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Calculus - Area under curves

Grade 12IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The definite integral ∫abf(x) dx\int_{a}^{b} f(x) \, dx represents the 'signed' area between the curve y=f(x)y = f(x) and the xx-axis from x=ax = a to x=bx = b.

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If the curve lies above the xx-axis (f(x)≥0f(x) \ge 0), the area is equal to the value of the definite integral.

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If the curve lies below the xx-axis (f(x)<0f(x) < 0), the integral will yield a negative value. To find the actual area, we take the absolute value: ∣∫abf(x) dx∣\left| \int_{a}^{b} f(x) \, dx \right|.

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Total area for a function that crosses the xx-axis is calculated by splitting the integral at the roots (x-intercepts) or by using the absolute value function: ∫ab∣f(x)∣ dx\int_{a}^{b} |f(x)| \, dx.

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Area between two curves f(x)f(x) and g(x)g(x) is found by integrating the difference between the 'upper' function and the 'lower' function: ∫ab(fupper(x)−glower(x)) dx\int_{a}^{b} (f_{upper}(x) - g_{lower}(x)) \, dx.

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In the IB AI course, the Graphic Display Calculator (GDC) is typically used to compute these integrals directly, especially for complex functions.

📐Formulae

Area=∫abf(x) dx(if f(x)≥0)\text{Area} = \int_{a}^{b} f(x) \, dx \quad \text{(if } f(x) \ge 0 \text{)}

Total Area=∫ab∣f(x)∣ dx\text{Total Area} = \int_{a}^{b} |f(x)| \, dx

Area between f(x) and g(x)=∫ab[f(x)−g(x)] dx\text{Area between } f(x) \text{ and } g(x) = \int_{a}^{b} [f(x) - g(x)] \, dx

Trapezoidal Rule: ∫aby dx≈12h[y0+yn+2(y1+y2+⋯+yn−1)]\text{Trapezoidal Rule: } \int_{a}^{b} y \, dx \approx \frac{1}{2}h [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})]

where h=b−an\text{where } h = \frac{b-a}{n}

💡Examples

Problem 1:

Find the area bounded by the curve f(x)=x2f(x) = x^2, the xx-axis, and the lines x=1x = 1 and x=3x = 3.

Solution:

Area=∫13x2 dx\text{Area} = \int_{1}^{3} x^2 \, dx Area=[x33]13\text{Area} = \left[ \frac{x^3}{3} \right]_{1}^{3} Area=(333)−(133)\text{Area} = \left( \frac{3^3}{3} \right) - \left( \frac{1^3}{3} \right) Area=9−13=263≈8.67 square units\text{Area} = 9 - \frac{1}{3} = \frac{26}{3} \approx 8.67 \text{ square units}

Explanation:

Since the function x2x^2 is always non-negative between x=1x=1 and x=3x=3, we simply evaluate the definite integral using the power rule for integration.

Problem 2:

Calculate the area of the region enclosed between the graphs of f(x)=x+2f(x) = x + 2 and g(x)=x2g(x) = x^2.

Solution:

  1. Find intersection points: x2=x+2  ⟹  x2−x−2=0  ⟹  (x−2)(x+1)=0x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 The points are x=−1x = -1 and x=2x = 2.
  2. Identify the upper curve: Between −1-1 and 22, x+2≥x2x+2 \ge x^2.
  3. Set up the integral: Area=∫−12(x+2−x2) dx\text{Area} = \int_{-1}^{2} (x + 2 - x^2) \, dx Area=[x22+2x−x33]−12\text{Area} = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} Area=(42+4−83)−(12−2+13)=4.5 square units\text{Area} = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) = 4.5 \text{ square units}

Explanation:

First, find where the two curves meet to determine the limits of integration. Then subtract the lower curve equation from the upper curve equation and integrate.