Integrals: NCERT Exercise 7.1 Concepts and Solutions

Welcome to the world of Integral Calculus! This chapter marks a significant shift from differentiation. If differentiation is about finding the rate of change, integration is about accumulating that change. Think of it as the reverse process. NCERT Exercise 7.1 is your foundational entry point into this fascinating topic. Here, you will learn about 'antiderivatives'—the core concept of indefinite integration. We will start by finding integrals using the method of inspection, which directly links back to your knowledge of derivatives. Then, you'll master the basic integration formulas for polynomial, trigonometric, and exponential functions. By the end of this guide, you will be able to confidently solve all the problems in Integrals Ex 7.1 and build a strong base for the more advanced topics in calculus that lie ahead. Let's begin this journey together!

Understanding Indefinite Integrals (Antiderivatives)

In differentiation, you are given a function F(x) and asked to find its derivative f(x). For example, if F(x) = x³, its derivative is f(x) = 3x². Integration, or finding the antiderivative, reverses this. You are given the derivative f(x) = 3x² and asked to find the original function F(x). But there's a small catch. The derivative of is 3x², but so is the derivative of x³ + 5, x³ - 100, or any function x³ + C (where C is a constant). Because the derivative of any constant is zero, we can't know the exact original constant. This is why we always add a 'Constant of Integration', + C, to our answer. The notation for an indefinite integral is:

∫f(x) dx = F(x) + C

Here:

  • is the integral symbol.
  • f(x) is the integrand (the function to be integrated).
  • dx indicates that the integration is with respect to the variable x.
  • F(x) is the antiderivative of f(x).
  • C is the constant of integration.

Fundamental Integration Formulas for Exercise 7.1

Power Rule
∫xⁿ dx = (xⁿ⁺¹ / (n+1)) + C, where n ≠ -1
Constant Rule
∫k dx = kx + C, where k is a constant.
Reciprocal Rule
∫(1/x) dx = log|x| + C
Exponential Rule
∫eˣ dx = eˣ + C and ∫aˣ dx = (aˣ / log a) + C
Trigonometric Formulas
∫sin(x) dx = -cos(x) + C ∫cos(x) dx = sin(x) + C ∫sec²(x) dx = tan(x) + C ∫csc²(x) dx = -cot(x) + C ∫sec(x)tan(x) dx = sec(x) + C ∫csc(x)cot(x) dx = -csc(x) + C
Linearity of Integration
∫[k f(x) ± g(x)] dx = k ∫f(x) dx ± ∫g(x) dx

Worked Examples for Integrals Ex 7.1

  • Question: Find the integral of ∫(4x³ - 6) dx Step 1: Apply Linearity. Break the integral into two separate integrals: ∫(4x³ - 6) dx = ∫4x³ dx - ∫6 dx Step 2: Use the Constant Multiple Rule. Move the constants outside the integral sign: = 4 ∫x³ dx - 6 ∫1 dx Step 3: Apply the Power Rule and Constant Rule. For the first term, use ∫xⁿ dx = xⁿ⁺¹/(n+1). Here, n=3. For the second term, use ∫k dx = kx. = 4 (x³⁺¹ / (3+1)) - 6x = 4 (x⁴ / 4) - 6x Step 4: Simplify and add the constant of integration. = x⁴ - 6x + C Final Answer: The integral of (4x³ - 6) is x⁴ - 6x + C.
  • Question: Find the integral of ∫(2x² - 3sin(x) + 5√x) dx Step 1: Apply Linearity. Separate the integral into three parts: ∫2x² dx - ∫3sin(x) dx + ∫5√x dx Step 2: Rewrite the terms and move constants out. Remember that √x = x¹/². = 2∫x² dx - 3∫sin(x) dx + 5∫x¹/² dx Step 3: Integrate each term using the appropriate formula. - For , use the power rule (n=2). - For sin(x), ∫sin(x) dx = -cos(x). - For x¹/², use the power rule (n=1/2). = 2(x³/3) - 3(-cos(x)) + 5(x¹/²⁺¹ / (1/2+1)) Step 4: Simplify and add the constant of integration. = (2/3)x³ + 3cos(x) + 5(x³/² / (3/2)) = (2/3)x³ + 3cos(x) + 5 * (2/3)x³/² = (2/3)x³ + 3cos(x) + (10/3)x³/² + C Final Answer: (2/3)x³ + 3cos(x) + (10/3)x³/² + C.
  • Question: Find the integral of ∫(x³ + 3x + 4) / √x dx Step 1: Simplify the integrand. Before integrating, divide each term in the numerator by the denominator, √x = x¹/². (x³ + 3x + 4) / x¹/² = x³/x¹/² + 3x¹/x¹/² + 4/x¹/² Using the law of exponents xᵃ/xᵇ = xᵃ⁻ᵇ: = x³⁻¹/² + 3x¹⁻¹/² + 4x⁻¹/² = x⁵/² + 3x¹/² + 4x⁻¹/² Step 2: Integrate the simplified expression. Now integrate term by term: ∫(x⁵/² + 3x¹/² + 4x⁻¹/²) dx = ∫x⁵/² dx + 3∫x¹/² dx + 4∫x⁻¹/² dx Step 3: Apply the Power Rule to each term. = [x⁵/²⁺¹ / (5/2+1)] + 3[x¹/²⁺¹ / (1/2+1)] + 4[x⁻¹/²⁺¹ / (-1/2+1)] = [x⁷/² / (7/2)] + 3[x³/² / (3/2)] + 4[x¹/² / (1/2)] Step 4: Simplify and add the constant of integration. = (2/7)x⁷/² + 3(2/3)x³/² + 4(2)x¹/² + C = (2/7)x⁷/² + 2x³/² + 8x¹/² + C Final Answer: (2/7)x⁷/² + 2x³/² + 8√x + C.

Exam Tip: Never Forget '+C'!

One of the most common mistakes students make in indefinite integration is forgetting to add the constant of integration, + C, at the end of their answer. Remember, the integral ∫f(x) dx represents a family of functions whose derivative is f(x), not just a single function. The + C accounts for all the possible vertical shifts of the antiderivative curve. In board exams, you can lose a crucial half or full mark for this omission, even if the rest of your integration is perfect. Make it a habit: as soon as you finish integrating, write + C.

Practice Questions with Solutions

  • Q: Find the antiderivative of f(x) = (x - 1/x)². A: Step 1: First, expand the expression. f(x) = x² - 2(x)(1/x) + (1/x)² = x² - 2 + 1/x² = x² - 2 + x⁻². Step 2: Now, integrate the simplified expression term by term: ∫(x² - 2 + x⁻²) dx = ∫x² dx - ∫2 dx + ∫x⁻² dx. Step 3: Apply the power rule ∫xⁿ dx = xⁿ⁺¹/(n+1) and the constant rule. (x³/3) - 2x + (x⁻¹/(-1)) + C. Step 4: Simplify the result. (x³/3) - 2x - 1/x + C. Final answer: The antiderivative is x³/3 - 2x - 1/x + C.
  • Q: Integrate ∫(ax² + bx + c) dx with respect to x. A: Step 1: Use the linearity property of integrals to separate the terms: ∫ax² dx + ∫bx dx + ∫c dx. Step 2: Move the constants (a, b, c) outside the integral signs: a∫x² dx + b∫x¹ dx + c∫1 dx. Step 3: Apply the power rule to and , and the constant rule to 1: a(x³/3) + b(x²/2) + c(x). Step 4: Combine the terms and add the constant of integration, C. (a/3)x³ + (b/2)x² + cx + C. Final answer: (a/3)x³ + (b/2)x² + cx + C.
  • Q: Find ∫(sec²(x) - 1) dx. A: Step 1: Recognize that sec²(x) - 1 = tan²(x). However, integrating tan²(x) directly is not a basic formula. It's easier to integrate sec²(x) and 1 separately. Step 2: Use the linearity property: ∫(sec²(x) - 1) dx = ∫sec²(x) dx - ∫1 dx. Step 3: Apply the standard integration formulas: ∫sec²(x) dx = tan(x) and ∫1 dx = x. Step 4: Combine the results and add the constant of integration: tan(x) - x + C. Final answer: tan(x) - x + C.
  • Q: If the derivative of a function is f'(x) = 4x³ - 3/x⁴ and f(2) = 0, find the function f(x). A: Step 1: To find f(x), we need to integrate f'(x). So, f(x) = ∫(4x³ - 3x⁻⁴) dx. Step 2: Integrate term by term: f(x) = 4∫x³ dx - 3∫x⁻⁴ dx = 4(x⁴/4) - 3(x⁻³/(-3)) + C. Step 3: Simplify the expression: f(x) = x⁴ + x⁻³ + C = x⁴ + 1/x³ + C. Step 4: Use the given condition f(2) = 0 to find C. f(2) = 2⁴ + 1/2³ + C = 16 + 1/8 + C = 0. This gives 129/8 + C = 0, so C = -129/8. Final answer: f(x) = x⁴ + 1/x³ - 129/8.

Frequently Asked Questions

What is the difference between an antiderivative and an indefinite integral?

They are very closely related. An antiderivative is a specific function `F(x)` whose derivative is `f(x)`. The indefinite integral `∫f(x) dx` represents the entire family of antiderivatives, written as `F(x) + C`.

Why is the constant of integration 'C' so important in indefinite integrals?

The constant `C` is crucial because the derivative of any constant is zero. This means there are infinitely many functions (e.g., `x²`, `x²+5`, `x²-10`) that all have the same derivative (`2x`). The `+C` represents all these possibilities.

What is the method of inspection for finding integrals?

The method of inspection is about mentally reversing the process of differentiation. You look at a function, say `cos(2x)`, and think, "Which function's derivative is `cos(2x)`?" You might guess `sin(2x)`, but its derivative is `2cos(2x)`. So you adjust your guess to `(1/2)sin(2x)`.