Area Between Curves with Multiple Intersections - AP Calculus AB

For $y=x^2$ and $y=2-x^2$ on $[-2,2]$ , which setup gives the total area between curves?
Answer: $\displaystyle \int_{-2}^{-1}\left[x^2-(2-x^2)\right]dx+\int_{-1}^{1}\left[(2-x^2)-x^2\right]dx+\int_{1}^{2}\left[x^2-(2-x^2)\right]dx$ 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= $x^2$ and y=2- $x^2$ within [-2,2]. From -2 to -1, y= $x^2$ is above y=2- $x^2$ . From -1 to 1, y=2- $x^2$ is above y= $x^2$ . From 1 to 2, y= $x^2$ is above y=2- $x^2$ 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.
For $y=x^4-x^2$ and $y=x^2$ on $[-1,1]$ , which setup gives the total area between curves?
Answer: $\displaystyle \int_{-1}^{0}\left[x^2-(x^4-x^2)\right]dx+\int_{0}^{1}\left[x^2-(x^4-x^2)\right]dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curves y = x⁴ - x² and y = x² intersect at x = 0 in [-1, 1], but y = x⁴ - x² < x² everywhere else. In [-1, 1], difference x⁴ - 2x² < 0, so lower < upper. The split at 0 is harmless with same positive integrand x² - (x⁴ - x²) = 2x² - x⁴. 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.
Let $y=1-x^2$ and $y=x^2-1$ on $[-2,2]$ . Which setup gives total area between curves?
Answer: $\displaystyle \int_{-2}^{-1}\left[(x^2-1)-(1-x^2)\right]dx+\int_{-1}^{1}\left[(1-x^2)-(x^2-1)\right]dx+\int_{1}^{2}\left[(x^2-1)-(1-x^2)\right]dx$ 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.
Let $y=\sin(2x)$ and $y=0$ on $[0,\pi]$ . Which setup gives the total area?
Answer: $\displaystyle \int_{0}^{\pi/2}\sin(2x)\,dx+\int_{\pi/2}^{\pi}-\sin(2x)\,dx$ 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.
For $y=\tan x$ and $y=0$ on $\left[-\tfrac\pi4,\tfrac\pi4\right]$ , which setup gives total area?
Answer: $\displaystyle \int_{-\pi/4}^{0}-\tan x\,dx+\int_{0}^{\pi/4}\tan x\,dx$ 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=tan x crosses y=0. From -π/4 to 0, y=0 is above y=tan x since tan x<0. From 0 to π/4, y=tan x is above y=0 since tan 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.
For $y=\sin x$ and $y=\cos x$ on $[0,2\pi]$ , which setup gives the total area between curves?
Answer: $\displaystyle \int_{0}^{\pi/4}(\cos x-\sin x)dx+\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)dx+\int_{5\pi/4}^{2\pi}(\cos x-\sin x)dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curves y = sin x and y = cos x intersect at x = π/4, 5π/4 within [0, 2π], dividing into three subintervals. In [0, π/4], cos is above sin; in [π/4, 5π/4], sin above cos; in [5π/4, 2π], cos above sin. This requires splits to ensure positive differences in each. 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.
Let $y=x^2-4$ and $y=0$ on $[-3,3]$ . Which setup gives total area between curves?
Answer: $\displaystyle \int_{-3}^{-2}(x^2-4)dx+\int_{-2}^{2}(4-x^2)dx+\int_{2}^{3}(x^2-4)dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-2 and x=2 because those are where y= $x^2$ -4 crosses y=0 within [-3,3]. From -3 to -2, y= $x^2$ -4 is above y=0 since $x^2$ -4>0 for |x|>2. From -2 to 2, y=0 is above y= $x^2$ -4 since $x^2$ -4<0. From 2 to 3, y= $x^2$ -4 is above y=0 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.
For $y=\cos x$ and $y=0$ on $[0,2\pi]$ , which setup gives the total area between curves?
Answer: $\displaystyle \int_{0}^{\pi/2}\cos x\,dx+\int_{\pi/2}^{3\pi/2}-\cos x\,dx+\int_{3\pi/2}^{2\pi}\cos x\,dx$ 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.
Let $y=\sin x$ and $y=0$ on $[0,2\pi]$ . Which setup gives the total area?
Answer: $\displaystyle \int_{0}^{\pi}\sin x\,dx+\int_{\pi}^{2\pi}-\sin x\,dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = sin x intersects y = 0 at x = 0, π, 2π, dividing [0, 2π] into subintervals. In [0, π], sin is above; in [π, 2π], below, requiring a split. The negative in the second integral 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.
Let $y=x^3-4x$ and $y=0$ on $[-2,2]$ . Which integral setup gives the total area?
Answer: $\displaystyle \int_{-2}^{0}(x^3-4x)\,dx+\int_{0}^{2}(4x-x^3)\,dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x³ - 4x intersects y = 0 at x = -2, 0, 2, dividing [-2, 2] into subintervals. In [-2, 0], the cubic is above, as at x = -1, (-1)³ - 4(-1) = 3 > 0. In [0, 2], it is below, as at x = 1, 1 - 4 = -3 < 0, requiring a split to ensure positive integrands. 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.
For $y=x^2-1$ and $y=0$ on $[-2,2]$ , which setup gives the total area between curves?
Answer: $\displaystyle \int_{-2}^{-1}(x^2-1)dx+\int_{-1}^{1}(1-x^2)dx+\int_{1}^{2}(x^2-1)dx$ 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 where y= $x^2$ -1 crosses y=0 within [-2,2]. From -2 to -1, y= $x^2$ -1 is above y=0 since $x^2$ -1>0 for |x|>1. From -1 to 1, y=0 is above y= $x^2$ -1 since $x^2$ -1<0. From 1 to 2, y= $x^2$ -1 is above y=0 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.
For $y=\sin x$ and $y=\tfrac12\sin x$ on $[0,2\pi]$ , which setup gives total area between curves?
Answer: $\displaystyle \int_{0}^{\pi}\left(\sin x-\tfrac12\sin x\right)dx+\int_{\pi}^{2\pi}\left(\tfrac12\sin x-\sin x\right)dx$ 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.
For $y=\ln x$ and $y=0$ on $\left[\tfrac1e,e\right]$ , which setup gives total area between curves?
Answer: $\displaystyle \int_{1/e}^{1}-\ln x\,dx+\int_{1}^{e}\ln x\,dx$ 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.
For $y=x^3$ and $y=x$ on $[-1,1]$ , which setup gives the total area between curves?
Answer: $\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}(x-x^3)\,dx$ 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.
For $y=\sin x$ and $y=-\sin x$ on $[0,2\pi]$ , which setup gives total area between curves?
Answer: $\displaystyle \int_{0}^{\pi}\big(\sin x-(-\sin x)\big)dx+\int_{\pi}^{2\pi}\big((-\sin x)-\sin x\big)dx$ 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=-sin x within [0,2π]. From 0 to π, y=sin x is above y=-sin x. From π to 2π, y=-sin x is above y=sin x. 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.
For $y=x(x-1)(x+1)$ and $y=0$ on $[-1,1]$ , which setup gives total area?
Answer: $\displaystyle \int_{-1}^{0}x(x-1)(x+1)\,dx+\int_{0}^{1}-x(x-1)(x+1)\,dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve $y = x(x-1)(x+1) = x^3 - x$ intersects $y = 0$ at $x = -1, 0, 1$ , dividing $[-1, 1]$ . In $[-1, 0]$ , $y > 0$ ; in $[0, 1]$ , $y < 0$ , 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.
Let $y=\sin x$ and $y=x-\pi$ on $[0,2\pi]$ . Which setup gives total area between curves?
Answer: $\displaystyle \int_{0}^{\pi}(\sin x-(x-\pi))dx+\int_{\pi}^{2\pi}((x-\pi)-\sin x)dx$ 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=x-π within [0,2π]. From 0 to π, y=sin x is above y=x-π since their difference is non-negative. From π to 2π, y=x-π is above y=sin x since their difference reverses. 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.
For $y=x^3-x$ and $y=0$ on $[-1,1]$ , which setup gives the total area between the curves?
Answer: $\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}-(x^3-x)\,dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x³ - x intersects y = 0 at x = -1, 0, 1, dividing [-1, 1] into subintervals. In [-1, 0], the cubic is above the x-axis, as at x = -0.5, (-0.5)³ - (-0.5) = 0.375 > 0. In [0, 1], it is below, as at x = 0.5, (0.5)³ - 0.5 = -0.375 < 0, necessitating a split for 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.
Let $y=x^4-5x^2+4$ and $y=0$ on $[-2,2]$ . Which setup gives total area?
Answer: $\displaystyle \int_{-2}^{-1}-(x^4-5x^2+4)dx+\int_{-1}^{1}(x^4-5x^2+4)dx+\int_{1}^{2}-(x^4-5x^2+4)dx$ This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x⁴ - 5x² + 4 intersects y = 0 at x = ±1, ±2, dividing [-2, 2] into subintervals. In [-2, -1] and [1, 2], y < 0; in [-1, 1], y > 0, requiring multiple splits. The negatives in outer integrals ensure 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.
Let $y=\ln x$ and $y=1$ on $[1,e^2]$ . Which setup gives total area between curves?
Answer: $\displaystyle \int_{1}^{e}(1-\ln x)\,dx+\int_{e}^{e^2}(\ln x-1)\,dx$ 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 = 1 at x = e within [1, e²], dividing into subintervals. In [1, e], ln x < 1; in [e, e²], > 1, requiring a split. This ensures positive differences. 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.

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