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) = 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) = 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, $f(x) = \sin x + 2$ 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.

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) = $e^x$ above g(x) = 1 + x, the area uses upper minus lower as f(x) - g(x). This structure applies since $e^x$ grows faster than the linear function, keeping the difference positive. Integrating from 0 to 1 accumulates the area from their touch at x=0 onward. A tempting distractor is choice A, using g(x) - f(x), but this integrates a negative function, not yielding positive area. 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.

This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. For f(x) = x + 2 and g(x) = x² on [0,2], with f(x) ≥ g(x), the integral is upper minus lower as f(x) - g(x). This is suitable because the line stays above the parabola in the interval, with positive difference. The setup from 0 to 2 covers the region up to their intersection at x=2. A tempting distractor is choice A, integrating g(x) - f(x), but this gives negative area, missing the positive requirement. 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.

This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. For f(x) = √x and g(x) = x on [0,1], with f(x) ≥ g(x), the area is found by integrating upper minus lower, which is √x - x. This applies because √x lies above x in the interval, making the difference positive and representative of the height at each x. The integral from 0 to 1 sums these heights, giving the total area without intersections requiring splits. A tempting distractor is choice C, integrating x - √x, but this gives a negative result since x < √x, not the positive area. 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.

This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. For f(x) = sin x and g(x) = 0 on [0, π], where f(x) ≥ g(x), the integral subtracts lower from upper as f(x) - g(x). This works because sin x is nonnegative over the interval, ensuring the difference measures positive heights. The limits from 0 to π cover the full hump above the x-axis without crossings. A tempting distractor is choice A, integrating g(x) - f(x), but this produces a negative area, which doesn't make sense for geometric area. 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.

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.

This problem involves finding the area between two curves defined as functions of x, a key skill in AP Calculus BC. On [0, π/2], with f(x) = cos x ≥ g(x) = cos² x, the integral uses upper minus lower as f(x) - g(x). This applies because cos x exceeds its square in the interval, with positive difference. The limits from 0 to π/2 cover the region without internal crossings. A tempting distractor is choice A, integrating g(x) - f(x), but this negatives the area, failing geometrically. 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.