This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = sin(2x) intersects y = 0 at x = 0, π/2, π, dividing [0, π] into subintervals. In [0, π/2], sin(2x) > 0; in [π/2, π], < 0, requiring a split. The negative in the second ensures positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-1 and x=1 because those are the intersection points of $y=1-x^2$ and $y=x^2$-1 within [-2,2]. From -2 to -1, $y=x^2$-1 is above $y=1-x^2$. From -1 to 1, $y=1-x^2$ is above $y=x^2$-1. From 1 to 2, $y=x^2$-1 is above $y=1-x^2$ again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curves y = (1/2)x³ - (3/2)x and y = x intersect at x = 0 within [-2, 2], as solving (1/2)x³ - (5/2)x = 0 gives x = 0 (other roots outside). In [-2, 0], the cubic is above the linear, as at x = -1, (1/2)(-1)³ - (3/2)(-1) = 1 > -1. In [0, 2], the linear is above, as at x = 1, (1/2)(1)³ - (3/2)(1) = -1 < 1, requiring a split to capture positive areas separately. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=π because that's where y=sin x crosses y=1/2 sin x within [0,2π], in addition to endpoints. From 0 to π, y=sin x is above y=1/2 sin x since sin x >0. From π to 2π, y=1/2 sin x is above y=sin x since sin x <0. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve $y = \cos x$ intersects $y = 0$ at $x = \pi/2$, $3\pi/2$ within $[0, 2\pi]$, dividing into three subintervals. In $[0, \pi/2]$ and $[3\pi/2, 2\pi]$, $\cos > 0$; in $[\pi/2, 3\pi/2]$, $\cos < 0$, requiring splits. The negative in the middle ensures positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = ln x intersects y = 0 at x = 1 within [1/e, e], dividing into subintervals. In [1/e, 1], ln x < 0; in [1, e], > 0, requiring a split. The negative in the first ensures positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = $e^x$ intersects y = 2 at x = ln 2 within [0, ln 4], dividing into subintervals. In [0, ln 2], $e^x$ < 2; in [ln 2, ln 4], > 2, requiring a split. This captures positive areas. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curves y = x³ - 4x and y = 2x intersect at x = -√6, 0, √6, dividing [-√6, √6] into subintervals. In [-√6, 0], the cubic is above, as verified by test points. In [0, √6], 2x is above, necessitating a split. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=0 because that's where $y=x^3$ crosses y=x within [-1,1], in addition to the endpoints. From -1 to 0, $y=x^3$ is above y=x since $x^3$ > x for x in (-1,0). From 0 to 1, y=x is above $y=x^3$ since x > $x^3$ for x in (0,1). One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x⁴ - x² intersects y = 0 at x = -1, 0, 1, but is below throughout [-1, 1]. In [-1, 1], y < 0, as x⁴ - x² = x²(x² - 1) < 0 for |x| < 1. The split at 0 is harmless since integrand same, giving positive area via x² - x⁴ > 0. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel (though none here). To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.