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.

The skill here is multi-interval area reasoning for computing areas between curves that intersect multiple times. The interval [-π,π] must be split because y = cos x intersects y = 0 at x = -π/2 and π/2, dividing regions of positive and negative values. Without splitting, net area would be computed, ignoring absolute contributions. Splitting at ±π/2 allows using cos x where positive and -cos x where negative. A tempting distractor like choice A integrates cos x directly, giving net zero over the symmetric interval. 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 requires multi-interval area reasoning to find the total area between a parabola and the x-axis where it crosses multiple times. The parabola f(x) = $x^2$ - 1 crosses g(x) = 0 at x = -1 and 1, dividing [-2, 2] into three subintervals with alternating signs. In [-2, -1] and [1, 2], f(x) is positive, but in [-1, 1], it is negative, requiring splits to compute absolute areas. Splitting at x = -1 and 1 allows adjusting the integrand to yield positive contributions in each part. A tempting distractor is the single integral of f(x) - g(x), which fails by giving net area with the negative middle region subtracting from the positives. Always find all intersection points, determine the upper function in each subinterval, and set up separate integrals of (upper - lower) for each to compute total area.

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 the interval [-1, 1], only x = 0 is relevant, creating two subintervals: [-1, 0] and [0, 1]. On [-1, 0], we need x - (x³ - x) = 2x - x³ since x ≥ x³ - x there; on [0, 1], we need (x³ - x) - x = x³ - 2x since x³ - x ≥ x there. Option A incorrectly uses (x³ - x) - x = x³ - 2x on the entire interval, missing that the curves switch positions at x = 0. When curves intersect within your interval, always split the integral at intersection points and determine which function is on top in each subinterval.

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.