Calculus

Introduction

Calculus is a significant part of the JEE Main Mathematics syllabus, encompassing concepts of differentiation, integration, and their applications. This study note aims to break down these complex ideas into digestible sections, ensuring a thorough understanding of each topic.

Differentiation

Differentiation deals with the concept of change. It is used to find the rate at which a quantity changes with respect to another.

Basics of Differentiation

• Definition: The derivative of a function $f(x)$ with respect to $x$ is defined as: $$ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} $$
• Notation: The derivative of $f(x)$ can be denoted as $f'(x)$, $\frac{df}{dx}$, or $D_x f$.

Rules of Differentiation

1. Power Rule: $$ \frac{d}{dx} [x^n] = nx^{n-1} $$

Example

Find the derivative of $f(x) = x^3$: $$ f'(x) = 3x^2 $$

1. Sum Rule: $$ \frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x) $$
2. Product Rule: $$ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) $$
3. Quotient Rule: $$ \frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2} $$
4. Chain Rule: $$ \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) $$

Tip

Memorize these rules and practice applying them to different functions to gain proficiency in differentiation.

Higher-Order Derivatives

• Second Derivative: The derivative of the first derivative: $$ f''(x) = \frac{d}{dx} [f'(x)] $$
• Higher-Order Derivatives: Continue differentiating to find third, fourth, etc., derivatives: $$ f^{(n)}(x) $$

Applications of Differentiation

1. Finding Tangents and Normals:
   • Slope of the tangent to $y = f(x)$ at $x = a$ is $f'(a)$.
   • Equation of the tangent: $y - f(a) = f'(a)(x - a)$.
2. Maxima and Minima:
   • Critical points where $f'(x) = 0$.
   • Use the second derivative test: If $f''(x) > 0$, it's a local minimum; if $f''(x)

< 0$, it's a local maximum.

Example

Find the local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 15$:

1. Find $f'(x)$: $$ f'(x) = 3x^2 - 12x + 9 $$

1. Set $f'(x) = 0$: $$ 3x^2 - 12x + 9 = 0 $$ $$ x^2 - 4x + 3 = 0 $$ $$ (x-3)(x-1) = 0 $$ So, $x = 1$ and $x = 3$.

1. Find $f''(x)$: $$ f''(x) = 6x - 12 $$ Evaluate $f''(x)$ at $x = 1$ and $x = 3$: $$ f''(1) = -6 \quad (\text{local maximum}) $$ $$ f''(3) = 6 \quad (\text{local minimum}) $$

Common Mistake

Not checking the second derivative can lead to incorrect conclusions about maxima and minima.

Integration

Integration is the reverse process of differentiation, used to find the area under curves.

Basics of Integration

• Definition: The integral of a function $f(x)$ with respect to $x$ is: $$ \int f(x) , dx $$
• Indefinite Integrals: Represent a family of functions with an arbitrary constant $C$: $$ \int f(x) , dx = F(x) + C $$

Rules of Integration

1. Power Rule: $$ \int x^n , dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) $$
2. Sum Rule: $$ \int [f(x) + g(x)] , dx = \int f(x) , dx + \int g(x) , dx $$
3. Integration by Parts: $$ \int u , dv = uv - \int v , du $$
4. Substitution Rule: If $x = g(t)$, then: $$ \int f(x) , dx = \int f(g(t)) g'(t) , dt $$

Tip

Use the substitution rule to simplify integrals involving composite functions.

Definite Integrals

• Definition: The definite integral of $f(x)$ from $a$ to $b$ is: $$ \int_{a}^{b} f(x) , dx $$
• Fundamental Theorem of Calculus: $$ \int_{a}^{b} f(x) , dx = F(b) - F(a) $$ where $F(x)$ is the antiderivative of $f(x)$.

Example

Evaluate $\int_{0}^{1} (3x^2 + 2x) , dx$:

1. Find the antiderivative: $$ \int (3x^2 + 2x) , dx = x^3 + x^2 + C $$

1. Apply the limits: $$ \left[ x^3 + x^2 \right]_{0}^{1} = (1^3 + 1^2) - (0^3 + 0^2) = 2 $$

Applications of Integration

1. Area under Curves: $$ \text{Area} = \int_{a}^{b} f(x) , dx $$
2. Volume of Solids of Revolution:
   • Using the disk method: $$ V = \pi \int_{a}^{b} [f(x)]^2 , dx $$
   • Using the shell method: $$ V = 2\pi \int_{a}^{b} x f(x) , dx $$

Note

Understanding the geometric interpretation of integrals is crucial for solving application problems.

Differential Equations

Differential equations involve functions and their derivatives and are used to model real-world phenomena.

Basics of Differential Equations

• Definition: An equation involving derivatives of a function: $$ \frac{dy}{dx} = f(x, y) $$

Solving Differential Equations

1. Separable Equations: $$ \frac{dy}{dx} = g(x)h(y) $$ Separate variables and integrate: $$ \int \frac{1}{h(y)} , dy = \int g(x) , dx $$

Example

Solve $\frac{dy}{dx} = xy$:

1. Separate variables: $$ \frac{1}{y} , dy = x , dx $$

1. Integrate both sides: $$ \int \frac{1}{y} , dy = \int x , dx $$ $$ \ln|y| = \frac{x^2}{2} + C $$ $$ y = e^{\frac{x^2}{2} + C} $$ $$ y = Ce^{\frac{x^2}{2}} $$

1. Linear Equations: $$ \frac{dy}{dx} + P(x)y = Q(x) $$ Use the integrating factor: $$ \mu(x) = e^{\int P(x) , dx} $$ Multiply through by $\mu(x)$ and integrate.

Tip

Practice solving various types of differential equations to understand the methods thoroughly.

Conclusion

Calculus is a fundamental component of the JEE Main Mathematics syllabus. Mastering differentiation, integration, and differential equations is crucial for success. Practice regularly, understand the underlying concepts, and apply them to solve practical problems.

Note

Consistent practice and understanding the geometric and physical interpretations of calculus concepts will enhance problem-solving skills.