📝 Net signed area under curve (25 MCQs)

📖 Explanation: The net signed area accounts for the sign of the function values. It is computed as the area where $f(x) \ge 0$ minus the area where $f(x) \le 0$. Option A is incorrect because it ignores the sign and sums absolute areas. Option C neglects negative regions entirely. Option D is incorrect because it ignores portions below the axis. Thus, the signed area represents a difference of areas, not a total sum.

📖 Explanation: The displacement is the integral of velocity, which is the net signed area. Compute $\int_0^4 (t^2 - 4t + 3) dt = [t^3/3 - 2t^2 + 3t]_0^4 = (64/3 - 32 + 12) - 0 = 64/3 - 20 = 4/3$. Wait, recalculating: $64/3 - 32 + 12 = 64/3 - 20 = 4/3$. I earlier made an error. Let's recompute carefully: $\int_0^4 (t^2 - 4t + 3) dt = [t^3/3 - 2t^2 + 3t]_0^4 = (64/3 - 32 + 12) - 0 = 64/3 - 20 = (64-60)/3 = 4/3$. So the correct answer is 4/3, not -4/3. The velocity is negative on (1,3) and positive elsewhere, so net displacement is positive.

📖 Explanation: By the additive property of definite integrals, $\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx = 5 + (-2) = 3$. This property holds regardless of the sign of the function. Option B (7) is the sum of absolute values, ignoring the negative sign. Option C (-3) is incorrect because it would require subtracting 5 from -2. Option D (-7) is the sum of absolute values with the wrong sign. The correct Easy of the additive property gives 3.

📖 Explanation: The integral of $\sin x$ from 0 to $2\pi$ is 0 because the positive area from 0 to $\pi$ cancels with the negative area from $\pi$ to $2\pi$. Option A is incorrect because total area would be 4, not 0. Option C is incorrect because it ignores the negative part. Option D is the total area, not net signed area. The net signed area is 0, reflecting that the curve has equal areas above and below the axis.

📖 Explanation: For a velocity function, the integral (net signed area) over a time interval gives the displacement, which is the change in position. Distance traveled would require integrating $|v(t)|$. Option A is incorrect because distance is the integral of speed, not velocity. Option C is incorrect because average speed is total distance divided by time. Option D is incorrect because acceleration is the derivative of velocity. Thus, the net signed area represents displacement.

📖 Explanation: For $f(x)=x$ on $[-1,1]$, the integral is 0 because the function is odd, but the function is not zero everywhere. Option B is an even function with positive integral. Option C has positive integral. Option D has positive integral. The student's claim is false because an integral being zero does not imply the function is zero; it only means the positive and negative areas balance. The correct counterexample is an odd function integrated over a symmetric interval.

📖 Explanation: $f(x) = x^3 - x = x(x-1)(x+1)$ is an odd function because $f(-x) = -f(x)$. For an odd function, the integral over a symmetric interval $[-a,a]$ is zero. Thus, $[-1,1]$ gives zero net signed area. Option A: $\int_0^2 (x^3 - x) dx = 4 - 2 = 2$. Option C: $\int_{-2}^0 (x^3 - x) dx = -4 + 2 = -2$. Option D: $\int_1^3 (x^3 - x) dx = (81/4 - 9/2) - (1/4 - 1/2) = (81/4 - 18/4) - (1/4 - 2/4) = 63/4 - (-1/4) = 64/4 = 16$. Only $[-1,1]$ gives zero.

📖 Explanation: A positive net signed area means that the integral is positive, which geometrically implies that the total area above the x-axis exceeds the total area below it. Option A is too strong; $f(x)$ could be negative in some places but the positive areas could dominate. Option C is about monotonicity, which is unrelated to signed area. Option D is incorrect because the function could cross the x-axis; the areas just need to balance out to a positive value.

📖 Explanation: Compute $\int_0^2 (x^2 - 1) dx = [x^3/3 - x]_0^2 = (8/3 - 2) - 0 = (8/3 - 6/3) = 2/3$. Wait, that's 2/3, but let's check: the function is negative on (0,1) and positive on (1,2). The net signed area is the difference, which is 2/3. Option A (0) would be if the positive and negative areas were equal, which is not the case. Option C (4/3) is the sum of absolute areas? No, total area is $\int_0^1 (1-x^2) dx + \int_1^2 (x^2-1) dx = 2/3 + 2/3 = 4/3$. So the net signed area is 2/3, not 4/3. Option D is wrong sign. So the correct answer is 2/3, which is option B? Let's recalc: $\int_0^2 (x^2-1) dx = (8/3 - 2) = 2/3$. So option B is 2/3. I'll correct the options. The correct answer is B.

📖 Explanation: The total area between $\sin x$ and the x-axis from 0 to $2\pi$ is $\int_0^{2\pi} |\sin x| dx = 2\int_0^{\pi} \sin x dx = 2[-\cos x]_0^{\pi} = 2(1 - (-1)) = 4$. Option A is the net signed area, not total. Option B is incorrect. Option D is incorrect. The total area is 4, which corresponds to two humps each of area 2.

📖 Explanation: Displacement is the integral of velocity, i.e., $\int_0^4 (t-2) dt = [t^2/2 - 2t]_0^4 = (8 - 8) - 0 = 0$. The particle moves 2 units in the negative direction from t=0 to t=2, and then 2 units in the positive direction from t=2 to t=4, returning to its starting position. Option A (-4) is the displacement if only negative motion is considered. Option C (4) is the total distance traveled. Option D (2) is incorrect. The net signed area is zero, so displacement is zero.

📖 Explanation: For integrable functions, the Riemann sum limit (and thus the net signed area) is independent of the choice of sample points. Option A is false because signed area can be negative. Option C is false because using absolute value gives total area, not signed area. Option D is false because if the function is negative everywhere, the signed area is negative, not equal to total area (which would be positive). The correct statement is B, highlighting a key property of the definite integral.

📖 Explanation: Using the additive property of integrals: $\int_0^5 f(x) dx = \int_0^2 f(x) dx + \int_2^5 f(x) dx = 3 + (-1) = 2$. Option B (4) is the sum of absolute values. Option C (-1) is just the second integral. Option D (3) is just the first integral. The correct answer is 2, demonstrating the additive nature of the definite integral over adjacent intervals.

📖 Explanation: For an odd function, $f(-x) = -f(x)$. The integral over a symmetric interval $[-a,a]$ is zero because the positive area on one side cancels the negative area on the other. Option A is true for even functions, not odd. Option B is false because it could be negative or zero. Option D is incorrect. The correct answer is C, which is a fundamental property of odd functions and signed area.

📖 Explanation: The triangle area above the axis is $\frac{1}{2} \times 2 \times 4 = 4$. The rectangle below the axis has area $3 \times (-2) = -6$. The net signed area is the sum: $4 + (-6) = -2$. Wait, that's -2. Let's recalc: Triangle area = 4 (positive). Rectangle area = 3 * (-2) = -6. Net = 4 - 6 = -2. So the correct answer is -2. Option A (4) is just the positive area. Option B (0) is not correct. Option D (6) is the absolute sum? No, 4+6=10, not 6. So the correct answer is -2, which is option C.

📖 Explanation: The student's computation is correct. $x^3$ is an odd function, so its integral over a symmetric interval is zero. The student evaluated the antiderivative correctly: $\int x^3 dx = x^4/4$, and $1/4 - 1/4 = 0$. Option A is incorrect because $x^3$ is odd, not even. Option C is incorrect because the integral is not $1/2$. Option D is incorrect because $x^3$ is continuous and hence integrable. The correct answer is B, and the student's reasoning is valid.

📖 Explanation: The net signed area is $\int_0^2 (2x - 1) dx = [x^2 - x]_0^2 = (4 - 2) - 0 = 2$. Geometrically, this is the area of the triangle above the axis minus the area below. The function crosses the x-axis at $x=0.5$. The area above is $\frac{1}{2} \times 1.5 \times 3 = 2.25$. The area below is $\frac{1}{2} \times 0.5 \times 1 = 0.25$. Net = 2.25 - 0.25 = 2. Option A (0) is incorrect. Option C (1) is incorrect. Option D (-1) is wrong sign. The correct answer is 2.

📖 Explanation: By definition, reversing the limits of integration changes the sign of the definite integral. Thus, $\int_b^a f(x) dx = -\int_a^b f(x) dx$, so $B = -A$, or $A = -B$. Option A is only true if the integral is zero. Option C is not necessarily true. Option D is not necessarily true. The correct answer is B, reflecting the antisymmetry property of definite integrals.

📖 Explanation: The net signed area is the sum of the signed areas: above the axis is +2, below is -2. Total = 2 + (-2) = 0. Option A (4) is the total distance (sum of absolute values). Option C (2) is only the positive part. Option D (-2) is only the negative part. The correct answer is 0, demonstrating that signed area accounts for direction, unlike total area.

📖 Explanation: The net signed area over [0,2] is the sum of the integrals over [0,1] and [1,2] by the additive property: $2 + (-3) = -1$. Option B (5) is the sum of absolute values. Option C (1) is the difference between the integrals. Option D (-5) is the sum of absolute values with wrong sign. The correct answer is -1, which means the area below the axis exceeds the area above by 1 unit.

📖 Explanation: The function is even and symmetric. The integral from -2 to 2 is $2\int_0^2 (|x| - 1) dx = 2\int_0^2 (x - 1) dx = 2([x^2/2 - x]_0^2) = 2((2 - 2) - 0) = 0$. Geometrically, the positive area from x=1 to 2 cancels the negative area from x=0 to 1 on both sides. Option B (2) is incorrect. Option C (4) is the total area if absolute value was used. Option D (-2) is the wrong sign. The net signed area is 0.

📖 Explanation: The statement is only true when $f(x) \ge 0$ on the interval. If $f$ takes negative values, then $|f(x)| = -f(x)$ in those regions, changing the integral. Option A is incorrect because linearity does not imply equality with absolute value. Option C is a correct statement but not a complete answer; the correct condition is nonnegativity. Option D is false because if $f$ is negative everywhere, $\int f < 0$ while $\int |f| > 0$, so it's not always less. The correct answer is B.

📖 Explanation: Compute $\int_0^{\pi} \cos x dx = [\sin x]_0^{\pi} = 0 - 0 = 0$. The positive area from 0 to $\pi/2$ cancels with the negative area from $\pi/2$ to $\pi$. Option B (1) is the area from 0 to $\pi/2$. Option C (2) is the total area. Option D (-1) is incorrect. The net signed area is 0, reflecting the symmetry of cosine about $\pi/2$.

📖 Explanation: For an even function, $f(-x) = f(x)$, so the integral over $[-a,a]$ is twice the integral over $[0,a]$. This is a key property of even functions and signed area. Option A is true for odd functions. Option C is incorrect because it misses the factor of 2. Option D is incorrect because it has the wrong sign. The correct answer is B.