Area Between Curves with Multiple Intersections - AP Calculus BC

Let $f(x)=x^3$ and $g(x)=x^2$ . Which setup gives total area between the curves on $[0,1]$ ?
Answer: $\displaystyle \int_{0}^{1}(x^2-x^3)\,dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curves f(x) = x³ and g(x) = x² intersect at x = 0 and 1, with g(x) above f(x) throughout (0, 1), so no internal crossing requires splitting beyond the single interval [0, 1]. Since they touch at endpoints and g > f inside, we integrate g - f = x² - x³ over [0, 1]. No further splitting is needed as the top function remains consistent. A tempting distractor like choice A fails because it integrates f - g, which is negative throughout, giving a negative value instead of positive total area. To find areas between intersecting curves generally, always locate intersection points, test which function is greater in each subinterval, and integrate the positive difference in each.
For $y=x^3$ and $y=x$ , which integral setup gives the total area between them on $[-1,1]$ ?
Answer: $\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}(x-x^3)\,dx$ This problem requires finding the area between curves that intersect within the given interval. The curves y = x³ and y = x intersect when x³ = x, giving x³ - x = 0, so x(x² - 1) = 0, yielding x = -1, 0, 1. Since we're integrating on [-1, 1], we need to split at x = 0. Testing x = -0.5: y₁ = (-0.5)³ = -0.125 and y₂ = -0.5, so x³ > x (less negative) on [-1, 0]; testing x = 0.5: y₁ = (0.5)³ = 0.125 and y₂ = 0.5, so x > x³ on [0, 1]. The correct setup uses (x³ - x) on [-1, 0] and (x - x³) on [0, 1]. Choice B incorrectly uses x - x³ throughout, which would give a negative result on [-1, 0]. The strategy is to identify all intersection points, then determine which function is greater on each resulting subinterval.
For $y=x^2$ and $y=|x|$ on $[-1,1]$ , which setup gives the total area between the curves?
Answer: $\displaystyle \int_{-1}^{0}(|x|-x^2)\,dx+\int_{0}^{1}(|x|-x^2)\,dx$ This problem requires finding the area between a parabola and the absolute value function. The curves y = x² and y = |x| intersect when x² = |x|. For x ≥ 0: x² = x gives x = 0, 1; for x < 0: x² = -x gives x = 0, -1. So intersections occur at x = -1, 0, 1, which are exactly our integration bounds. Testing x = -0.5: y₁ = 0.25 and y₂ = 0.5, so |x| > x² on [-1, 0]; testing x = 0.5: y₁ = 0.25 and y₂ = 0.5, so |x| > x² on [0, 1]. Since |x| is above x² on both intervals, the correct setup uses (|x| - x²) throughout. Choice E incorrectly reverses the order on one interval, suggesting a sign change that doesn't occur. When dealing with absolute value functions, carefully analyze the piecewise nature and test points in each region.
For $y=\sin x$ and $y=0$ on $[0,2\pi]$ , which setup gives the total area between the curves?
Answer: $\displaystyle \int_{0}^{\pi}(\sin x-0)\,dx+\int_{\pi}^{2\pi}(0-\sin x)\,dx$ This problem requires finding the area between the sine function and the x-axis over one complete period. The function y = sin x crosses y = 0 at x = 0, π, 2π on the interval [0, 2π]. Testing the sign: sin(π/2) = 1 > 0 on (0, π) and sin(3π/2) = -1 < 0 on (π, 2π). Therefore, sin x is above the x-axis on [0, π] and below on [π, 2π]. The correct area setup uses (sin x - 0) on [0, π] where sine is positive, and (0 - sin x) on [π, 2π] where sine is negative. Choice D incorrectly reverses these, using (0 - sin x) on [0, π], which would give negative area where we need positive. For periodic functions, the key is identifying where the function changes sign and using the appropriate order of subtraction on each interval.
Let $f(x)=x^3-4x$ and $g(x)=0$ . What is the correct setup for total area between $f$ and $g$ on $[-2,2]$ ?
Answer: $\displaystyle \int_{-2}^{0}(x^3-4x)\,dx+\int_{0}^{2}(4x-x^3)\,dx$ This problem tests finding area between a curve and the x-axis when the curve has multiple x-intercepts. The function f(x) = x³ - 4x = x(x² - 4) has zeros at x = -2, 0, 2, which are the endpoints and one interior point of our interval. For x ∈ [-2, 0], testing x = -1 gives f(-1) = -1 + 4 = 3 > 0, so f(x) > 0; for x ∈ [0, 2], testing x = 1 gives f(1) = 1 - 4 = -3 < 0, so f(x) < 0. The area calculation requires f(x) - 0 = f(x) on [-2, 0] and 0 - f(x) = -f(x) on [0, 2]. Choice B fails because it doesn't account for the sign change of f(x) at x = 0. When a function changes sign, split the integral at zeros and ensure each piece contributes positive area.
For $f(x)=x^3-x$ and $g(x)=0$ , which integrals give the total area between the curve and $x$ -axis on $[-1,1]$ ?
Answer: $\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}(x-x^3)\,dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve f(x) = x³ - x intersects the x-axis (g(x) = 0) at x = -1, 0, and 1, dividing the interval [-1, 1] into subintervals [-1, 0] and [0, 1]. In [-1, 0], f(x) > 0, so we integrate f(x) = x³ - x. In [0, 1], f(x) < 0, so we integrate -f(x) = x - x³ to get positive area. A tempting distractor like choice A fails because it integrates f(x) over the entire interval without adjusting for sign changes, resulting in negative areas in [0, 1] that reduce the total to net area. To find areas between intersecting curves generally, always locate intersection points, test which function is greater in each subinterval, and integrate the positive difference in each.
For $y=\cos x$ and $y=0$ on $[0,2\pi]$ , which setup gives the total area between the curves?
Answer: $\displaystyle \int_{0}^{\pi/2}\cos x\,dx+\int_{\pi/2}^{3\pi/2}-\cos x\,dx+\int_{3\pi/2}^{2\pi}\cos x\,dx$ This problem requires finding the area between y = cos x and y = 0 on [0, 2π], accounting for multiple sign changes of the cosine function. The cosine function equals zero at x = π/2 and x = 3π/2 within [0, 2π]. On [0, π/2], cos x > 0; on [π/2, 3π/2], cos x < 0; and on [3π/2, 2π], cos x > 0 again. To find total area, we need positive contributions from all three regions: integrate cos x where it's positive and -cos x where it's negative. Choice A incorrectly integrates cos x over the entire interval, which would give a net signed area rather than total area. The correct approach splits at both zeros and adjusts signs appropriately: ∫[0 to π/2]cos x dx + ∫[π/2 to 3π/2](-cos x)dx + ∫[3π/2 to 2π]cos x dx.
For $y=x^3$ and $y=x$ , what is the correct setup for the total area between the curves on $[-1,1]$ ?
Answer: $\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}(x-x^3)\,dx$ This problem tests the skill of finding area between curves that intersect multiple times within the given interval. The curves y = x³ and y = x intersect when x³ = x, which gives x(x² - 1) = 0, so x = -1, 0, 1. Since we're integrating on [-1, 1], we need to split at the interior intersection x = 0. For x ∈ [-1, 0], testing x = -0.5 shows x³ = -0.125 > -0.5 = x, so x³ > x; for x ∈ [0, 1], testing x = 0.5 shows x³ = 0.125 < 0.5 = x, so x < x³. Choice A incorrectly uses a single integral that would yield zero due to symmetry, missing the actual enclosed area. To find total area between curves, identify all intersection points, split the interval at these points, and ensure each integral uses (upper curve - lower curve).
Let $f(x)=x^3$ and $g(x)=x^2$ . What is the correct setup for the total area between $f$ and $g$ on $[0,1]$ ?
Answer: $\displaystyle \int_{0}^{1}(x^2-x^3)\,dx$ This problem requires finding area between two curves on an interval where one is consistently above the other. The curves f(x) = x³ and g(x) = x² intersect when x³ = x², which gives x²(x - 1) = 0, so x = 0, 1. These are exactly our interval endpoints, with no interior intersections. For any x ∈ (0, 1), we have x² > x³ since multiplying the inequality x < 1 by the positive number x² gives x³ < x². Therefore, g(x) > f(x) throughout (0, 1), and the area is ∫₀¹ (x² - x³) dx. Choice A reverses the order, which would give a negative result since x³ < x² on this interval. When curves intersect only at endpoints, check which is on top using a test point in the interior.
Let $y=x^3$ and $y=x$ . Which integral setup gives the total area between the curves on $[-1,1]$ ?
Answer: $\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}(x-x^3)\,dx$ Finding the area between curves with multiple intersections requires identifying where the curves cross and splitting the integral accordingly. The curves y = x³ and y = x intersect when x³ = x, giving x³ - x = 0, so x(x² - 1) = 0, yielding x = -1, 0, 1. Since we're integrating over [-1, 1], we must split at x = 0 (the interior intersection point). On [-1, 0], we have x³ ≥ x (both are negative, and |x³| < |x|), so we integrate x³ - x; on [0, 1], we have x ≥ x³, so we integrate x - x³. Option A incorrectly assumes x ≥ x³ on the entire interval, which fails for x ∈ [-1, 0] where both functions are negative. The key strategy is to test a point in each subinterval to determine which function is larger, ensuring you always subtract the lower function from the upper function.
Between $y=\sin x$ and $y=0$ on $[0,2\pi]$ , which integral setup gives the total enclosed area?
Answer: $\displaystyle \int_{0}^{\pi}\sin x\,dx+\int_{\pi}^{2\pi}-\sin x\,dx$ Finding the area between y = sin x and y = 0 on [0, 2π] requires recognizing that sin x crosses the x-axis, creating regions where the function switches sign. The sine function equals zero at x = 0, π, and 2π within this interval. On [0, π], sin x > 0 (the function is above the x-axis), while on [π, 2π], sin x < 0 (the function is below the x-axis). To find total area, we need positive contributions from both regions, so we integrate sin x on [0, π] where it's positive, and -sin x on [π, 2π] where sin x is negative. Choice A incorrectly integrates sin x over the entire interval, which would give ∫[0 to 2π]sin x dx = 0 due to cancellation of positive and negative areas. The correct setup splits at x = π where sin x changes sign: ∫[0 to π]sin x dx + ∫[π to 2π](-sin x)dx.
For $y=x^3-x$ and $y=x$ , what integral setup gives the total area between the curves on $[-1,1]$ ?
Answer: $\displaystyle \int_{-1}^{0}\big[x-(x^3-x)\big]dx+\int_{0}^{1}\big[(x^3-x)-x\big]dx$ This problem requires finding the area between curves with multiple intersections, which demands splitting the integral at intersection points. The curves y = x³ - x and y = x intersect when x³ - x = x, giving x³ - 2x = 0, so x(x² - 2) = 0, yielding x = -√2, 0, √2. On [-1, 1], only x = 0 is an intersection point. Testing x = -0.5: y₁ = (-0.5)³ - (-0.5) = 0.375 and y₂ = -0.5, so x > x³ - x on [-1, 0]; testing x = 0.5: y₁ = (0.5)³ - 0.5 = -0.375 and y₂ = 0.5, so x³ - x > x on [0, 1]. Choice A incorrectly uses (x³ - x) - x throughout, ignoring the sign change at x = 0. The key strategy is to find all intersection points within the interval, then test which function is greater on each subinterval.
For $y=x^4-1$ and $y=0$ , which integral setup gives the total area between curves on $[-2,2]$ ?
Answer: $\displaystyle \int_{-2}^{-1}(x^4-1)dx+\int_{-1}^{1}(1-x^4)dx+\int_{1}^{2}(x^4-1)dx$ Finding the area between y = x⁴ - 1 and y = 0 requires identifying where this quartic function crosses the x-axis on [-2, 2]. Setting x⁴ - 1 = 0 gives x⁴ = 1, so x = ±1, creating three distinct regions. On [-2, -1], testing x = -1.5 gives y = (-1.5)⁴ - 1 = 5.0625 - 1 = 4.0625 > 0, so the curve is above the x-axis. On [-1, 1], testing x = 0 gives y = 0⁴ - 1 = -1 < 0, so the curve is below the x-axis. On [1, 2], testing x = 1.5 gives y = (1.5)⁴ - 1 = 5.0625 - 1 = 4.0625 > 0, so the curve is above the x-axis again. Choice A incorrectly integrates (x⁴ - 1) throughout, counting the middle region as negative area. The correct setup ensures all area contributions are positive: ∫[-2 to -1](x⁴ - 1)dx + ∫[-1 to 1](1 - x⁴)dx + ∫[1 to 2](x⁴ - 1)dx.
On $[-1,2]$ , curves $y=x^2$ and $y=x$ intersect twice; which setup gives the total area between them?
Answer: $\displaystyle \int_{-1}^{0}(x^2-x)\,dx+\int_{0}^{1}(x-x^2)\,dx+\int_{1}^{2}(x^2-x)\,dx$ The skill here is multi-interval area reasoning for computing areas between curves that intersect multiple times. The interval [-1,2] must be split because y = $x^2$ and y = x intersect at x = 0 and 1, switching dominance in [-1,0], [0,1], and [1,2]. Without splitting at both points, a single integrand would mix signs and give incorrect net area. Splitting allows identification of the upper function in each segment for positive differences. A tempting distractor like choice A uses ( $x^2$ - x) throughout, resulting in net area with potential negative parts. To handle such problems generally, locate all intersection points, divide the domain into subintervals accordingly, determine which curve is above in each, and sum the integrals of (upper - lower) over each subinterval.
For $y=\sin x$ and $y=0$ on $[0,2\pi]$ , what is the correct setup for the total area between the curve and the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi}\sin x\,dx+\int_{\pi}^{2\pi}(-\sin x)\,dx$ This problem involves finding the total area between a periodic function and the x-axis over multiple periods. The sine function equals zero at x = 0, π, 2π on the given interval, and we know sin x > 0 on (0, π) and sin x < 0 on (π, 2π). To find total area, we need ∫₀^π sin x dx for the region above the x-axis and ∫_π^{2π} |sin x| dx = ∫_π^{2π} (-sin x) dx for the region below. Choice A would give ∫₀^{2π} sin x dx = 0, completely missing the geometric area because the positive and negative regions cancel. For periodic functions, identify where the function changes sign and split the integral to ensure all contributions are positive.
Let $y=x^3$ and $y=x^2$ . Which setup gives the total area between the curves on $[0,1]$ ?
Answer: $\displaystyle \int_{0}^{1}(x^2-x^3)\,dx$ This problem involves finding the area between two curves on a single interval where no interior intersections occur, making it a straightforward application of the upper-minus-lower principle. The curves y = x³ and y = x² intersect when x³ = x², giving x³ - x² = 0, so x²(x - 1) = 0, yielding x = 0, 1. These are exactly our integration bounds, so we only need to determine which function is on top within (0, 1). Testing x = 0.5: y₁ = (0.5)³ = 0.125 and y₂ = (0.5)² = 0.25, so x² > x³ on (0, 1), requiring the integral of x² - x³. Option A incorrectly uses x³ - x², which would give a negative result since x³ < x² on this interval. When curves intersect only at the endpoints of your integration interval, you need just one integral with the correct upper-minus-lower ordering throughout.
Let $f(x)=x^3-x$ and $g(x)=x$ . What is the correct setup for the total area between $f$ and $g$ on $[-1,1]$ ?
Answer: $\displaystyle \int_{-1}^{0}\big[(x^3-x)-x\big]dx+\int_{0}^{1}\big[x-(x^3-x)\big]dx$ This problem requires finding the area between curves with multiple intersections, which demands splitting the integral at intersection points. The curves f(x) = x³ - x and g(x) = x intersect when x³ - x = x, which simplifies to x³ - 2x = 0, giving x(x² - 2) = 0, so x = -√2, 0, √2. On the interval [-1, 1], only x = 0 is an intersection point. Testing x = -0.5: f(-0.5) = -0.375 and g(-0.5) = -0.5, so f > g on [-1, 0]; testing x = 0.5: f(0.5) = -0.375 and g(0.5) = 0.5, so g > f on [0, 1]. The tempting choice A would give a negative result since it doesn't account for the changing relationship between the curves. When curves intersect within your interval, always split the integral at intersection points and ensure the integrand is (upper - lower) on each subinterval.
For $y=x^3-4x$ and $y=0$ , what is the correct total area setup on $[-2,2]$ ?
Answer: $\displaystyle \int_{-2}^{0}\big(0-(x^3-4x)\big)\,dx+\int_{0}^{2}\big((x^3-4x)-0\big)\,dx$ This problem requires finding the area between a curve and the x-axis when the curve crosses the axis multiple times, necessitating a split integral. The curve y = x³ - 4x intersects y = 0 when x³ - 4x = 0, giving x(x² - 4) = 0, so x = -2, 0, 2. Since these are exactly our integration bounds, we split at x = 0. On [-2, 0], testing x = -1 gives y = (-1)³ - 4(-1) = -1 + 4 = 3 > 0, so the curve is above the x-axis and we integrate (x³ - 4x) - 0. On [0, 2], testing x = 1 gives y = 1³ - 4(1) = -3 < 0, so the curve is below the x-axis and we integrate 0 - (x³ - 4x). Option A incorrectly uses x³ - 4x throughout, which would give a net signed area of zero due to symmetry, not the total area. When finding area between a curve and the x-axis, always split at x-intercepts and use |f(x)| or upper-minus-lower in each region.
For $y=\sin x$ and $y=\cos x$ on $[0,2\pi]$ , what is the correct setup for the total enclosed area?
Answer: $\displaystyle \int_{0}^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx+\int_{5\pi/4}^{2\pi}(\cos x-\sin x)\,dx$ This problem requires finding the area between trigonometric curves that intersect multiple times over the given interval. The curves y = sin x and y = cos x intersect when sin x = cos x, which occurs when tan x = 1, giving x = π/4, 5π/4 on [0, 2π]. We must determine which function is on top in each subinterval: for x ∈ [0, π/4], cos x > sin x; for x ∈ [π/4, 5π/4], sin x > cos x; for x ∈ [5π/4, 2π], cos x > sin x. Choice A incorrectly assumes sin x is always above cos x, which would give a negative area on some subintervals. For periodic functions with multiple intersections, carefully track which function is larger on each subinterval to ensure all integrands are positive.
Let $y=x^4-x^2$ and $y=0$ on $[-1,1]$ . Which setup gives the total area between the curves?
Answer: $\displaystyle \int_{-1}^{0}\big[0-(x^4-x^2)\big]dx+\int_{0}^{1}\big[0-(x^4-x^2)\big]dx$ This problem involves finding the area between a quartic function and the x-axis. The curve y = x⁴ - x² crosses y = 0 when x⁴ - x² = 0, giving x²(x² - 1) = 0, so x = -1, 0, 1. Since we're integrating on [-1, 1], we need to check the sign of the function. Testing x = 0.5: y = (0.5)⁴ - (0.5)² = 0.0625 - 0.25 = -0.1875 < 0. By symmetry and continuity, x⁴ - x² < 0 on (-1, 0) and (0, 1), meaning the curve is entirely below the x-axis on [-1, 1]. The correct setup uses 0 - (x⁴ - x²) throughout to get positive area. Choice A would give negative area since it doesn't account for the curve being below the x-axis. When a function doesn't change sign at x = 0, don't assume you need to split there—test the function's sign first.

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