big(\tfrac{x^2+1}{2}-x\big)\,dx∫41​(2x2+1​−x)dx
  4. ∫14(x2+12+x) dx\displaystyle \int_{1}^{4}\big(\tfrac{x^2+1}{2}+x\big)\,dx∫14​(2x2+1​+x)dx
  5. ∫41(x−x2+12) dx\displaystyle \int_{4}^{1}\big(x-\tfrac{x^2+1}{2}\big)\,dx∫41​(x−2x2+1​)dx

Explanation: 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 + 1)/2 is the upper curve and g(x) = x is the lower curve on [1, 4], so the integrand is (x² + 1)/2 - x. Integrating from 1 to 4 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.

Question 4

Between x=−1x=-1x=−1 and x=2x=2x=2, f(x)=x2+2x+3f(x)=x^2+2x+3f(x)=x2+2x+3 is above g(x)=x+1g(x)=x+1g(x)=x+1; which integral represents the area?

  1. ∫−12((x2+2x+3)−(x+1)) dx\displaystyle \int_{-1}^{2}\big((x^2+2x+3)-(x+1)\big)\,dx∫−12​((x2+2x+3)−(x+1))dx (correct answer)
  2. ∫2−1((x2+2x+3)−(x+1)) dx\displaystyle \int_{2}^{-1}\big((x^2+2x+3)-(x+1)\big)\,dx∫2−1​((x2+2x+3)−(x+1))dx
  3. ∫−12((x+1)−(x2+2x+3)) dx\displaystyle \int_{-1}^{2}\big((x+1)-(x^2+2x+3)\big)\,dx∫−12​((x+1)−(x2+2x+3))dx
  4. ∫−12((x2+2x+3)+(x+1)) dx\displaystyle \int_{-1}^{2}\big((x^2+2x+3)+(x+1)\big)\,dx∫−12​((x2+2x+3)+(x+1))dx
  5. ∫2−1((x+1)−(x2+2x+3)) dx\displaystyle \int_{2}^{-1}\big((x+1)-(x^2+2x+3)\big)\,dx∫2−1​((x+1)−(x2+2x+3))dx

Explanation: 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.

Question 5

For 0≤x≤20\le x\le 20≤x≤2, f(x)=x2+1f(x)=x^2+1f(x)=x2+1 and g(x)=3x+1g(x)=3x+1g(x)=3x+1 with g(x)≥f(x)g(x)\ge f(x)g(x)≥f(x). Which integral gives the area between them?

  1. ∫02[(x2+1)−(3x+1)] dx\displaystyle \int_{0}^{2}\big[(x^2+1)-(3x+1)\big]\,dx∫02​[(x2+1)−(3x+1)]dx
  2. ∫02[(3x+1)−(x2+1)] dx\displaystyle \int_{0}^{2}\big[(3x+1)-(x^2+1)\big]\,dx∫02​[(3x+1)−(x2+1)]dx (correct answer)
  3. ∫20[(3x+1)−(x2+1)] dx\displaystyle \int_{2}^{0}\big[(3x+1)-(x^2+1)\big]\,dx∫20​[(3x+1)−(x2+1)]dx
  4. ∫03[(3x+1)−(x2+1)] dx\displaystyle \int_{0}^{3}\big[(3x+1)-(x^2+1)\big]\,dx∫03​[(3x+1)−(x2+1)]dx
  5. ∫02[(3x+1)+(x2+1)] dx\displaystyle \int_{0}^{2}\big[(3x+1)+(x^2+1)\big]\,dx∫02​[(3x+1)+(x2+1)]dx

Explanation: 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.

Question 6

For 0≤x≤10\le x\le 10≤x≤1, f(x)=exf(x)=e^xf(x)=ex is above g(x)=1+xg(x)=1+xg(x)=1+x. Which integral gives the area between them?

  1. ∫01[(1+x)−ex]dx\displaystyle \int_{0}^{1}\big[(1+x)-e^x\big]dx∫01​[(1+x)−ex]dx
  2. ∫10[ex−(1+x)]dx\displaystyle \int_{1}^{0}\big[e^x-(1+x)\big]dx∫10​[ex−(1+x)]dx
  3. ∫01[ex+(1+x)]dx\displaystyle \int_{0}^{1}\big[e^x+(1+x)\big]dx∫01​[ex+(1+x)]dx
  4. ∫0e[ex−(1+x)]dx\displaystyle \int_{0}^{e}\big[e^x-(1+x)\big]dx∫0e​[ex−(1+x)]dx
  5. ∫01[ex−(1+x)]dx\displaystyle \int_{0}^{1}\big[e^x-(1+x)\big]dx∫01​[ex−(1+x)]dx (correct answer)

Explanation: 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.

Question 7

For −1≤x≤1-1\le x\le 1−1≤x≤1, f(x)=2−x2f(x)=2-x^2f(x)=2−x2 is above g(x)=xg(x)=xg(x)=x. Which integral gives the area between the curves?

  1. ∫−11[x−(2−x2)]dx\displaystyle \int_{-1}^{1}\big[x-(2-x^2)\big]dx∫−11​[x−(2−x2)]dx
  2. ∫1−1[(2−x2)−x]dx\displaystyle \int_{1}^{-1}\big[(2-x^2)-x\big]dx∫1−1​[(2−x2)−x]dx
  3. ∫−11[(2−x2)−x]dx\displaystyle \int_{-1}^{1}\big[(2-x^2)-x\big]dx∫−11​[(2−x2)−x]dx (correct answer)
  4. ∫−12[(2−x2)−x]dx\displaystyle \int_{-1}^{2}\big[(2-x^2)-x\big]dx∫−12​[(2−x2)−x]dx
  5. ∫−11[(2−x2)+x]dx\displaystyle \int_{-1}^{1}\big[(2-x^2)+x\big]dx∫−11​[(2−x2)+x]dx

Explanation: 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.

Question 8

For 2≤x≤52\le x\le 52≤x≤5, f(x)=ln⁡xf(x)=\ln xf(x)=lnx lies above g(x)=ln⁡2g(x)=\ln 2g(x)=ln2. Which integral gives the area between the graphs?

  1. ∫25[(ln⁡2)−ln⁡x]dx\displaystyle \int_{2}^{5}\big[(\ln 2)-\ln x\big]dx∫25​[(ln2)−lnx]dx
  2. ∫52[ln⁡x−ln⁡2]dx\displaystyle \int_{5}^{2}\big[\ln x-\ln 2\big]dx∫52​[lnx−ln2]dx
  3. ∫25[ln⁡x−ln⁡2]dx\displaystyle \int_{2}^{5}\big[\ln x-\ln 2\big]dx∫25​[lnx−ln2]dx (correct answer)
  4. ∫05[ln⁡x−ln⁡2]dx\displaystyle \int_{0}^{5}\big[\ln x-\ln 2\big]dx∫05​[lnx−ln2]dx
  5. ∫25[ln⁡x+ln⁡2]dx\displaystyle \int_{2}^{5}\big[\ln x+\ln 2\big]dx∫25​[lnx+ln2]dx

Explanation: To find the area between curves, we integrate the difference of the upper function minus the lower function. The problem states that f(x) = ln x lies above g(x) = ln 2 (a horizontal line) on [2, 5]. The correct setup is ∫₂⁵[ln x - ln 2]dx, subtracting the lower constant function from the upper logarithmic function. Choice A reverses this order to [ln 2 - ln x], which would give a negative result since ln x > ln 2 for x > 2. The key insight is that even when one function is constant, the same upper-minus-lower principle applies for finding enclosed area.

Question 9

For −3≤x≤−1-3\le x\le -1−3≤x≤−1, f(x)=1x+5f(x)=\frac{1}{x}+5f(x)=x1​+5 lies above g(x)=4g(x)=4g(x)=4; which integral gives the area between them?

  1. ∫−3−1(4−(1x+5)) dx\displaystyle \int_{-3}^{-1}\big(4-(\tfrac{1}{x}+5)\big)\,dx∫−3−1​(4−(x1​+5))dx
  2. ∫−1−3((1x+5)−4) dx\displaystyle \int_{-1}^{-3}\big((\tfrac{1}{x}+5)-4\big)\,dx∫−1−3​((x1​+5)−4)dx
  3. ∫−3−1((1x+5)+4) dx\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)+4\big)\,dx∫−3−1​((x1​+5)+4)dx
  4. ∫−3−1((1x+5)−4) dx\displaystyle \int_{-3}^{-1}\big((\tfrac{1}{x}+5)-4\big)\,dx∫−3−1​((x1​+5)−4)dx (correct answer)
  5. ∫−1−3(4−(1x+5)) dx\displaystyle \int_{-1}^{-3}\big(4-(\tfrac{1}{x}+5)\big)\,dx∫−1−3​(4−(x1​+5))dx

Explanation: 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.

Question 10

On [1,5][1,5][1,5], f(x)=5x+1f(x)=\frac{5}{x}+1f(x)=x5​+1 lies above g(x)=1g(x)=1g(x)=1; which integral gives the area between the curves?

  1. ∫15((5x+1)−1) dx\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)-1\big)\,dx∫15​((x5​+1)−1)dx (correct answer)
  2. ∫15(1−(5x+1)) dx\displaystyle \int_{1}^{5}\big(1-(\tfrac{5}{x}+1)\big)\,dx∫15​(1−(x5​+1))dx
  3. ∫51((5x+1)−1) dx\displaystyle \int_{5}^{1}\big((\tfrac{5}{x}+1)-1\big)\,dx∫51​((x5​+1)−1)dx
  4. ∫15((5x+1)+1) dx\displaystyle \int_{1}^{5}\big((\tfrac{5}{x}+1)+1\big)\,dx∫15​((x5​+1)+1)dx
  5. ∫51(1−(5x+1)) dx\displaystyle \int_{5}^{1}\big(1-(\tfrac{5}{x}+1)\big)\,dx∫51​(1−(x5​+1))dx

Explanation: 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.

Question 11

On [2,4][2,4][2,4], f(x)=x2f(x)=x^2f(x)=x2 lies above g(x)=4x−4g(x)=4x-4g(x)=4x−4; which integral represents the area between the curves?

  1. ∫24((4x−4)−x2) dx\displaystyle \int_{2}^{4}\big((4x-4)-x^2\big)\,dx∫24​((4x−4)−x2)dx
  2. ∫24(x2−(4x−4)) dx\displaystyle \int_{2}^{4}\big(x^2-(4x-4)\big)\,dx∫24​(x2−(4x−4))dx (correct answer)
  3. ∫42(x2−(4x−4)) dx\displaystyle \int_{4}^{2}\big(x^2-(4x-4)\big)\,dx∫42​(x2−(4x−4))dx
  4. ∫24(x2+(4x−4)) dx\displaystyle \int_{2}^{4}\big(x^2+(4x-4)\big)\,dx∫24​(x2+(4x−4))dx
  5. ∫42((4x−4)−x2) dx\displaystyle \int_{4}^{2}\big((4x-4)-x^2\big)\,dx∫42​((4x−4)−x2)dx

Explanation: 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.

Question 12

Between x=−1x=-1x=−1 and x=3x=3x=3, f(x)=x+5f(x)=x+5f(x)=x+5 is above g(x)=2x+1g(x)=2x+1g(x)=2x+1; which integral represents the area between them?

  1. ∫3−1((x+5)−(2x+1)) dx\displaystyle \int_{3}^{-1}\big((x+5)-(2x+1)\big)\,dx∫3−1​((x+5)−(2x+1))dx
  2. ∫−13((2x+1)−(x+5)) dx\displaystyle \int_{-1}^{3}\big((2x+1)-(x+5)\big)\,dx∫−13​((2x+1)−(x+5))dx
  3. ∫−13((x+5)+(2x+1)) dx\displaystyle \int_{-1}^{3}\big((x+5)+(2x+1)\big)\,dx∫−13​((x+5)+(2x+1))dx
  4. ∫−13((x+5)−(2x+1)) dx\displaystyle \int_{-1}^{3}\big((x+5)-(2x+1)\big)\,dx∫−13​((x+5)−(2x+1))dx (correct answer)
  5. ∫3−1((2x+1)−(x+5)) dx\displaystyle \int_{3}^{-1}\big((2x+1)-(x+5)\big)\,dx∫3−1​((2x+1)−(x+5))dx

Explanation: 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.

Question 13

Between x=−πx=-\pix=−π and x=0x=0x=0, f(x)=3−cos⁡xf(x)=3-\cos xf(x)=3−cosx is above g(x)=1+sin⁡xg(x)=1+\sin xg(x)=1+sinx; which integral gives the area?

  1. ∫−π0((3−cos⁡x)−(1+sin⁡x)) dx\displaystyle \int_{-\pi}^{0}\big((3-\cos x)-(1+\sin x)\big)\,dx∫−π0​((3−cosx)−(1+sinx))dx (correct answer)
  2. ∫0−π((3−cos⁡x)−(1+sin⁡x)) dx\displaystyle \int_{0}^{-\pi}\big((3-\cos x)-(1+\sin x)\big)\,dx∫0−π​((3−cosx)−(1+sinx))dx
  3. ∫−π0((1+sin⁡x)−(3−cos⁡x)) dx\displaystyle \int_{-\pi}^{0}\big((1+\sin x)-(3-\cos x)\big)\,dx∫−π0​((1+sinx)−(3−cosx))dx
  4. ∫−π0((3−cos⁡x)+(1+sin⁡x)) dx\displaystyle \int_{-\pi}^{0}\big((3-\cos x)+(1+\sin x)\big)\,dx∫−π0​((3−cosx)+(1+sinx))dx
  5. ∫0−π((1+sin⁡x)−(3−cos⁡x)) dx\displaystyle \int_{0}^{-\pi}\big((1+\sin x)-(3-\cos x)\big)\,dx∫0−π​((1+sinx)−(3−cosx))dx

Explanation: 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.

Question 14

On [0,3][0,3][0,3], f(x)=x+2f(x)=\sqrt{x}+2f(x)=x​+2 is above g(x)=x3g(x)=\frac{x}{3}g(x)=3x​; which integral represents the area between them?

  1. ∫30((x+2)−x3) dx\displaystyle \int_{3}^{0}\big((\sqrt{x}+2)-\tfrac{x}{3}\big)\,dx∫30​((x​+2)−3x​)dx
  2. ∫03(x3−(x+2)) dx\displaystyle \int_{0}^{3}\big(\tfrac{x}{3}-(\sqrt{x}+2)\big)\,dx∫03​(3x​−(x​+2))dx
  3. ∫03((x+2)−x3) dx\displaystyle \int_{0}^{3}\big((\sqrt{x}+2)-\tfrac{x}{3}\big)\,dx∫03​((x​+2)−3x​)dx (correct answer)
  4. ∫03((x+2)+x3) dx\displaystyle \int_{0}^{3}\big((\sqrt{x}+2)+\tfrac{x}{3}\big)\,dx∫03​((x​+2)+3x​)dx
  5. ∫30(x3−(x+2)) dx\displaystyle \int_{3}^{0}\big(\tfrac{x}{3}-(\sqrt{x}+2)\big)\,dx∫30​(3x​−(x​+2))dx

Explanation: 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.

Question 15

For 1≤x≤21\le x\le 21≤x≤2, f(x)=x+1xf(x)=x+\frac{1}{x}f(x)=x+x1​ lies above g(x)=2g(x)=2g(x)=2; which integral gives the area between them?

  1. ∫12(2−(x+1x)) dx\displaystyle \int_{1}^{2}\big(2-(x+\tfrac{1}{x})\big)\,dx∫12​(2−(x+x1​))dx
  2. ∫21((x+1x)−2) dx\displaystyle \int_{2}^{1}\big((x+\tfrac{1}{x})-2\big)\,dx∫21​((x+x1​)−2)dx
  3. ∫12((x+1x)+2) dx\displaystyle \int_{1}^{2}\big((x+\tfrac{1}{x})+2\big)\,dx∫12​((x+x1​)+2)dx
  4. ∫21(2−(x+1x)) dx\displaystyle \int_{2}^{1}\big(2-(x+\tfrac{1}{x})\big)\,dx∫21​(2−(x+x1​))dx
  5. ∫12((x+1x)−2) dx\displaystyle \int_{1}^{2}\big((x+\tfrac{1}{x})-2\big)\,dx∫12​((x+x1​)−2)dx (correct answer)

Explanation: 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 + 1/x is the upper curve and g(x) = 2 is the lower curve on [1, 2], so the integrand is (x + 1/x) - 2. Integrating from 1 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.

Question 16

Between x=−2x=-2x=−2 and x=2x=2x=2, f(x)=x2+2f(x)=x^2+2f(x)=x2+2 lies above g(x)=∣x∣g(x)=|x|g(x)=∣x∣; which integral gives the area setup?

  1. ∫−22(∣x∣−(x2+2)) dx\displaystyle \int_{-2}^{2}\big(|x|-(x^2+2)\big)\,dx∫−22​(∣x∣−(x2+2))dx
  2. ∫−22((x2+2)−∣x∣) dx\displaystyle \int_{-2}^{2}\big((x^2+2)-|x|\big)\,dx∫−22​((x2+2)−∣x∣)dx (correct answer)
  3. ∫2−2((x2+2)−∣x∣) dx\displaystyle \int_{2}^{-2}\big((x^2+2)-|x|\big)\,dx∫2−2​((x2+2)−∣x∣)dx
  4. ∫−22((x2+2)+∣x∣) dx\displaystyle \int_{-2}^{2}\big((x^2+2)+|x|\big)\,dx∫−22​((x2+2)+∣x∣)dx
  5. ∫2−2(∣x∣−(x2+2)) dx\displaystyle \int_{2}^{-2}\big(|x|-(x^2+2)\big)\,dx∫2−2​(∣x∣−(x2+2))dx

Explanation: 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.

Question 17

On 1≤x≤e1\le x\le e1≤x≤e, f(x)=4xf(x)=\tfrac{4}{x}f(x)=x4​ and g(x)=1g(x)=1g(x)=1. Which integral represents the area between the curves?

  1. ∫1e[1−4x]dx\displaystyle \int_{1}^{e}\big[1-\tfrac{4}{x}\big]dx∫1e​[1−x4​]dx
  2. ∫e1[4x−1]dx\displaystyle \int_{e}^{1}\big[\tfrac{4}{x}-1\big]dx∫e1​[x4​−1]dx
  3. ∫1e[4x−1]dx\displaystyle \int_{1}^{e}\big[\tfrac{4}{x}-1\big]dx∫1e​[x4​−1]dx (correct answer)
  4. ∫0e[4x−1]dx\displaystyle \int_{0}^{e}\big[\tfrac{4}{x}-1\big]dx∫0e​[x4​−1]dx
  5. ∫1e[4x+1]dx\displaystyle \int_{1}^{e}\big[\tfrac{4}{x}+1\big]dx∫1e​[x4​+1]dx

Explanation: To find the area between curves, we integrate the difference between the upper and lower functions over the given interval. At x=1: f(1)=4/1=4 and g(1)=1, so f(x)=4/x is above g(x)=1. Since 4/x decreases as x increases, we need to check if they intersect: 4/x=1 gives x=4, which is outside [1,e]. Therefore f(x)>g(x) throughout [1,e]. The area formula is ∫[upper-lower]dx, yielding ∫₁ᵉ[4/x-1]dx. Choice A reverses the subtraction to 1-4/x, which would give a negative area since 4/x>1 on [1,e]. Choice D starts at x=0, but 4/x is undefined at x=0, making this integral improper. When setting up area integrals, verify which function is larger and use proper domain restrictions.

Question 18

For −1≤x≤1-1\le x\le 1−1≤x≤1, f(x)=2−x2f(x)=2-x^2f(x)=2−x2 and g(x)=xg(x)=xg(x)=x. Which integral gives the area between the curves?

  1. ∫−11[x−(2−x2)] dx\displaystyle \int_{-1}^{1}\big[x-(2-x^2)\big]\,dx∫−11​[x−(2−x2)]dx
  2. ∫−11[(2−x2)−x] dx\displaystyle \int_{-1}^{1}\big[(2-x^2)-x\big]\,dx∫−11​[(2−x2)−x]dx (correct answer)
  3. ∫1−1[(2−x2)−x] dx\displaystyle \int_{1}^{-1}\big[(2-x^2)-x\big]\,dx∫1−1​[(2−x2)−x]dx
  4. ∫−11[(2−x2)+x] dx\displaystyle \int_{-1}^{1}\big[(2-x^2)+x\big]\,dx∫−11​[(2−x2)+x]dx
  5. ∫01[(2−x2)−x] dx\displaystyle \int_{0}^{1}\big[(2-x^2)-x\big]\,dx∫01​[(2−x2)−x]dx

Explanation: This problem requires finding the area between two curves by integrating the difference of the upper and lower functions. To identify which is on top, test a point like x=0: f(0)=2-0²=2 and g(0)=0, so f(x)=2-x² is above g(x)=x throughout [-1,1]. The area formula is ∫[upper-lower]dx, giving us ∫₋₁¹[(2-x²)-x]dx. Choice A reverses the subtraction, which would produce a negative area since x < 2-x² on this interval. Choice D adds the functions instead of subtracting, calculating total area under both curves rather than the area between them. When setting up area integrals, always check which function has larger values and subtract lower from upper.

Question 19

For −π2≤x≤π2-\tfrac{\pi}{2}\le x\le \tfrac{\pi}{2}−2π​≤x≤2π​, f(x)=cos⁡xf(x)=\cos xf(x)=cosx and g(x)=0g(x)=0g(x)=0. Which integral gives the area between the curves?

  1. ∫−π/2π/2[0−cos⁡x]dx\displaystyle \int_{-\pi/2}^{\pi/2}\big[0-\cos x\big]dx∫−π/2π/2​[0−cosx]dx
  2. ∫−π/2π/2[cos⁡x−0]dx\displaystyle \int_{-\pi/2}^{\pi/2}\big[\cos x-0\big]dx∫−π/2π/2​[cosx−0]dx (correct answer)
  3. ∫π/2−π/2[cos⁡x−0]dx\displaystyle \int_{\pi/2}^{-\pi/2}\big[\cos x-0\big]dx∫π/2−π/2​[cosx−0]dx
  4. ∫−π/2π/2[cos⁡x+0]dx\displaystyle \int_{-\pi/2}^{\pi/2}\big[\cos x+0\big]dx∫−π/2π/2​[cosx+0]dx
  5. ∫0π/2[cos⁡x−0]dx\displaystyle \int_{0}^{\pi/2}\big[\cos x-0\big]dx∫0π/2​[cosx−0]dx

Explanation: This problem asks for the area between the cosine curve and the x-axis (where g(x)=0). On the interval [-π/2,π/2], cos(x) ranges from 0 at the endpoints to 1 at x=0, so cos(x)≥0 throughout, meaning f(x) is above g(x). The area formula requires ∫[upper-lower]dx, giving us ∫₋π/₂^π/₂[cos x-0]dx. Choice A incorrectly computes 0-cos x, which would yield a negative area since cos x>0 on this interval. Choice E only integrates from 0 to π/2, missing half the region since the problem specifies the full interval [-π/2,π/2]. For area between curves, always use the complete interval and subtract lower from upper function.

Question 20

On 0≤x≤π0\le x\le \pi0≤x≤π, f(x)=sin⁡x+2f(x)=\sin x+2f(x)=sinx+2 and g(x)=1g(x)=1g(x)=1. Which integral represents the area between the curves?

  1. ∫0π[1−(sin⁡x+2)]dx\displaystyle \int_{0}^{\pi}\big[1-(\sin x+2)\big]dx∫0π​[1−(sinx+2)]dx
  2. ∫π0[(sin⁡x+2)−1]dx\displaystyle \int_{\pi}^{0}\big[(\sin x+2)-1\big]dx∫π0​[(sinx+2)−1]dx
  3. ∫0π[(sin⁡x+2)−1]dx\displaystyle \int_{0}^{\pi}\big[(\sin x+2)-1\big]dx∫0π​[(sinx+2)−1]dx (correct answer)
  4. ∫0π[(sin⁡x+2)+1]dx\displaystyle \int_{0}^{\pi}\big[(\sin x+2)+1\big]dx∫0π​[(sinx+2)+1]dx
  5. ∫−ππ[(sin⁡x+2)−1]dx\displaystyle \int_{-\pi}^{\pi}\big[(\sin x+2)-1\big]dx∫−ππ​[(sinx+2)−1]dx

Explanation: To find the area between curves, we integrate the difference between the upper and lower functions over the specified interval. Since sin(x) ranges from 0 to 1 on [0,π], we have f(x)=sin(x)+2 ranging from 2 to 3, while g(x)=1 is constant, so f(x) is always above g(x). The area formula requires ∫[upper-lower]dx, yielding ∫₀^π[(sin x+2)-1]dx. Choice A incorrectly subtracts in the wrong order, giving 1-(sin x+2) which would be negative since sin x+2 > 1. Choice E uses interval [-π,π] instead of [0,π], which changes the problem entirely. For area between curves, verify which function is larger throughout the interval, then integrate (upper minus lower).