📝 Area Between Two Curves (30 MCQs)

📖 Explanation: The core condition for the formula $A = \int_a^b [f(x) - g(x)] \, dx$ is that $f(x) \geq g(x)$ over the entire interval. This ensures the integrand represents the positive vertical distance between the curves, and thus the integral yields a positive area. The functions need not be nonnegative (the formula works even if both are negative, as long as the inequality holds), they need not intersect at the endpoints, and they can be any continuous functions. A common misconception is to assume the functions must be above the x-axis, but the formula's validity relies solely on the relative position of the two curves.

📖 Explanation: On the interval $(0,1)$, $\sqrt{x}$ is greater than $x^2$. For example, at $x = 0.25$, $\sqrt{0.25} = 0.5$ and $0.25^2 = 0.0625$. Thus, $g(x)$ is the upper boundary and $f(x)$ is the lower boundary. Students might incorrectly assume that $x^2$ is always above $\sqrt{x}$ based on their behavior for $x > 1$, but on $(0,1)$ the relationship is reversed. The area between them is found by integrating $[\sqrt{x} - x^2] \, dx$.

📖 Explanation: The student's approach is incorrect because the relative positions of the curves change. On $[0,1]$, $x \ge x^2$, but on $[1,2]$, $x^2 \ge x$. Simply integrating $x - x^2$ from 0 to 2 would subtract the area where $x^2 > x$, leading to an incorrect (and possibly negative) result. The correct method is to split the integral at the intersection point $x=1$: $\int_0^1 (x - x^2) \, dx + \int_1^2 (x^2 - x) \, dx$. This is a classic error in applying the area formula without checking the inequality throughout the interval.

📖 Explanation: To find the area, we first find the points of intersection by setting $x^2 = 2x - x^2$, which gives $2x^2 - 2x = 0$, or $2x(x-1)=0$. Thus, $x=0$ and $x=1$. On the interval $[0,1]$, $2x - x^2 \ge x^2$. The area is $A = \int_0^1 [(2x - x^2) - x^2] \, dx = \int_0^1 (2x - 2x^2) \, dx = [x^2 - \frac{2}{3}x^3]_0^1 = 1 - \frac{2}{3} = \frac{1}{3}$. The lower curve is $y=x^2$, and the upper is the parabola opening downward.

📖 Explanation: On $[0, \pi/4]$, $\cos x \ge \sin x$, so the upper curve is $y = \cos x$. On $[\pi/4, \pi/2]$, $\sin x \ge \cos x$, so the upper curve is $y = \sin x$. Therefore, the area must be split at the intersection point $x = \pi/4$. The correct integral is $\int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$. Students often make the mistake of integrating the difference in one order across the entire interval, forgetting that the upper and lower functions swap at the intersection.

📖 Explanation: The curve intersects the x-axis where $x(x^2 - 6x + 8) = 0$, which gives $x=0, 2, 4$. On $[0,2]$, the curve is above the x-axis (e.g., at $x=1$, $y=3$). On $[2,4]$, the curve is below the x-axis (e.g., at $x=3$, $y=-3$). The area is the sum of the absolute values of the integrals: $\int_0^2 (x^3 - 6x^2 + 8x) \, dx + \int_2^4 -(x^3 - 6x^2 + 8x) \, dx$. Evaluating gives $[x^4/4 - 2x^3 + 4x^2]_0^2 = 4$ and $-[x^4/4 - 2x^3 + 4x^2]_2^4 = 4$. The total area is $8$. The integral from 0 to 4 is 0, which would be a common error if one doesn't account for the sign change.

📖 Explanation: The definite integral of the vertical distance between two curves gives a positive area only when the integrand is nonnegative. A negative result indicates that the assumed upper function $f(x)$ is actually below the lower function $g(x)$ on at least part, or all, of the interval. The correct procedure is to ensure the integrand is positive. This could involve splitting the interval at intersection points or swapping the functions. The limits of integration and the variable are usually determined by the problem; the constant of integration is irrelevant for definite integrals.

📖 Explanation: Let the parabola be $f(x)$. The area $A$ between $f(x)$ and the x-axis (y=0) is $\int_a^b [f(x) - 0] \, dx$. The area between $f(x)$ and the line $y=1$ is $\int_a^b [f(x) - 1] \, dx$, assuming $f(x) \ge 1$ on the interval. This is equal to $\int_a^b f(x) \, dx - \int_a^b 1 \, dx = A - (b-a)$. The area is reduced by the area of a rectangle of height 1 and width (b-a) between the x-axis and the line y=1. This question tests the concept that the integrand is the *difference* of the functions, and a vertical shift changes the integrand.

📖 Explanation: Set $x^2 - 2x = 3$ to find intersections: $x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0$, so $x = -1, 3$. On $[-1, 3]$, the line $y=3$ is above the parabola. The area is $\int_{-1}^3 [3 - (x^2 - 2x)] \, dx = \int_{-1}^3 (-x^2 + 2x + 3) \, dx$. The antiderivative is $-x^3/3 + x^2 + 3x$. Evaluating from -1 to 3: $(-9 + 9 + 9) - (1/3 + 1 - 3) = 9 - (-5/3) = 32/3$. The parabola opens upward, and its vertex is below the line, so this region forms a closed bounded area.

📖 Explanation: Finding intersections requires solving $|x| = 2 - x^2$. For $x \ge 0$, $x = 2 - x^2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0 \Rightarrow x=1$. By symmetry, for $x < 0$, the intersection is at $x = -1$. On $[-1, 1]$, $2 - x^2 \ge |x|$. The area is $\int_{-1}^1 (2 - x^2 - |x|) \, dx = 2 \int_0^1 (2 - x^2 - x) \, dx = 2[2x - x^3/3 - x^2/2]_0^1 = 2(2 - 1/3 - 1/2) = 2(7/6) = 7/3$. This combines symmetry, absolute values, and integration. The symmetry is key to simplifying the calculation.

📖 Explanation: The formula $\int_a^b [f(x) - g(x)] \, dx$ gives the area between two curves regardless of their position relative to the x-axis, as long as the curve $y=f(x)$ is above $y=g(x)$. For example, if both are negative but $f(x) \ge g(x)$, the difference $f(x) - g(x)$ is still the positive vertical distance between them. The integral of this positive difference yields the positive area. This concept is often misunderstood; students may incorrectly think the functions must be above the x-axis. The area is simply the integral of the difference of the y-values, which is independent of vertical shifts.

📖 Explanation: The region described is complex because the lower boundary changes. The line $y = x-2$ intersects the x-axis at $x=2$. The parabola $y = \sqrt{x}$ intersects the line when $\sqrt{x} = x-2$, which gives $x=4$. The region is bounded above by $y=\sqrt{x}$. Below, it's the x-axis from $x=0$ to $x=2$, and the line $y=x-2$ from $x=2$ to $x=4$. Therefore, the area is best found by splitting at $x=2$: $\int_0^2 \sqrt{x} \, dx + \int_2^4 [\sqrt{x} - (x-2)] \, dx$. Integrating with respect to y is also possible but requires finding the x-intervals in terms of y. A common error is to use a single integrand for the whole interval.

📖 Explanation: These curves intersect when $y^2 = y+2 \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y-2)(y+1) = 0$, giving $y = -1, 2$. The corresponding x-values are 1 and 4. When integrating with respect to $y$, the right boundary is $x = y+2$ and the left boundary is $x = y^2$. Thus, the area is $\int_{-1}^2 [(y+2) - y^2] \, dy$. The function $y+2 \ge y^2$ on this interval. Integrating with respect to $x$ would require splitting the region at $x=1$, as the left boundary changes. This question assesses understanding of the axis of integration and the choice of the easier method.

📖 Explanation: On the interval $[0, \ln 2]$, $e^{2x} \ge e^x$. The area between them is $\int_0^{\ln 2} (e^{2x} - e^x) \, dx$. The antiderivative is $\frac{e^{2x}}{2} - e^x$. Evaluating from 0 to $\ln 2$: $(\frac{e^{2\ln 2}}{2} - e^{\ln 2}) - (\frac{1}{2} - 1) = (\frac{4}{2} - 2) - (-\frac{1}{2}) = (2 - 2) + \frac{1}{2} = \frac{1}{2}$. The careful evaluation of the exponential at the bounds is crucial; $e^{\ln 2} = 2$ and $e^{2\ln 2} = 4$. A common mistake is mishandling the exponential rules.

📖 Explanation: If $f(x)$ is positive, the area under $f(x) + 3$ from 0 to 2 is $\int_0^2 (f(x) + 3) \, dx = \int_0^2 f(x) \, dx + \int_0^2 3 \, dx = 4 + 6 = 10$. This is because the area between $y = f(x)+3$ and the x-axis is the area under $f(x)$ plus the area of a rectangle of height 3 and width 2. This is a direct Medium of the linearity of the integral. Students might mistakenly think the area increases by a factor, or that the original area is insufficient information. The key is that a vertical shift of the curve changes the area by a constant amount that is easy to compute.

📖 Explanation: The total area between $y=x^2$ and $y=4$ is found where $x^2 = 4$, giving $x = \pm 2$. The area is $\int_{-2}^2 (4 - x^2) \, dx = 2 \int_0^2 (4 - x^2) \, dx = 2[4x - x^3/3]_0^2 = 2(8 - 8/3) = 32/3$. Half of this area is $16/3$. The line $y=k$ intersects the parabola at $x = \pm \sqrt{k}$. The area above the line is $\int_{-\sqrt{k}}^{\sqrt{k}} (k - x^2) \, dx = 2[kx - x^3/3]_0^{\sqrt{k}} = 2(k^{3/2} - k^{3/2}/3) = \frac{4}{3} k^{3/2}$. Setting this equal to $16/3$ gives $k^{3/2} = 4$, so $k = 4^{2/3} = \sqrt[3]{16}$. This problem requires setting up a more complex area equation and solving for the parameter $k$.

📖 Explanation: Find intersections: $x^2 = 2 - x \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0$. In the first quadrant, the intersection is at $x=1$. The curves also intersect the axes: $y=x^2$ passes through (0,0), and $y=2-x$ has intercepts (0,2) and (2,0). The region is bounded by the y-axis, the parabola, and the line. The area is $\int_0^1 [(2-x) - x^2] \, dx = [2x - x^2/2 - x^3/3]_0^1 = 2 - 1/2 - 1/3 = 7/6$. The line is above the parabola on this interval. This problem requires careful identification of the region boundaries in the first quadrant.

📖 Explanation: The absolute value of the integral $\left| \int_a^b [f(x) - g(x)] \, dx \right|$ is not generally the area between the curves. This expression gives the absolute value of the *net signed area*, which can be zero or less than the actual area if the curves cross. For example, if $f(x) = x$ and $g(x) = x^2$ on $[0,1]$, the integral is $1/6$, and its absolute value is $1/6$, which is the area. But if the curves cross, like $f(x) = x^2$ and $g(x) = x$ on $[0,2]$, the integral is $\int_0^2 (x^2 - x) \, dx = -2/3$, and its absolute value is $2/3$, while the actual area is $\int_0^1 (x - x^2) \, dx + \int_1^2 (x^2 - x) \, dx = 1/6 + 5/6 = 1$.

📖 Explanation: This question mixes the concept of area between curves with the method of disks for volume. The volume of revolution about the x-axis is $V = \pi \int_0^1 [f(x)]^2 \, dx$. The area $A$ is $\int_0^1 f(x) \, dx$. These two integrals are not generally related in a simple way. To find the volume, one would need to compute a new integral involving the square of the function. The correct answer is that the volume cannot be determined from $A$ alone, as the shape of the function matters. This distinguishes the concept of area from volume, which is a common point of confusion.

📖 Explanation: The area between the curves from $a$ to $c$ is simply $\int_a^c [f(x) - g(x)] \, dx$. Since the integral from $a$ to $b$ is the sum of the integrals from $a$ to $c$ and $c$ to $b$ (additivity of the definite integral), the integral from $a$ to $c$ is also $\int_a^b [f(x) - g(x)] \, dx - \int_c^b [f(x) - g(x)] \, dx$. Both expressions are equivalent. This tests the fundamental property of definite integrals: $\int_a^b = \int_a^c + \int_c^b$.

📖 Explanation: The area between an even and an odd function over a symmetric interval $[-1,1]$ is not necessarily symmetric. The area on $[-1,0]$ is $\int_{-1}^0 [f(x) - g(x)] \, dx$. Let $u = -x$, then $dx = -du$, and when $x=-1$, $u=1$; when $x=0$, $u=0$. The integral becomes $\int_{1}^{0} [f(-u) - g(-u)] (-du) = \int_0^1 [f(u) + g(u)] \, du$. This is not generally equal to $\int_0^1 [f(u) - g(u)] \, du = 2$. For example, if $f(x) = 2$ and $g(x) = x$, the area on $[0,1]$ is 2. The area on $[-1,0]$ is $\int_{-1}^0 (2 - x) \, dx = [2x - x^2/2]_{-1}^0 = 0 - (-2 - 1/2) = 2.5$, which is not 2. The integrand changes in a non-trivial way due to the parity properties.

📖 Explanation: The calculator computed $\int_0^{2\pi} (\sin x - \cos x) \, dx$, which is the net signed area. The area between the curves is the sum of the absolute values of the area segments where they cross. On $[0, 2\pi]$, $\sin x$ and $\cos x$ intersect at $\pi/4$ and $5\pi/4$. The functions swap positions, so the integral over the entire interval cancels out. The correct area is $\int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) \, dx + \int_{5\pi/4}^{2\pi} (\cos x - \sin x) \, dx$, which is not zero. This is a common error when relying on numerical integration without considering the geometry.

📖 Explanation: On the interval $[0,2]$, $20 - 3x^2$ is greater than $e^x$. For example, at $x=0$, $20 > 1$, and at $x=2$, $8 > 7.39$. Therefore, the integrand $(20 - 3x^2) - e^x$ is positive, so the area is $\int_0^2 [(20 - 3x^2) - e^x] \, dx$. Since the integrand is positive, its absolute value is the same, making option C also correct. This question highlights that while the absolute value is always correct, it's often unnecessary if the order of the functions is known. The trick is recognizing that both expressions are equivalent under the given conditions.

📖 Explanation: The area between $y = x^2 + c$ and $y = c$ is $\int_{-1}^1 (x^2 + c - c) \, dx = \int_{-1}^1 x^2 \, dx = \frac{2}{3}$. This is independent of $c$. Now, the area between $y = x^2 + c$ and $y = c+2$ is $\int_{-1}^1 [(x^2 + c) - (c+2)] \, dx = \int_{-1}^1 (x^2 - 2) \, dx$. This is $\frac{2}{3} - 4 = -\frac{10}{3}$. Since the integrand is negative, the area is the absolute value, $\frac{10}{3}$. But wait, the line $y = c+2$ is above the parabola $y=x^2+c$ on $[-1,1]$ if $x^2 \le 2$, which is true. So the area is $\int_{-1}^1 (2 - x^2) \, dx = 4 - \frac{2}{3} = \frac{10}{3}$. The correct answer is $\frac{10}{3}$. The vertical shift of the lower curve changes the area by the area of the rectangle between the two horizontal lines.

📖 Explanation: Find intersections: $\sqrt{x} = x^3$. Squaring gives $x = x^6 \Rightarrow x^6 - x = 0 \Rightarrow x(x^5 - 1) = 0$. Thus, $x=0$ and $x=1$. On $[0,1]$, $\sqrt{x} \ge x^3$. The area is $\int_0^1 (\sqrt{x} - x^3) \, dx = [\frac{2}{3} x^{3/2} - \frac{x^4}{4}]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{8-3}{12} = \frac{5}{12}$. This is a standard Medium where the functions are chosen to produce a nice result. Students should be comfortable with integrating power functions and evaluating at the bounds.

📖 Explanation: The region is bounded by the parabola $y = 4 - x^2$, the x-axis ($y=0$), and the vertical line $x=1$. The parabola intersects the x-axis at $x = \pm 2$. In the first quadrant, the relevant section is from $x=0$ to $x=2$. However, the condition $x=1$ and the first quadrant bound the region to the left of $x=1$. The area is $\int_0^1 (4 - x^2) \, dx$. The line $x=1$ acts as the right boundary, not the x-intercept. This tests the ability to correctly identify the bounds from a description of a region, which is more subtle than just finding intersection points.

📖 Explanation: The area between $y = f(x)$ and $y = k$ is $\int_a^b [f(x) - k] \, dx$. Since $f(x) \ge 0$ and $k \le 0$, this is $\int_a^b f(x) \, dx - \int_a^b k \, dx = A - k(b-a)$. Because $k$ is negative, $-k(b-a)$ is positive, so the area is $A + |k|(b-a)$. The area between the curve and the x-axis is $A$. The line $y=k$ is below the x-axis, so the region between the curve and this line is larger than the area under the curve. It adds the area of a rectangle of height $|k|$ and width $(b-a)$.

📖 Explanation: First, find the area of the region (this was done in a previous question: $A = 1/3$). The volume of revolution about the x-axis is found using the washer method: $V = \pi \int_0^1 [(2x - x^2)^2 - (x^2)^2] \, dx$. Simplify the integrand: $(4x^2 - 4x^3 + x^4) - x^4 = 4x^2 - 4x^3$. Thus, $V = \pi \int_0^1 (4x^2 - 4x^3) \, dx = \pi [\frac{4x^3}{3} - x^4]_0^1 = \pi (\frac{4}{3} - 1) = \frac{\pi}{3}$. The correct answer is $\frac{\pi}{3}$. This question cleverly combines area between curves with the volume of revolution, but the area itself is a distractor. The student must recognize that the volume requires integrating the squares of the functions.