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Integrals - Basic properties of definite integrals and evaluation of definite integrals

Grade 12CBSE

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

🔑Concepts

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Definition of Definite Integral: The definite integral ∫abf(x)dx\int_{a}^{b} f(x) dx is the limit of a sum that represents the net signed area bounded by the curve y=f(x)y = f(x), the xx-axis, and the vertical lines x=ax = a and x=bx = b. Visually, if the function lies above the xx-axis, the area is calculated as positive; if it lies below, the area is negative. The final value is the algebraic sum of these areas.

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Fundamental Theorem of Calculus: This concept provides the practical method for evaluation. If F(x)F(x) is the antiderivative (or primitive) of a continuous function f(x)f(x), then the integral from aa to bb is calculated as F(b)−F(a)F(b) - F(a). This links the process of differentiation and integration into a single inverse relationship.

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Change of Variable (Dummy Variable): The value of a definite integral is independent of the name of the variable of integration. Thus, ∫abf(x)dx=∫abf(t)dt\int_{a}^{b} f(x) dx = \int_{a}^{b} f(t) dt. Graphically, the shape of the function and the bounds on the axis remain identical regardless of whether we label the horizontal axis xx, tt, or uu.

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Additivity of the Interval: A definite integral can be split at any point cc such that ∫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 represented by dividing the total region into two adjacent sub-regions at the line x=cx = c. This property is essential for evaluating integrals of piecewise or modulus functions.

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King's Property (Reflection Property): The property ∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx is one of the most useful tools for simplification. Geometrically, it reflects the graph of the function across the midpoint of the interval [a,b][a, b], which leaves the total area under the curve unchanged. It is often used to eliminate complex denominators or logarithmic terms.

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Symmetry in Symmetric Intervals: For an integral evaluated from −a-a to aa, we check the parity of the function. If f(x)f(x) is odd (f(−x)=−f(x)f(-x) = -f(x)), the graph has rotational symmetry about the origin, causing the areas on opposite sides of the yy-axis to cancel out to 00. If f(x)f(x) is even (f(−x)=f(x)f(-x) = f(x)), the graph is mirrored across the yy-axis, so the total area is simply twice the area from 00 to aa.

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Property of 2a2a: The integral ∫02af(x)dx\int_{0}^{2a} f(x) dx can be reduced to 2∫0af(x)dx2\int_{0}^{a} f(x) dx if f(2a−x)=f(x)f(2a-x) = f(x). If f(2a−x)=−f(x)f(2a-x) = -f(x), the value of the integral is 00. Visually, this identifies whether the second half of the interval [a,2a][a, 2a] mirrors the area of the first half [0,a][0, a] or negates it.

📐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=∫acf(x)dx+∫cbf(x)dx\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx

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

∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx

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

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

∫02af(x)dx=2∫0af(x)dx if f(2a−x)=f(x)\int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dx \text{ if } f(2a-x) = f(x)

💡Examples

Problem 1:

Evaluate the definite integral: ∫12(3x2+4x+5)dx\int_{1}^{2} (3x^2 + 4x + 5) dx

Solution:

Step 1: Find the antiderivative of the function f(x)=3x2+4x+5f(x) = 3x^2 + 4x + 5. Using the power rule, F(x)=x3+2x2+5xF(x) = x^3 + 2x^2 + 5x. Step 2: Apply the Fundamental Theorem of Calculus by substituting the limits: I=[x3+2x2+5x]12I = [x^3 + 2x^2 + 5x]_{1}^{2}. Step 3: Calculate the value at the upper limit (x=2x=2): F(2)=(2)3+2(2)2+5(2)=8+8+10=26F(2) = (2)^3 + 2(2)^2 + 5(2) = 8 + 8 + 10 = 26. Step 4: Calculate the value at the lower limit (x=1x=1): F(1)=(1)3+2(1)2+5(1)=1+2+5=8F(1) = (1)^3 + 2(1)^2 + 5(1) = 1 + 2 + 5 = 8. Step 5: Subtract the lower limit value from the upper limit value: I=26−8=18I = 26 - 8 = 18.

Explanation:

This example demonstrates the basic evaluation of a definite integral using the Fundamental Theorem of Calculus. The process involves finding the indefinite integral and then finding the difference between the values at the upper and lower bounds.

Problem 2:

Evaluate I=∫0π2sin⁡xsin⁡x+cos⁡xdxI = \int_{0}^{\frac{\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. Here a=π2a = \frac{\pi}{2}. Step 2: Rewrite the integral as I=∫0π2sin⁡(π2−x)sin⁡(π2−x)+cos⁡(π2−x)dxI = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx. Step 3: Use the identities sin⁡(π2−x)=cos⁡x\sin(\frac{\pi}{2}-x) = \cos x and cos⁡(π2−x)=sin⁡x\cos(\frac{\pi}{2}-x) = \sin x. Now, I=∫0π2cos⁡xcos⁡x+sin⁡xdxI = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx. Step 4: Add the original integral and the modified integral: 2I=∫0π2(sin⁡xsin⁡x+cos⁡x+cos⁡xcos⁡x+sin⁡x)dx2I = \int_{0}^{\frac{\pi}{2}} (\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}) dx. Step 5: Simplify the integrand: 2I=∫0π2sin⁡x+cos⁡xsin⁡x+cos⁡xdx=∫0π21dx2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{0}^{\frac{\pi}{2}} 1 dx. Step 6: Evaluate: 2I=[x]0π2=π2−02I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0. Therefore, I=π4I = \frac{\pi}{4}.

Explanation:

This example utilizes 'King's Property' to simplify a complex trigonometric fraction. By adding the reflected version of the integral to itself, the integrand simplifies to 11, making the calculation straightforward.