Integration and Differentiation Techniques: A Comprehensive Guide

In this detailed guide, we will explore the concept of integration and differentiation, focusing on the Newton Leibnitz rule. This rule is fundamental in calculus and is widely used in various mathematical applications.

Introduction to Integration and Differentiation

Integration and differentiation are two fundamental processes in calculus. Integration is the process of finding the area under a curve, while differentiation involves finding the rate of change of a function.

The Newton Leibnitz Rule

The Newton Leibnitz rule, also known as the Fundamental Theorem of Calculus, connects integration and differentiation. It states that if a function f(x) is continuous on the interval [a, b], then the integral of f(x) from a to b can be computed using the antiderivative of f(x) as follows:

[ int_a^b f(x) , dx F(b) - F(a) ]

where F(x) is the antiderivative of f(x).

Example Problem

Let's solve the following problem using the Newton Leibnitz rule.

Given:

[ 2f(2x) 2xleft[ (x^2-1)^2-1right]-left[x^2-1right] 2x(x^2-1)^2 - x^2 - 2x(1) ]

We need to find the value of ( int_{-2}^2 f(x) , dx ).

Step 1: Differentiation

Start by differentiating both sides of the given equation:

[ 2f(2x) 2xleft[ (x^2-1)^2-1right]-left[x^2-1right] ]

To differentiate the right-hand side, apply the chain rule of differentiation to (f(2x)).

[ frac{d}{dx} [2f(2x)] 2 cdot 2 cdot f'(2x) 4f'(2x) ]

However, on the right-hand side, we only differentiate the expression (2x(x^2-1)^2 - x^2 - 2x(1)).

For simplicity, let's take a step to find f(x) from the given equation and then integrate.

Step 2: Integration

Now, we will integrate the given expression from (-2) to (2):

[ f(2) - f(-2) frac{1}{2}left{ left[ 2(1^2-1)^2 - 1^2 - 2(1) right] - left[ -2(-1^2-1)^2 - (-1)^2 - 2(-1) right] right} ]

Let's evaluate the terms step-by-step:

[ 2(1^2-1)^2 - 1^2 - 2(1) 2(0)^2 - 1 - 2 -3 ] [ -2(-1^2-1)^2 - (-1)^2 - 2(-1) -2(-2)^2 - 1 2 -8 - 1 2 -7 ]

Now substitute these values back into the expression:

[ f(2) - f(-2) frac{1}{2} left( -3 - (-7) right) frac{1}{2} left( -3 7 right) frac{1}{2} times 4 2 ]

However, there seems to be a mistake in the initial problem statement. Let's re-evaluate the expression more carefully:

Note: Given the correct problem should be as follows:

[ 2f(2x) 2xleft[ (x^2-1)^2-1right]-left[x^2-1right] 2x(x^2-1)^2 - x^2 - 2x(1) Rightarrow 2f(x) 2x(x^2-1) - x^2 - 2x ]

Integrating from -2 to 2:

[ int_{-2}^2 f(x) dx frac{1}{2} left[ 2(1^2-1) - 1^2 - 2 - 2(1) right] - left[ -2(-1^2-1) - (-1)^2 2(-1) - 2(1) right] ]

This will simplify as:

[ int_{-2}^2 f(x) dx frac{1}{2} left( -1 - (-1) right) 6 ]

Therefore, the final answer is:

[ int_{-2}^2 f(x) dx 6 ]

Conclusion

Understanding and applying the Newton Leibnitz rule is crucial for solving various calculus problems. The example demonstrated here highlights the importance of careful differentiation and integration to achieve the correct solution.

Keywords: integration, differentiation, Newton Leibnitz Rule