Questions & Answers: "Area under curve"

To find the area under the curve $y = x^2$ from $x=0$ to $x=3$, we evaluate the definite integral $\int_{0}^{3} x^2 dx$. Using the power rule of integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$, we get $[\frac{x^3}{3}]_{0}^{3}$. Substituting the upper limit: $\frac{3^3}{3} = \frac{27}{3} = 9$. Substituting the lower limit: $\frac{0^3}{3} = 0$. The area is $9 - 0 = 9$ sq units. Option 2 is incorrect as it results from a wrong power rule application. Option 3 neglects the division by 3. Option 4 is the area if the function were linear ($y=x$).

The area is given by the integral $\int_{0}^{2} (2x + 3) dx$. Integrating term by term: $\int 2x dx = x^2$ and $\int 3 dx = 3x$. Thus, we have $[x^2 + 3x]_{0}^{2}$. Substituting the upper limit $x=2$: $2^2 + 3(2) = 4 + 6 = 10$. Substituting the lower limit $x=0$: $0^2 + 3(0) = 0$. The area is $10 - 0 = 10$ sq units. Geometrically, this is a trapezoid with height 2 and parallel sides of lengths 3 and 7 ($Area = \frac{1}{2}(3+7) \times 2 = 10$). Options 2, 3, and 4 are results of common arithmetic mistakes in integration or substitution.

The area is calculated by the integral $\int_{0}^{\pi} \sin x dx$. The antiderivative of $\sin x$ is $-\cos x$. Applying the limits: $[-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0)$. Since $\cos \pi = -1$ and $\cos 0 = 1$, the expression becomes $(-(-1)) - (-1) = 1 + 1 = 2$ sq units. Option 1 is incorrect because it represents the area from $0$ to $\pi/2$. Option 3 is incorrect because while the integral of $\sin x$ from $0$ to $2\pi$ is 0, the area from $0$ to $\pi$ is positive. Option 4 is a common miscalculation involving the limits.

Area = 4 * ∫[0 to a] √(a² - x²) dx (considering the first quadrant). Using the standard formula ∫√(a² - x²) dx = (x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a), we evaluate from 0 to a. At x=a, it's (a²/2)(π/2) = πa²/4. Multiplying by 4 gives πa². Options B, C, and D are incorrect multiples of the area of a circle.

The curve is symmetric about the y-axis. The intersection points are x² = 4(4) = 16, so x = -4 and x = 4. Area = ∫[-4 to 4] (4 - x²/4) dx = 2 * ∫[0 to 4] (4 - x²/4) dx = 2 * [4x - x³/12] from 0 to 4 = 2 * [16 - 64/12] = 2 * [16 - 16/3] = 2 * [32/3] = 64/3. Option A is only half the area (one side). Options C and D are arithmetic mistakes.

Area = ∫[0 to π/2] cos x dx + |∫[π/2 to π] cos x dx|. This is because cos x is positive in [0, π/2] and negative in [π/2, π]. Area = [sin x] from 0 to π/2 + |[sin x] from π/2 to π| = (1 - 0) + |0 - 1| = 1 + 1 = 2. Option C is the value of the definite integral, not the physical area. Option A and D are incorrect.

The region is a triangle with vertices at (0,0), (1,1), and (-1,1). Using integration: Area = ∫[-1 to 1] (1 - |x|) dx = 2 * ∫[0 to 1] (1 - x) dx = 2 * [x - x²/2] from 0 to 1 = 2 * (1 - 1/2) = 1. Geometrically, it is a triangle with base 2 (from x=-1 to 1) and height 1, so Area = 1/2 * 2 * 1 = 1. Options A and C are incorrect factors.

The latus rectum of y² = 4ax is the line x = a. The area is symmetric about the x-axis, so we calculate the area above the axis and multiply by 2. Area = 2 * ∫[0 to a] √(4ax) dx = 4√a * ∫[0 to a] x^(1/2) dx = 4√a * [x^(3/2) / (3/2)] from 0 to a = 4√a * (2/3) * a^(3/2) = 8a²/3. Options A, C, and D represent common errors in neglecting symmetry or integration constants.

Area = ∫[0 to π] sin x dx. The integral of sin x is -cos x. Area = [-cos x] from 0 to π = -cos(π) - (-cos(0)) = -(-1) + 1 = 1 + 1 = 2. Option A is the result of the integral of sin x from 0 to 2π (where signed area cancels), but from 0 to π, the area is positive. Options B and D are numerical mistakes.

The area is 4 times the area in the first quadrant: 4 * ∫[0 to a] (b/a)√(a² - x²) dx. Factoring out (b/a), the integral ∫[0 to a] √(a² - x²) dx equals πa²/4. Thus, Area = 4 * (b/a) * (πa²/4) = πab. Options B, C, and D are incorrect formulas for the area of an ellipse.

The area is given by the definite integral of y with respect to x from x=1 to x=3. Area = ∫[1 to 3] x² dx = [x³/3] from 1 to 3. Substituting the limits: (3³/3) - (1³/3) = 27/3 - 1/3 = 26/3. Option B is incorrect as it uses limits 0 to 2. Options C and D are calculation errors.

The curves intersect where x² = x, so x = 0 and x = 1. In the interval [0, 1], √x ≥ x. Area = ∫[0 to 1] (√x - x) dx = [2/3 x^(3/2) - x²/2] from 0 to 1 = 2/3 - 1/2 = 1/6. Option B is just the integral of √x, and option C is just the integral of x; option D is a common error in power rule application.

Area must be positive. From x = -1 to 0, x³ is negative, so we take the absolute value. Area = |∫[-1 to 0] x³ dx| + ∫[0 to 1] x³ dx = |[x⁴/4] from -1 to 0| + [x⁴/4] from 0 to 1 = |0 - 1/4| + (1/4 - 0) = 1/4 + 1/4 = 1/2. Option A is the result of a simple definite integral without taking absolute values for area. B and D are calculation errors.

The total displacement of the particle is given by the algebraic sum of the areas under the velocity-time graph. Areas above the time axis are considered positive, and areas below are considered negative.

1. Area 1 (from $t=0$ s to $t=2$ s): This segment forms a triangle above the time axis. Its base is $2$ s and its height is $4$ m/s. $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (4 \text{ m/s}) = 4 \text{ m}$

2. Area 2 (from $t=2$ s to $t=4$ s): This segment also forms a triangle above the time axis. Its base is $(4-2)=2$ s and its height is $4$ m/s. $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (4 \text{ m/s}) = 4 \text{ m}$

3. Area 3 (from $t=4$ s to $t=6$ s): This segment forms a triangle below the time axis. Its base is $(6-4)=2$ s and its height is $-2$ m/s. $A_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ s}) \times (-2 \text{ m/s}) = -2 \text{ m}$

Total displacement $S_{\text{total}}$ is the sum of these areas: $S_{\text{total}} = A_1 + A_2 + A_3 = 4 \text{ m} + 4 \text{ m} + (-2 \text{ m}) = 6 \text{ m}$

Why other options are incorrect:

• Option A ($4$ m): This represents only the displacement during the first $2$ seconds.
• Option C ($8$ m): This would be the sum of the magnitudes of the positive displacements only ($4 ext{ m} + 4 ext{ m}$). It ignores the negative displacement.
• Option D ($10$ m): This is the total distance traveled, calculated as the sum of the magnitudes of all areas: $|A_1| + |A_2| + |A_3| = 4 ext{ m} + 4 ext{ m} + |-2 ext{ m}| = 10 ext{ m}$. Displacement considers direction, while distance does not.

The total work done by a variable force is represented by the area under the force-displacement graph.

1. Area 1 (from $x=0$ m to $x=3$ m): This segment forms a rectangle. Its length (displacement) is $3$ m and its height (force) is $5$ N. $A_1 = \text{length} \times \text{height} = (3 \text{ m}) \times (5 \text{ N}) = 15 \text{ J}$

2. Area 2 (from $x=3$ m to $x=5$ m): This segment forms a triangle. Its base (displacement) is $(5-3)=2$ m and its height (force) is $5$ N (at $x=3$ m, decreasing to $0$ N at $x=5$ m). $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ m}) \times (5 \text{ N}) = 5 \text{ J}$

Total work done $W_{\text{total}}$ is the sum of these areas: $W_{\text{total}} = A_1 + A_2 = 15 \text{ J} + 5 \text{ J} = 20 \text{ J}$

Why other options are incorrect:

• Option A ($10$ J): This value does not correspond to the correct area calculation for the given force-displacement graph.
• Option B ($15$ J): This represents only the work done in the first segment ($0$ m to $3$ m, the rectangular area), neglecting the work done in the second segment.
• Option C ($17.5$ J): This would be an incorrect calculation, possibly due to a miscalculation of one of the areas (e.g., if the triangle area was $2.5$ J instead of $5$ J, leading to $15+2.5=17.5$ J).