Problems based on Newton-Leibnitz Rule

The Newton-Leibnitz rule is a fundamental theorem in calculus that connects the concepts of differentiation and integration. It states that if a function $f$ is continuous on the closed interval $[a, b]$ and has an antiderivative $F$ on that interval, then the definite integral of $f$ from $a$ to $b$ is equal to the difference between the values of $F$ at $b$ and $a$.

Newton-Leibnitz Rule Formula

The formula for the Newton-Leibnitz rule is given by:

$$ \int_{a}^{b} f(x) \, dx = F(b) - F(a) $$

where $F'(x) = f(x)$.

Important Points

• The function $f(x)$ must be continuous on the interval $[a, b]$.
• $F(x)$ is an antiderivative of $f(x)$, which means $F'(x) = f(x)$.
• The rule simplifies the process of finding the definite integral of a function.

Differences and Important Points

Aspect	Differentiation	Integration
Fundamental Concept	Derivative of a function	Antiderivative of a function
Newton-Leibnitz Rule	Not directly applicable	Directly provides the value of an integral
Operation	Local (at a point)	Global (over an interval)
Result	Slope of the tangent line	Net area under the curve
Notation	$f'(x)$ or $\frac{df}{dx}$	$\int f(x) \, dx$

Examples

Example 1: Basic Application

Find the definite integral of $f(x) = x^2$ from $a = 1$ to $b = 3$ using the Newton-Leibnitz rule.

Solution:

First, find an antiderivative $F(x)$ of $f(x) = x^2$.

$$ F(x) = \frac{x^3}{3} + C $$

Using the Newton-Leibnitz rule:

$$ \int_{1}^{3} x^2 \, dx = F(3) - F(1) = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3} $$

Example 2: Using Properties of Definite Integrals

Evaluate the integral $\int_{0}^{2\pi} \sin(x) \, dx$.

Solution:

An antiderivative of $\sin(x)$ is $F(x) = -\cos(x)$. Applying the Newton-Leibnitz rule:

$$ \int_{0}^{2\pi} \sin(x) \, dx = F(2\pi) - F(0) = -\cos(2\pi) + \cos(0) = -1 + 1 = 0 $$

Example 3: Piecewise Functions

Evaluate the integral $\int_{-1}^{1} |x| \, dx$.

Solution:

The function $|x|$ is continuous but not differentiable at $x = 0$. We split the integral into two parts:

$$ \int_{-1}^{1} |x| \, dx = \int_{-1}^{0} (-x) \, dx + \int_{0}^{1} x \, dx $$

Find the antiderivatives:

$$ F(x) = \begin{cases} -\frac{x^2}{2} & \text{for } x < 0 \ \frac{x^2}{2} & \text{for } x \geq 0 \end{cases} $$

Apply the Newton-Leibnitz rule to each part:

$$ \int_{-1}^{0} (-x) \, dx = F(0) - F(-1) = 0 - \left(-\frac{(-1)^2}{2}\right) = \frac{1}{2} $$

$$ \int_{0}^{1} x \, dx = F(1) - F(0) = \frac{1^2}{2} - 0 = \frac{1}{2} $$

Adding both parts:

$$ \int_{-1}^{1} |x| \, dx = \frac{1}{2} + \frac{1}{2} = 1 $$

Example 4: Indefinite Integral as an Antiderivative

Find the definite integral of $f(x) = e^x$ from $a = 0$ to $b = 1$.

Solution:

The antiderivative of $e^x$ is $F(x) = e^x + C$. Using the Newton-Leibnitz rule:

$$ \int_{0}^{1} e^x \, dx = F(1) - F(0) = e^1 - e^0 = e - 1 $$

In conclusion, the Newton-Leibnitz rule is a powerful tool for evaluating definite integrals. Understanding how to apply this rule and recognizing when a function is continuous and differentiable are key to solving problems based on this rule.