To find the area between curves, we must integrate the difference (upper - lower) over the interval. Comparing f(x)=eˣ and g(x)=1+x on [0,1], at x=0: f(0)=1 and g(0)=1 (they start together), but at x=0.5: f(0.5)≈1.649 and g(0.5)=1.5, so f(x) is above g(x). Since eˣ grows faster than the linear function 1+x, f remains above g throughout [0,1]. The area is ∫[f(x)-g(x)]dx = ∫[eˣ-(1+x)]dx from 0 to 1. Choice B would give the negative of the area since it reverses the subtraction. Always verify which function is on top before setting up your integral.

This problem involves finding the area between two curves, f(x) = 3x and g(x) = x³, over the interval [0, √3] using integration, a key skill in AP Calculus BC. To compute the area, we integrate the difference between the upper curve and the lower curve over the given interval. Here, f(x) lies above g(x) on (0, √3), since 3x > x³ when x² < 3. Therefore, the integral setup is ∫ from 0 to √3 of (f(x) - g(x)) dx, ensuring a positive area. A tempting distractor is integrating g(x) - f(x), but this gives a negative result instead of the area. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.

This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. On [0,1], with f(x) = ln(x+1) ≥ g(x) = x/2, the area integral uses upper minus lower as f(x) - g(x). This structure holds as the logarithm starts equal but stays above due to its concavity. Integrating from 0 to 1 sums the positive differences effectively. A tempting distractor is choice A, using g(x) - f(x), but it results in negative area, not the desired positive value. In general, to set up the area between two curves y = upper(x) and y = lower(x) from a to b, use ∫_$a^b$ (upper - lower) dx, first confirming which is upper via graphing or comparison.

To find the area between curves, we must integrate the difference between the upper and lower functions. The problem states g(x) ≥ f(x) on [-1,2], making g(x) = x + 5 the upper curve and f(x) = x³ + 5 the lower curve. The area is ∫[g(x) - f(x)]dx = ∫[(x + 5) - (x³ + 5)]dx = ∫[x - x³]dx from -1 to 2, which corresponds to choice C. Choice A incorrectly uses f(x) - g(x), reversing the subtraction order and yielding a negative result. Always integrate (upper function - lower function) to ensure positive area.

This problem asks for the area between two curves, requiring integration of the difference between upper and lower functions. Given f(x) = 4/x and g(x) = 1 with f(x) ≥ g(x) on [1,4], f(x) is the upper curve. The area is ∫[f(x) - g(x)]dx = ∫[4/x - 1]dx from 1 to 4, matching choice C. Choice A reverses the order to g(x) - f(x) = 1 - 4/x, which is negative on [1,4] and would give negative area. Remember: area between curves equals ∫[upper - lower]dx over the interval.

This is an area between curves problem where we need to integrate the difference between the upper and lower functions. Since f(x) = 4 - x² and g(x) = 0 with f(x) ≥ g(x) on [-2,2], f(x) is the upper curve. The area is ∫[f(x) - g(x)]dx = ∫[4 - x²]dx from -2 to 2, corresponding to choice D. Choice A reverses the subtraction to g(x) - f(x) = -(4 - x²), which would yield a negative value. When finding area between curves, always integrate (upper - lower) to ensure a positive result.

This area between curves problem requires careful identification of which function is on top. Given that y = eˣ is the upper curve and y = 1 + x is the lower curve on [0, 1], we compute the area as ∫₀¹[eˣ - (1 + x)]dx. This follows the standard formula for area between curves: integrate the difference (upper - lower) over the given interval. Choice A reverses the subtraction, computing (1 + x) - eˣ, which would give a negative result. To avoid this error, always verify the relative positions of the curves before setting up the integral.

To find the area between trigonometric curves, we apply the same principle: integrate (upper - lower). Since y = 3cos x is above y = sin x on [-π/2, π/2], the area integral is ∫₋π/₂^π/₂[3cos x - sin x]dx. This matches choice D, which correctly orders the subtraction to ensure a positive result. Choice A incorrectly computes (sin x - 3cos x), reversing the subtraction and yielding a negative integrand. When working with trig functions, it's especially helpful to sketch or evaluate at key points to confirm which function is on top.

This problem involves finding the area between two curves, f(x) = 2 - x² and g(x) = x, over the interval [-1, 1] using integration, a key skill in AP Calculus BC. To compute the area, we integrate the difference between the upper curve and the lower curve over the given interval. Here, f(x) lies above g(x) throughout [-1, 1], as f - g = 2 - x² - x remains positive, verified by its minimum value. Therefore, the integral setup is ∫ from -1 to 1 of (f(x) - g(x)) dx, providing the enclosed area. A tempting distractor is integrating g(x) - f(x), but this would give a negative value instead of the positive area. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.

This problem involves finding the area between two curves, $f(x) = \sin x + 2$ and $g(x) = 2$, over the interval $[0, \pi]$ using integration, a key skill in AP Calculus BC. To compute the area, we integrate the difference between the upper curve and the lower curve over the given interval. Here, $f(x)$ lies above $g(x)$ on $(0, \pi)$, since $\sin x > 0$ in that interval, making $f(x) > 2$. Therefore, the integral setup is $\int_0^\pi(f(x) - g(x)) , dx$, which simplifies to $\int_0^\pi \sin x , dx$ and gives a positive area. A tempting distractor is integrating $g(x) - f(x)$, but this results in a negative value, not the area. Always identify the upper and lower functions over the interval and integrate upper minus lower for area calculations.