Integrals Ex 7.9 Class 12 NCERT: Evaluating Definite Integrals

The Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus is the cornerstone of evaluating definite integrals. It provides a direct link between the definite integral of a function and its antiderivative. The theorem states:

If 'f' is a continuous function defined on a closed interval [a, b] and 'F' is an antiderivative of 'f' (meaning F'(x) = f(x)), then the definite integral of 'f' from 'a' to 'b' is given by:

∫ₐᵇ f(x) dx = F(b) - F(a)

Let's break this down:

• f(x): This is the function you are integrating (the integrand).
• [a, b]: This is the interval of integration, where 'a' is the lower limit and 'b' is the upper limit.
• F(x): This is the antiderivative, or the indefinite integral, of f(x). You find this using the standard integration rules you've learned.
• F(b) - F(a): This is the core calculation. You evaluate the antiderivative at the upper limit ('b') and subtract the value of the antiderivative at the lower limit ('a'). The result is a single numerical value, representing the net area between the curve f(x) and the x-axis from x=a to x=b. This theorem bypasses the need for infinite sums, making calculations significantly easier.

How to Evaluate a Definite Integral (Step-by-Step)

1. Step 1: Find the Antiderivative F(x) — Given the integral ∫ₐᵇ f(x) dx, your first task is to find the indefinite integral of f(x). Let's call this F(x). Remember, F(x) = ∫ f(x) dx. For this step, you can ignore the constant of integration '+ C' because it will cancel out in the final subtraction.
2. Step 2: Set up the Evaluation Bracket — Once you have the antiderivative F(x), write it within square brackets with the limits of integration on the right side. This is standard notation: [F(x)]ₐᵇ. This notation signifies that you are about to evaluate F(x) at the limits 'a' and 'b'.
3. Step 3: Apply the Limits and Subtract — Substitute the upper limit 'b' into F(x) to get F(b). Then, substitute the lower limit 'a' into F(x) to get F(a). The final answer is the difference between these two values: F(b) - F(a). Be very careful with your arithmetic and signs during this final step.

Worked Examples for Integrals Ex 7.9

• Example 1: Evaluate ∫₂³ x³ dx Step 1: Find the Antiderivative. The function is f(x) = x³. The antiderivative is F(x) = ∫ x³ dx = x⁴/4. Step 2: Set up the Evaluation Bracket. We write this as [x⁴/4]₂³. Step 3: Apply the Limits and Subtract. Now we calculate F(3) - F(2). F(3) = 3⁴/4 = 81/4 F(2) = 2⁴/4 = 16/4 = 4 The integral's value is F(3) - F(2) = (81/4) - (16/4) = 65/4. Final Answer: ∫₂³ x³ dx = 65/4.
• Example 2: Evaluate ∫₀^(π/4) sec²(x) dx Step 1: Find the Antiderivative. The function is f(x) = sec²(x). We know from standard formulas that the integral of sec²(x) is tan(x). So, F(x) = tan(x). Step 2: Set up the Evaluation Bracket. We write this as [tan(x)]₀^(π/4). Step 3: Apply the Limits and Subtract. Now we calculate F(π/4) - F(0). F(π/4) = tan(π/4) = 1 F(0) = tan(0) = 0 The integral's value is F(π/4) - F(0) = 1 - 0 = 1. Final Answer: ∫₀^(π/4) sec²(x) dx = 1.
• Example 3: Evaluate ∫₀¹ dx / (1 + x²) Step 1: Find the Antiderivative. The function is f(x) = 1 / (1 + x²). This is a standard integral form. The antiderivative is F(x) = ∫ dx / (1 + x²) = tan⁻¹(x). Step 2: Set up the Evaluation Bracket. We write this as [tan⁻¹(x)]₀¹. Step 3: Apply the Limits and Subtract. Now we calculate F(1) - F(0). F(1) = tan⁻¹(1) = π/4 F(0) = tan⁻¹(0) = 0 The integral's value is F(1) - F(0) = π/4 - 0 = π/4. Final Answer: ∫₀¹ dx / (1 + x²) = π/4.

Exam Traps and Important Pointers

1. Order of Subtraction: Always subtract the value at the lower limit from the value at the upper limit: F(b) - F(a). Reversing this (F(a) - F(b)) is a common mistake that will give you the correct magnitude but the wrong sign.

2. Handling Substitution: If you use the substitution method to find the antiderivative (e.g., let u = g(x)), you must do one of two things:

• Change the Limits (Recommended): Convert your original limits a and b (which are in terms of x) to new limits in terms of u. The new lower limit will be g(a) and the upper limit will be g(b). Then, evaluate the integral with the new variable and new limits.
• Substitute Back: Alternatively, find the antiderivative in terms of u, then substitute g(x) back in for u, and finally apply the original limits a and b. The first method is generally faster and less prone to errors.

3. Arithmetic Precision: Definite integrals often result in fractions or expressions involving π and e. Be very careful with your calculations, especially when dealing with negative signs and subtracting terms.

Practice Questions with Solutions

• Q: Evaluate ∫₁² (4x³ - 5x² + 6x + 9) dx A: Step 1: Find the antiderivative of the polynomial. F(x) = ∫ (4x³ - 5x² + 6x + 9) dx = 4(x⁴/4) - 5(x³/3) + 6(x²/2) + 9x = x⁴ - (5/3)x³ + 3x² + 9x. Step 2: Evaluate F(2) and F(1). F(2) = (2)⁴ - (5/3)(2)³ + 3(2)² + 9(2) = 16 - 40/3 + 12 + 18 = 46 - 40/3 = (138 - 40)/3 = 98/3. F(1) = (1)⁴ - (5/3)(1)³ + 3(1)² + 9(1) = 1 - 5/3 + 3 + 9 = 13 - 5/3 = (39 - 5)/3 = 34/3. Step 3: Subtract F(1) from F(2). Value = F(2) - F(1) = (98/3) - (34/3) = 64/3. Final answer: 64/3
• Q: Evaluate ∫₀^(π/2) cos²(x) dx A: Step 1: Simplify the integrand. We cannot directly integrate cos²(x). Use the identity cos(2x) = 2cos²(x) - 1, which gives cos²(x) = (1 + cos(2x))/2. So, we need to evaluate ∫₀^(π/2) (1/2)(1 + cos(2x)) dx. Step 2: Find the antiderivative. F(x) = (1/2) ∫ (1 + cos(2x)) dx = (1/2) [x + sin(2x)/2]. Step 3: Apply the limits 0 and π/2. F(π/2) = (1/2) [π/2 + sin(2 * π/2)/2] = (1/2) [π/2 + sin(π)/2] = (1/2) [π/2 + 0] = π/4. F(0) = (1/2) [0 + sin(0)/2] = (1/2) [0 + 0] = 0. Step 4: Subtract. Value = F(π/2) - F(0) = π/4 - 0 = π/4. Final answer: π/4
• Q: Evaluate ∫₀¹ dx / √(1 - x²) A: Step 1: Identify the standard integral form. The integrand 1/√(1 - x²) is the derivative of sin⁻¹(x). Step 2: Find the antiderivative. F(x) = ∫ dx / √(1 - x²) = sin⁻¹(x). Step 3: Apply the limits 0 and 1. F(1) = sin⁻¹(1) = π/2. F(0) = sin⁻¹(0) = 0. Step 4: Subtract F(0) from F(1). Value = F(1) - F(0) = π/2 - 0 = π/2. Final answer: π/2
• Q: Evaluate ∫₂³ dx / (x² - 1) A: Step 1: Use partial fractions to simplify the integrand. We can write 1/(x²-1) as 1/((x-1)(x+1)). Let 1/((x-1)(x+1)) = A/(x-1) + B/(x+1). Solving this gives A = 1/2 and B = -1/2. So, the integral is ∫₂³ [1/2(x-1) - 1/2(x+1)] dx. Step 2: Find the antiderivative. F(x) = (1/2) [∫ dx/(x-1) - ∫ dx/(x+1)] = (1/2) [ln|x-1| - ln|x+1|] = (1/2) ln|(x-1)/(x+1)|. Step 3: Apply the limits 2 and 3. F(3) = (1/2) ln|(3-1)/(3+1)| = (1/2) ln|2/4| = (1/2) ln(1/2). F(2) = (1/2) ln|(2-1)/(2+1)| = (1/2) ln|1/3| = (1/2) ln(1/3). Step 4: Subtract F(2) from F(3). Value = F(3) - F(2) = (1/2) [ln(1/2) - ln(1/3)] = (1/2) ln[(1/2)/(1/3)] = (1/2) ln(3/2). Final answer: (1/2) ln(3/2)

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