Solving the Definite Integral \(\int^5_{-5}5xdx\)

We are asked to find the value of the definite integral \(\int^5_{-5}5xdx\). This is a standard problem involving definite integrals and their properties.

Understanding the Given Definite Integral Problem

The problem requires us to evaluate the definite integral of the function \(f(x) = 5x\) from \(x = -5\) to \(x = 5\). Finding the value of this Definite Integral involves understanding either its geometric interpretation, fundamental theorem of calculus, or properties of integrals.

Identifying the Function Type: Odd or Even Function

The function we are integrating is \(f(x) = 5x\). Functions can be classified as even or odd, which is particularly useful when dealing with definite integrals over symmetric intervals.

• An even function satisfies \(f(-x) = f(x)\) for all \(x\) in its domain.
• An odd function satisfies \(f(-x) = -f(x)\) for all \(x\) in its domain.

Let's test our function \(f(x) = 5x\):

Substitute \(-x\) for \(x\):

\(f(-x) = 5(-x) = -5x\)

Comparing \(f(-x)\) with \(f(x)\), we see that \(-5x = -f(x)\). Therefore, \(f(x) = 5x\) is an odd function.

Applying the Property of Definite Integrals with Symmetric Limits

The limits of integration for our problem are from -5 to 5. This is a symmetric interval around zero, of the form \([-a, a]\) where \(a=5\).

There is a powerful property for definite integration over symmetric limits:

• If \(f(x)\) is an even function, then \(\int^a_{-a}f(x)dx = 2\int^a_0f(x)dx\).
• If \(f(x)\) is an odd function, then \(\int^a_{-a}f(x)dx = 0\).

Since we determined that \(f(x) = 5x\) is an odd function and the limits are symmetric (from -5 to 5), we can directly use the property for odd functions. The value of the integral \(\int^5_{-5}5xdx\) must be 0.

Direct Calculation of the Definite Integral

Alternatively, we can calculate the Definite Integral using the Fundamental Theorem of Calculus. This involves finding the antiderivative of the function and evaluating it at the limits of integration.

The antiderivative of \(5x\) is \(\frac{5x^{1+1}}{1+1} + C = \frac{5x^2}{2} + C\). For definite integrals, we don't need the constant \(C\).

So, the antiderivative is \(F(x) = \frac{5x^2}{2}\).

Now we evaluate \(F(x)\) at the upper limit (5) and the lower limit (-5) and subtract the results:

\begin{equation*} \int^5_{-5}5xdx = \left[\frac{5x^2}{2}\right]^5_{-5} \end{equation*}

Evaluate at the upper limit:

\(F(5) = \frac{5(5)^2}{2} = \frac{5(25)}{2} = \frac{125}{2}\)

Evaluate at the lower limit:

\(F(-5) = \frac{5(-5)^2}{2} = \frac{5(25)}{2} = \frac{125}{2}\)

Subtract the lower limit evaluation from the upper limit evaluation:

\begin{equation*} \int^5_{-5}5xdx = F(5) - F(-5) = \frac{125}{2} - \frac{125}{2} = 0 \end{equation*}

Both the property of odd functions over symmetric limits and direct integration confirm that the value of the Definite Integral \(\int^5_{-5}5xdx\) is 0.