Problem 15 - College Algebra and Calculus: An Applied Approach

Question

$$ \int_{1 / 2}^{\infty} \frac{1}{\sqrt{2 x-1}} d x $$

Step-by-Step Solution

Answer

The integral from \(1 / 2\) to \(\infty\) of \(\frac{1}{\sqrt{2x-1}} dx\) is \(\infty\).

Step 1: Substitution

Let's start by setting \(u = 2x - 1\). Before moving forward with this integral, we need to convert every instance of \(x\) in the original integral to \(u\). This involves not only replacing \(x\), but also \(dx\). This can be done by differentiating \(u\) with respect to \(x\), to find \(dx\). Here, \(du = 2 dx\) or \(dx = du / 2\). This way, \(\frac{1}{\sqrt{2x-1}} dx\) becomes \(\frac{1}{2\sqrt{u}} du\).

Step 2: Calculate the New Integral

The integral is now in a form that we can deal with using basic rules of integration. So, integrating \(\frac{1}{2\sqrt{u}}\), we have \(\frac{1}{2} \int \frac{1}{\sqrt{u}} du\). The integral of \(u^{-1 / 2}\) can be done using the power rule, \(\int u^n du = \frac{1}{n + 1} u^{n + 1} + C\), where \(n = -1/2\). The power increases by one, and the coefficient becomes \(\frac{1}{-1/2 + 1}\). This gives us \(\frac{1}{2} \cdot 2u^{1 / 2}\). Here, the constants cancel out, and we are left with \(u^{1 / 2}\) and the limits of the integration change to \(u(1/2) = 0\) and \(u(\infty) = \infty\). This gives us \(\int_{0}^{\infty} u^{1/2} du\).

Step 3: Backsubstitution

Now, we have to substitute \(u\) back in the place of \(x\). Thus, we have \(u^{1/2}\) is equal to \((2x-1)^{1/2}\) and taking the integral from 0 to infinity gives us the result.

Step 4: Final Evaluation

Last thing remaining is to evaluate at the limits. So, it's \(((2x-1)^{1/2})|_{0}^{\infty}\) which gives us \(\infty - 0 = \infty\).

Key Concepts

Substitution Method

The substitution method is a powerful tool for tackling complex integrals by simplifying them into a more manageable form. It involves replacing a troublesome part of the integral with a new variable, typically denoted by \( u \). In this example, the integral \( \int_{1/2}^{\infty} \frac{1}{\sqrt{2x-1}} \, dx \) can be simplified by letting \( u = 2x - 1 \).- This substitution transforms the integral by making it easier to handle: converting \( x \) and \( dx \) to \( u \) and \( du \).- To replace \( dx \), we find \( du = 2 \, dx \), simplifying to \( dx = \frac{du}{2} \).- Thus, the integral's expression becomes \( \frac{1}{2\sqrt{u}} \, du \), which is easier to integrate.By using the substitution method, complicated integrals can often be reduced to basic forms that can be readily solved, aiding in understanding and computation.

Definite Integrals

Definite integrals are used to find the accumulation of quantities, like area under curves, over a specified interval. From \( a \) to \( b \), the definite integral of a function \( f(x) \) is expressed as \( \int_a^b f(x) \, dx \).- In the original exercise, the problem defines a definite integral with new limits when substitution is applied.- Our original function, \( \frac{1}{\sqrt{2x-1}} \), initially ranges from \( x = \frac{1}{2} \) to \( x = \infty \).- After substitution, these become \( u = 0 \) to \( u = \infty \), transforming the integral to \( \int_0^{\infty} u^{1/2} \, du \).This shift in limits often makes calculation simpler and reveals whether an integral is convergent or divergent. The final result computes the complete area from start to end, encapsulating the function's behavior across given intervals.

Limits of Integration

The limits of integration establish the range over which integration occurs. In problems involving substitution, these limits must be adjusted to match the new variable. Here, the original integral \( \int_{1/2}^{\infty} \) changes as we substitute \( u = 2x - 1 \).- For the lower limit: when \( x = \frac{1}{2} \), \( u = 2 \times \frac{1}{2} - 1 = 0 \).- For the upper limit: as \( x \to \infty \), \( u = 2 \times \infty - 1 = \infty \).- These new limits \( u: 0 \to \infty \) redefine how we evaluate the integral.Through this adjustment in limits, the process respects the substitution and maintains accuracy in areas covered by the integral. Remember that replacing variables during integration requires a direct reevaluation of limits, ensuring alignment with the transformed function.