Question

Use symmetry to explain why $$\int_{-4}^{4}\left(5 x^{4}+3 x^{3}+2 x^{2}+x+1\right) d x=2 \int_{0}^{4}\left(5 x^{4}+2 x^{2}+1\right) d x$$

Short answer

The even part of the given polynomial is $5x^4 + 2x^2 + 1$, and the odd part is $3x^3 + x$. 2) What is the value of the integral of the odd part from $-4$ to $4$ and why? The integral of the odd part from $-4$ to $4$ is equal to zero. This is because, for odd functions, the definite integral from $-a$ to $a$ is equal to zero. 3) How did the even function properties help to prove the given identity? For even functions, the definite integral from $-a$ to $a$ equals two times the definite integral from $0$ to $a$. By applying this property to the even part, we were able to show the given identity holds.

Step-by-step solution

Identify odd and even parts of the polynomial

We need to separate the polynomial elements such that we have odd and even parts. The even parts are those functions that satisfy f(x) = f(-x) and odd parts satisfy f(x) = -f(-x). The expression is: $$ 5x^4 + 3x^3 + 2x^2 + x + 1 $$ The even parts are the terms with even exponents as they have the mentioned symmetry: \(5x^4\), \(2x^2\), and the constant \(1\). The odd part contains the terms with odd exponents: \(3x^3\) and \(x\). So, we have: Even part: \(5x^4 + 2x^2 + 1\) Odd part: \(3x^3 + x\)

Apply the symmetry property for odd functions

Now, we know that for odd functions, the definite integral from \(-a\) to \(a\) equals zero, like: $$ \int_{-a}^{a} f_{odd}(x) dx = 0 $$ In this case, our odd part is \(f_{odd}(x) = 3x^3 + x\), if we calculate the integral of the odd part from \(-4\) to \(4\), it results in zero: $$ \int_{-4}^{4} (3x^3 + x) dx = 0 $$

Apply symmetry property for even functions and final equivalence

We can rewrite the integral of the entire polynomial as the sum of integrals of even and odd parts: $$ \int_{-4}^{4}(5 x^{4}+3 x^{3}+2 x^{2}+x+1) dx = \int_{-4}^{4}(5 x^{4} + 2 x^2 + 1) dx + \int_{-4}^{4} (3 x^3 + x) dx $$ As we already proved that the integral of the odd part is zero: $$ \int_{-4}^{4}(5 x^{4}+3 x^{3}+2 x^{2}+x+1) dx = \int_{-4}^{4}(5 x^{4} + 2 x^2 + 1) dx $$ Now, we know that for even functions, the definite integral from \(-a\) to \(a\) equals two times the definite integral from \(0\) to \(a\): $$ \int_{-a}^{a} f_{even}(x) dx = 2 \int_{0}^{a} f_{even}(x)dx $$ Applying this property to our even part, we get: $$ \int_{-4}^{4}(5 x^{4} + 2 x^2 + 1) dx = 2 \int_{0}^{4}(5 x^{4} + 2 x^2 + 1) dx $$ Thus, we have shown that using symmetry, the given identity holds: $$ \int_{-4}^{4}\left(5 x^{4}+3 x^{3}+2 x^{2}+x+1\right) d x = 2 \int_{0}^{4}\left(5 x^{4}+2 x^{2}+1\right) d x $$

Key concepts

Definite Integrals

A definite integral has both a start and an endpoint, and it represents the signed area under the curve of a function on a graph within those bounds. The integral is 'definite' because it has clear boundaries, indicated by the lower limit \(a\) and the upper limit \(b\) in the notation \(\int_{a}^{b} f(x) dx\).

For example, in the problem \(\int_{-4}^{4}(5 x^{4}+3 x^{3}+2 x^{2}+x+1) dx\), we are interested in the area from \(x = -4\) to \(x = 4\). This integral is called 'definite' since these limits are specified. When we solve it, the result is a numerical value, unlike in indefinite integrals, which have an arbitrary constant \(C\).

Understanding how to evaluate definite integrals efficiently involves recognizing patterns and properties that simplify computation, such as symmetry in even and odd functions, that can significantly cut down the work required in finding the solution.

Odd and Even Functions

In the world of functions, 'odd' and 'even' have special meanings. These terms refer to symmetry: an even function \(f(x)\) is identical to \(f(-x)\) and typically has mirror symmetry across the y-axis, while an odd function satisfies the relationship \(f(x) = -f(-x)\) and has rotational symmetry around the origin.

Visualizing Even and Odd Functions

A classic example of an even function is \(y = x^{2}\), which looks the same if we flip it around the y-axis. For odd functions, think of \(y = x^{3}\) - if you rotate the graph 180 degrees around the origin, it looks unchanged.

The exercise involves recognizing that \(3x^3\) and \(x\) are odd functions and their integral over a symmetric interval around zero will cancel out, \(\int_{-a}^{a} f_{odd}(x) dx = 0\). On the other hand, the even functions, including terms like \(5x^4\), \(2x^2\), and the constant term, contribute to the area under the curve and affect the integral's value.

Integral Properties

One of the key properties of integrals that can ease calculations is the behavior of integrals of even and odd functions over symmetrical limits. Specifically, the integral of an odd function from \( -a \) to \( a \) is always zero because the areas above and below the x-axis cancel each other out.

Even Functions and Symmetry

For even functions, however, symmetry helps simplify the integral, and we only need to calculate the area from \(0\) to \(a\) and then double it to get the full integral from \( -a \) to \( a \) as shown in the relationship: \(\int_{-a}^{a} f_{even}(x) dx = 2 \int_{0}^{a} f_{even}(x) dx\).

In the given example, knowing this property allowed us to separate the integrals of even and odd parts, showing that the integral of the original function from \( -4 \) to \( 4 \) is double the integral of the even function from \( 0 \) to \( 4 \). This technique of using symmetry is a valuable tool, streamlining the calculation process and helping to avoid unnecessary work.