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Calculus - Definite Integrals and their Properties

Grade 12ICSE

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

🔑Concepts

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Fundamental Theorem of Calculus: This theorem bridges differentiation and integration. It states that if f(x)f(x) is continuous on [a,b][a, b] and F(x)F(x) is the antiderivative, then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a). Visually, this represents the net signed area bounded by the graph of y=f(x)y=f(x), the x-axis, and the vertical lines x=ax=a and x=bx=b.

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Area Interpretation: A definite integral ∫abf(x) dx\int_a^b f(x) \, dx represents the area under the curve if f(x)≥0f(x) \ge 0. If the curve dips below the x-axis, the area below the axis is considered negative. Geometrically, the integral calculates the 'net' area, which is the sum of areas above the x-axis minus the areas below it.

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Change of Variable Property: The value of a definite integral is independent of the variable of integration, meaning ∫abf(x) dx=∫abf(t) dt\int_a^b f(x) \, dx = \int_a^b f(t) \, dt. This confirms that the integration variable is a 'dummy variable' and the result depends only on the function and the limits.

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Splitting the Interval: For any real number cc such that a<c<ba < c < b, the integral can be split as ∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx. This is visually equivalent to dividing the region under a curve into two adjacent vertical strips and adding their areas together; it is particularly useful for modulus or piecewise functions.

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The King's Property: One of the most powerful properties in solving ICSE problems is ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx. Visually, this corresponds to reflecting the graph of the function across the midpoint of the interval [a,b][a, b], which leaves the total area under the curve unchanged.

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Symmetry for Even and Odd Functions: For an integral with symmetric limits ∫−aaf(x) dx\int_{-a}^a f(x) \, dx, if f(x)f(x) is an even function (f(−x)=f(x)f(-x) = f(x)), the integral is 2∫0af(x) dx2\int_0^a f(x) \, dx, representing a mirror-image area across the y-axis. If f(x)f(x) is an odd function (f(−x)=−f(x)f(-x) = -f(x)), the integral is 00, as the area on the left of the y-axis perfectly cancels out the area on the right.

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Property of 2a2a: The integral ∫02af(x) dx\int_0^{2a} f(x) \, dx can be simplified to 2∫0af(x) dx2\int_0^a f(x) \, dx if f(2a−x)=f(x)f(2a-x) = f(x), or it becomes 00 if f(2a−x)=−f(x)f(2a-x) = -f(x). This is useful for trigonometric functions where the graph repeats or reflects its shape over half the interval.

📐Formulae

∫abf(x) dx=[F(x)]ab=F(b)−F(a)\int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)

∫abf(x) dx=−∫baf(x) dx\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx

∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx

∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx

∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx (if f(−x)=f(x)f(-x) = f(x))

∫−aaf(x) dx=0\int_{-a}^a f(x) \, dx = 0 (if f(−x)=−f(x)f(-x) = -f(x))

∫02af(x) dx=∫0af(x) dx+∫0af(2a−x) dx\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(2a-x) \, dx

💡Examples

Problem 1:

Evaluate I=∫0π/2sin⁡xsin⁡x+cos⁡x dxI = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx

Solution:

Step 1: Apply the property ∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx. I=∫0π/2sin⁡(π/2−x)sin⁡(π/2−x)+cos⁡(π/2−x) dxI = \int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx I=∫0π/2cos⁡xcos⁡x+sin⁡x dxI = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx

Step 2: Add the original integral and the modified integral. 2I=∫0π/2sin⁡xsin⁡x+cos⁡x dx+∫0π/2cos⁡xsin⁡x+cos⁡x dx2I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx + \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx 2I=∫0π/2sin⁡x+cos⁡xsin⁡x+cos⁡x dx2I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx 2I=∫0π/21 dx2I = \int_0^{\pi/2} 1 \, dx

Step 3: Solve for II. 2I=[x]0π/2=π2−02I = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 I=π4I = \frac{\pi}{4}

Explanation:

This approach uses the 'King's Property' to create a second integral with the same denominator. When added, the numerators combine to match the denominator, simplifying the integrand to 1.

Problem 2:

Evaluate ∫−12∣x3−x∣ dx\int_{-1}^2 |x^3 - x| \, dx

Solution:

Step 1: Find the critical points where x3−x=0x^3 - x = 0. x(x2−1)=0  ⟹  x=0,1,−1x(x^2 - 1) = 0 \implies x = 0, 1, -1.

Step 2: Determine the sign of f(x)=x3−xf(x) = x^3 - x in the sub-intervals. For x∈(−1,0)x \in (-1, 0), x3−xx^3 - x is positive. For x∈(0,1)x \in (0, 1), x3−xx^3 - x is negative. For x∈(1,2)x \in (1, 2), x3−xx^3 - x is positive.

Step 3: Split the integral based on these signs. I=∫−10(x3−x) dx+∫01−(x3−x) dx+∫12(x3−x) dxI = \int_{-1}^0 (x^3 - x) \, dx + \int_0^1 -(x^3 - x) \, dx + \int_1^2 (x^3 - x) \, dx

Step 4: Integrate each part. I=[x44−x22]−10−[x44−x22]01+[x44−x22]12I = [\frac{x^4}{4} - \frac{x^2}{2}]_{-1}^0 - [\frac{x^4}{4} - \frac{x^2}{2}]_0^1 + [\frac{x^4}{4} - \frac{x^2}{2}]_1^2 I=(0−(14−12))−((14−12)−0)+((164−42)−(14−12))I = (0 - (\frac{1}{4} - \frac{1}{2})) - ((\frac{1}{4} - \frac{1}{2}) - 0) + ((\frac{16}{4} - \frac{4}{2}) - (\frac{1}{4} - \frac{1}{2})) I=14+14+(2+14)=114I = \frac{1}{4} + \frac{1}{4} + (2 + \frac{1}{4}) = \frac{11}{4}

Explanation:

Since the integrand contains an absolute value, we must identify where the expression inside the modulus changes sign and split the integral into intervals where the function is purely positive or negative.