This problem requires finding the area between curves with multiple intersections, which demands splitting the integral at intersection points. The curves y = x³ - 3x and y = x intersect when x³ - 3x = x, giving x³ - 4x = 0, so x(x² - 4) = 0, yielding intersections at x = -2, 0, and 2. On [-2, 0], we need to determine which function is on top: testing x = -1 gives y = (-1)³ - 3(-1) = 2 for the cubic and y = -1 for the line, so x³ - 3x > x on this interval. On [0, 2], testing x = 1 gives y = 1³ - 3(1) = -2 for the cubic and y = 1 for the line, so x < x³ - 3x on this interval. Choice A incorrectly uses (x³ - 3x) - x throughout without checking which function is on top in each subinterval. The correct approach splits at x = 0 and uses (top - bottom) in each piece: ∫[-2 to 0][(x³ - 3x) - x]dx + ∫[0 to 2][x - (x³ - 3x)]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.

This problem involves finding the area between a curve and the x-axis with multiple intersections, requiring interval splitting. The function f(x) = x³ - 4x crosses g(x) = 0 when x³ - 4x = 0, giving x(x² - 4) = 0, so x = -2, 0, 2. Testing x = -1: f(-1) = -1 + 4 = 3 > 0, so f(x) > 0 on [-2, 0]; testing x = 1: f(1) = 1 - 4 = -3 < 0, so f(x) < 0 on [0, 2]. The area setup requires (x³ - 4x) - 0 on [-2, 0] where f is above g, and 0 - (x³ - 4x) on [0, 2] where g is above f. Choice A incorrectly integrates x³ - 4x without accounting for the sign change, which would give net signed area rather than total area. When finding area between curves, always split at intersection points and use (upper function) - (lower function) on each interval.

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.

This problem involves finding the area between a cosine curve and the x-axis over a full period. The function y = cos x equals 0 when x = π/2, 3π/2 on [0, 2π]. Testing the sign: cos(0) = 1 > 0, cos(π) = -1 < 0, and cos(2π) = 1 > 0. Therefore, cos x is above the x-axis on [0, π/2] and [3π/2, 2π], and below on [π/2, 3π/2]. The correct setup uses (cos x - 0) where cosine is positive and (0 - cos x) where cosine is negative. Choice C incorrectly splits at π instead of π/2 and 3π/2, which would mix positive and negative regions in each integral. For periodic functions, identify all zeros in the given interval and determine the sign of the function on each resulting subinterval.

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.

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.

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.

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.

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).