Understanding the Definite Integral Problem

We are asked to evaluate the definite integral \( \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right){\rm{dx}} \) given that \( {\rm{f}}\left( {\rm{x}} \right) \) and \( {\rm{g}}\left( {\rm{x}} \right) \) are continuous functions satisfying two specific conditions:

• Condition 1: \( {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} – {\rm{x}} \right) \)
• Condition 2: \( {\rm{g}}\left( {\rm{x}} \right) + {\rm{g}}\left( {\rm{a}} – {\rm{x}} \right) = 2 \)

To solve this problem, we will use a fundamental property of definite integrals over the interval \( [0, a] \).

Applying Definite Integral Properties

The key property for integrals from \( 0 \) to \( {\rm{a}} \) is:

Let the given integral be \( I \). So, \( I = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right){\rm{dx}} \).

Using the property with \( {\rm{h}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right) \), we apply the substitution \( {\rm{x}} \rightarrow {\rm{a}} – {\rm{x}} \):

Using the Given Conditions

We are given the condition \( {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} – {\rm{x}} \right) \). This tells us that \( {\rm{f}}\left( {\rm{a}} – {\rm{x}} \right) \) is equal to \( {\rm{f}}\left( {\rm{x}} \right) \).

We are also given the condition \( {\rm{g}}\left( {\rm{x}} \right) + {\rm{g}}\left( {\rm{a}} – {\rm{x}} \right) = 2 \). From this, we can isolate \( {\rm{g}}\left( {\rm{a}} – {\rm{x}} \right) \):

Now, substitute these expressions for \( {\rm{f}}\left( {\rm{a}} – {\rm{x}} \right) \) and \( {\rm{g}}\left( {\rm{a}} – {\rm{x}} \right) \) back into the transformed integral for \( I \):

Simplifying and Solving the Integral

Next, we expand the integrand (the function inside the integral) by distributing \( {\rm{f}}\left( {\rm{x}} \right) \):

Using the linearity property of definite integrals, which allows us to split the integral of a difference into the difference of integrals:

We can take the constant factor 2 out of the first integral:

Notice that the second integral on the right side is the original integral \( I \). So, we have an equation for \( I \):

Now, we solve this algebraic equation for \( I \). Add \( I \) to both sides of the equation:

Finally, divide both sides by 2:

Conclusion

The value of the integral \( \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right){\rm{dx}} \) is equal to \( \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} \).

Revision Table - Key Integral Concepts

Additional Information - Understanding Function Properties

The conditions given on the functions \( {\rm{f}}\left( {\rm{x}} \right) \) and \( {\rm{g}}\left( {\rm{x}} \right) \) are important. The condition \( {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} – {\rm{x}} \right) \) means that the function \( {\rm{f}} \) has symmetry about the vertical line \( {\rm{x}} = {\rm{a}}/2 \). For example, \( \cos(\rm{x}) \) on \( [0, \pi] \) satisfies \( \cos(\rm{x}) = \cos(\pi - \rm{x}) \).

The condition \( {\rm{g}}\left( {\rm{x}} \right) + {\rm{g}}\left( {\rm{a}} – {\rm{x}} \right) = 2 \) indicates that the average of the function values at \( {\rm{x}} \) and \( {\rm{a}} – {\rm{x}} \) is 1. For instance, \( {\rm{g}}\left( {\rm{x}} \right) = 1 \) satisfies this condition: \( 1 + 1 = 2 \). Another example is \( {\rm{g}}\left( {\rm{x}} \right) = \sin\left(\frac{\pi x}{a}\right) + 1 \) on \( [0, a] \). Here, \( \sin\left(\frac{\pi x}{a}\right) + 1 + \sin\left(\frac{\pi (a-x)}{a}\right) + 1 = \sin\left(\frac{\pi x}{a}\right) + 1 + \sin\left(\pi - \frac{\pi x}{a}\right) + 1 = \sin\left(\frac{\pi x}{a}\right) + 1 + \sin\left(\frac{\pi x}{a}\right) + 1 = 2\sin\left(\frac{\pi x}{a}\right) + 2 \). This example does not fit the condition exactly unless \( \sin\left(\frac{\pi x}{a}\right) = 0 \) for all x, which is not possible. A function like \( g(x) = cx/a + (2-c) \) would satisfy it if \( c + (2-c) = 2 \). For instance, \( g(x) = x/a + 1 \): \( x/a + 1 + (a-x)/a + 1 = (x+a-x)/a + 2 = a/a + 2 = 1+2=3 \). This example also doesn't work universally. Let's consider a simpler structure. If \( g(x) = 1 + h(x) \) where \( h(x) + h(a-x) = 0 \), i.e., \( h \) is odd about \( a/2 \). For example, \( g(x) = 1 + \cos\left(\frac{\pi x}{a} + \frac{\pi}{2}\right) \). Then \( g(a-x) = 1 + \cos\left(\frac{\pi (a-x)}{a} + \frac{\pi}{2}\right) = 1 + \cos\left(\pi - \frac{\pi x}{a} + \frac{\pi}{2}\right) = 1 + \cos\left(\frac{3\pi}{2} - \frac{\pi x}{a}\right) = 1 - \sin\left(\frac{\pi x}{a}\right) \). So \( g(x)+g(a-x) = 1+\cos(\frac{\pi x}{a} + \frac{\pi}{2}) + 1 - \sin(\frac{\pi x}{a}) = 2 - \sin(\frac{\pi x}{a}) - \sin(\frac{\pi x}{a}) \) using \( \cos(\theta + \pi/2) = -\sin(\theta) \). This is getting complicated. A simpler example for \( g(x) + g(a-x) = 2 \) is \( g(x) = cx + d \). Then \( cx+d + c(a-x)+d = cx+d+ca-cx+d = ca+2d \). For this to be 2, \( ca+2d=2 \). This must hold for any \(a\). If \( c=0 \), then \( 2d=2 \implies d=1 \), so \( g(x)=1 \). If \( a=1 \), \( c+2d=2 \). If \( a=2 \), \( 2c+2d=2 \implies c+d=1 \). From \( c+2d=2 \) and \( c+d=1 \), subtracting gives \( d=1 \) and \( c=0 \). So only \( g(x)=1 \) works for linear functions. The conditions on \( f(x) \) and \( g(x) \) are key to transforming the integral using the property.