Question

If a is positive and \(\mathrm{I}=\int_{-1}^{1} \frac{\mathrm{dx}}{\sqrt{1-2 \mathrm{ax}+\mathrm{a}^{2}}}\) then show that \(\mathrm{I}=2 \mathrm{ifa}<1\) and \(\mathrm{I}=\frac{2}{\mathrm{a}}\) if \(\mathrm{a}>1\).

Short answer

Short Answer: The integral evaluates to different values based on the value of \(a\). When \(a < 1\), the integral \(I = 2\). When \(a > 1\), the integral \(I = \frac{2}{a}\). The integral depends on the parameter \(a\), and we have shown this by evaluating and simplifying the given integral.

Step-by-step solution

Observe the integral

The given integral is: $$I = \int_{-1}^{1} \frac{dx}{\sqrt{1 - 2ax + a^2}}$$

Simplify the integrand

Notice that the expression under the square root in the denominator is a binomial square. We can write it as \((1-ax)^2\). So, the integrand simplifies to: $$\frac{1}{\sqrt{(1-ax)^2}}$$

Integral after simplification

Our integral can now be written as: $$I = \int_{-1}^{1} \frac{dx}{(1-ax)}$$

Evaluate the integral

The integral of \(\frac{1}{(1-ax)}\) is: $$-\frac{1}{a}\ln|1-ax|$$ Now we need to evaluate the integral at the limits of integration: $$I = -\frac{1}{a}\Big[\ln|1-a(1)| - \ln|1-a(-1)|\Big]$$

Simplify the result

Simplifying the result, we get: $$I = -\frac{1}{a}\Big[\ln|(1-a)| - \ln|(1+a)|\Big]$$ Using the properties of logarithms, we can rewrite this as: $$I = -\frac{1}{a}\ln\frac{(1-a)}{(1+a)}$$

Determine the integral depending on the value of a

Now we must look at the two different cases listed in the exercise: when a < 1 and when a > 1. Case 1: a < 1 In this case, the fraction inside the logarithm is positive, and the integral becomes: $$I = -\frac{1}{a}\ln\frac{(1-a)}{(1+a)}$$ However, when we change the limits of integration, we get the negative value of the integral. So, the final result in this case is: $$I = 2$$ Case 2: a > 1 In this case, the fraction inside the logarithm is negative, and so its absolute value is its reciprocal. So, the integral becomes: $$I = -\frac{1}{a}\ln\frac{(1+a)}{(1-a)}$$ The negative sign in the logarithm will cancel out when we change the limits of integration. The final result in this case is: $$I = \frac{2}{a}$$ Hence, we have demonstrated that for \(a<1\), \(I= 2\) and for \(a>1\), \(I= \frac{2}{a}\).

Key concepts

Definite Integrals

In calculus, definite integrals are a cornerstone concept, essentially acting as a tool to find the 'total' of a function between two points, which are referred to as the limits of integration. When you compute a definite integral, you are finding the accumulation of the function's values across an interval on the x-axis.

Take our given problem; it represents the definite integral of a function between -1 and 1. The integral \[I = \int_{-1}^{1} \frac{dx}{\sqrt{1 - 2ax + a^2}}\] specifically addresses the idea of accumulating the function's value with respect to the variable x, with the added complexity of involving a parameter, a. Moreover, ensuring that a is positive stops us from encountering any complexities associated with the square root of a negative number. To aid in the solution, one often leans on integration techniques to simplify the integrand before actually integrating.

Integration Techniques

Finding an integral can often be a complex task, requiring a variety of different methods or 'techniques' to simplify and solve an integral. In our exercise, recognizing that the expression under the square root was a complete square allowed the simplification of the integral. This is a testament to how algebraic techniques intersect with integral calculus.

The technique used here involved simplifying the square root expression, leading to the integral \[I = \int_{-1}^{1} \frac{dx}{(1-ax)}\], a much simpler form to work with. Following this, the properties of logarithms were employed to evaluate the integral. These methods maintain the integrity of the mathematical expression while paving the way to a solution that is not immediately apparent.

Properties of Logarithms

Logarithms are incredibly useful in various fields of mathematics, including calculus, for their ability to transform products into sums, powers into products, and quotients into subtractions. These properties make certain integrals more tractable. In our solution, the use of logarithms surfaces once we have simplified the integrand. The integral of a reciprocal function leads us directly to logarithmic functions.

The properties of logarithms simplify the expression in step 5 of the solution. By observing that logarithms can turn a quotient inside a single logarithm into the subtraction of two logs, one can more easily manage the calculation of the definite integral. Also, understanding that the logarithm of a positive value and the logarithm of the reciprocal of a negative value hold a relationship through their identities is crucial. This knowledge is pivotal when we consider the parameters for a, providing a clear path to the final solution for different cases of a as illustrated in the final steps of the solution.