Problem 41 - Essential Calculus

Question

Consider the function \(f\) that is continuous on the interval [-5,5] and for which \(\int_{0}^{5} f(x) d x=4\) Evaluate each integral. (a) \(\int_{0}^{5}[f(x)+2] d x\) (b) \(\int_{-2}^{3} f(x+2) d x\) (c) \(\int_{-5}^{5} f(x) d x(f\) is even. \()\) (d) \(\int_{-5}^{5} f(x) d x(f\) is odd. \()\)

Step-by-Step Solution

Answer

(a) 14, (b) 4, (c) 8, (d) 0

Step 1: Problem (a)

The integral \(\int_{0}^{5}[f(x)+2] dx\) can be separated due to linearity of the integral. So, it equals \(\int_{0}^{5} f(x) dx+ \int_{0}^{5} 2 dx = 4+2*5 = 14.\)

Step 2: Problem (b)

To calculate \(\int_{-2}^{3} f(x+2) dx\), we should note that the function f(x) is shifted by 2 to the left. The effect on the integral is that the bounds are shifted. So, we end up with the integral \(\int_{-2+2}^{3+2} f(x) dx = \int_{0}^{5} f(x) dx = 4\).

Step 3: Problem (c)

If f(x) is an even function, then \(\int_{-a}^{a} f(x) dx = 2* \int_{0}^{a} f(x) dx\). Applying this to our situation, \(\int_{-5}^{5} f(x) dx = 2* \int_{0}^{5} f(x) dx = 2*4 = 8\).

Step 4: Problem (d)

If f(x) is an odd function, then \(\int_{-a}^{a} f(x) dx = 0\). So, \(\int_{-5}^{5} f(x) dx = 0\).

Key Concepts

Continuity

In mathematics, continuity is a fundamental concept that describes a situation where small changes in the input lead to small changes in the output. A function is continuous on an interval if there are no abrupt changes or jumps in its graph over that interval.
Continuity ensures that a function's graph can be drawn without lifting the pen from the paper.
Mathematically, a function \( f \) is continuous over a closed interval \([a, b]\) if it is continuous for every point \( c \) in the interval and:\( \lim_{x \to c} f(x) = f(c) \).
This implies three things:

• The function \( f(x) \) exists at \( c \).
• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \).

Continuity plays a crucial role in integral calculus, as it often guarantees the existence of definite integrals over an interval. It ensures that the integral accurately sums the "area under the curve" of the function.

Even and Odd Functions

Understanding whether a function is even or odd can simplify the process of integration. This classification depends on the function's symmetry.
An even function is symmetric about the y-axis, meaning that \( f(x) = f(-x) \) for all x in the function's domain.
Consequently, integrals of even functions across symmetric limits \([-a, a]\) simplify to \(2 \times \int_{0}^{a} f(x) \, dx \).
For example, if \( f(x) \) is even, then \( \int_{-5}^{5} f(x) \, dx = 2 \times \int_{0}^{5} f(x) \, dx \).
An odd function has a different kind of symmetry: \( f(x) = -f(-x) \).
This means the function is symmetric about the origin.
For odd functions, any definite integral over a symmetric interval results in zero: \( \int_{-a}^{a} f(x) \, dx = 0 \). Both of these properties are useful shortcuts in evaluating integrals.

Definite Integrals

Definite integrals are a crucial concept in calculus. They represent the net area under a curve within a certain interval and provide a numerical value.
The notation \( \int_{a}^{b} f(x) \, dx \) is used to denote a definite integral from \( a \) to \( b \). This corresponds to the signed areas under a function's graph between \( x = a \) and \( x = b \).
Here's a quick breakdown of how to interpret a definite integral:

• \( a \) and \( b \) are the limits of integration.
• \( f(x) \) is the function being integrated.
• \( dx \) signifies integration with respect to \( x \).

If \( f(x) \) is continuous over \([a, b]\), the definite integral exists and provides the measure of accumulated quantities, such as areas, over the interval. In practical terms, this is often visualized as the total area between the function's curve and the x-axis, accounting for areas above the x-axis being positive and those below being negative.

Linearity of Integration

The linearity of integration is a powerful property that allows for the simplification of integrals. It's based on two main principles:

• **Addition**: The integral of the sum of two functions is the sum of their integrals. If \( g(x) \) and \( h(x) \) are functions, then \( \int (g(x) + h(x)) \, dx = \int g(x) \, dx + \int h(x) \, dx \).
• **Scalar Multiplication**: A constant factor can be pulled out of the integral: \( \int c \cdot g(x) \, dx = c \cdot \int g(x) \, dx \).

These properties make working with integrals much easier.
For instance, evaluating the integral \( \int_{0}^{5} [f(x) + 2] \, dx \) can be separated into two simpler integrals: \( \int_{0}^{5} f(x) \, dx + \int_{0}^{5} 2 \, dx \).
These linearity rules stem from the fundamental properties of sums and scalar multiplication and apply to both definite and indefinite integrals.
Understanding these rules helps in simplifying complex integrals into manageable computations.