Problem 74 - Calculus Early Transcendentals

Question

The following integrals require a preliminary step such as long division or a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{d y}{y(\sqrt{a}-\sqrt{y})}, \text { for } a>0 .(\text {Hint: Let } u=\sqrt{y} .)$$

Step-by-Step Solution

Answer

In order to evaluate the integral \(\int \frac{dy}{y(\sqrt{a}-\sqrt{y})}\), we applied the change of variables and let \(u = \sqrt{y}\). We then applied the method of partial fractions to rewrite the integrand, which allowed us to simplify the expression and perform the integration. The final result is: \(\int \frac{dy}{y(\sqrt{a}-\sqrt{y})} = \ln \left|\frac{\sqrt{y}}{\sqrt{a}-\sqrt{y}}\right| + C\)

Step 1: Change of Variables

Let \(u = \sqrt{y}\), so \(y = u^2\). Also, we need to find \(dy\) in terms of \(u\). Differentiating both sides of \(y = u^2\) with respect to \(y\), we get: $$\frac{d y}{d u}=2 u$$ Now, multiply both sides by \(du\) to get: $$d y=2 u d u$$

Step 2: Substitute the Change of Variables

Substitute \(u = \sqrt{y}\) and \(dy = 2ud u\) into the given integral, we get: $$\int \frac{d y}{y(\sqrt{a}-\sqrt{y})} = \int \frac{2 u d u}{u^2(\sqrt{a}-u)}$$ The \(u\) in the numerator and denominator will cancel out, so the integral simplifies to: $$\int \frac{2 d u}{u(\sqrt{a}-u)}$$

Step 3: Partial Fractions

Rewrite the integrand using partial fractions so that it can be integrated more easily: $$\frac{2}{u(\sqrt{a}-u)} = \frac{A}{u} + \frac{B}{\sqrt{a}-u}$$ Clear the denominator to solve for \(A\) and \(B\): $$2 = A (\sqrt{a}-u) + B u$$ Set \(u = 0\) to find \(A\): $$2 = A \sqrt{a} \Rightarrow A = \frac{2}{\sqrt{a}}$$ Set \(u = \sqrt{a}\) to find \(B\): $$2 = B \sqrt{a} \Rightarrow B = \frac{2}{\sqrt{a}}$$ Now we can rewrite the integrand: $$\int \frac{2 d u}{u(\sqrt{a}-u)} = \int \left(\frac{1}{u} + \frac{1}{\sqrt{a}-u}\right) d u$$

Step 4: Integration

Now, integrate each term separately: $$\int \left(\frac{1}{u} + \frac{1}{\sqrt{a}-u}\right) d u = \int \frac{1}{u} d u + \int \frac{1}{\sqrt{a}-u} d u$$ The first integral is the natural log of the absolute value of \(u\): $$\int \frac{1}{u} d u = \ln |u| + C_1$$ For the second integral, apply substitution with \(v = (\sqrt{a}-u)\) and \(d v=-d u\): $$\int \frac{1}{\sqrt{a}-u} d u = -\int \frac{1}{v} d v = -\ln |v| + C_2$$ Combine the two integrals: $$\int \left(\frac{1}{u} + \frac{1}{\sqrt{a}-u}\right) d u = \ln |u| - \ln |v| + C$$

Step 5: Substitute Back

Replace \(u\) and \(v\) with the original variables, so \(u=\sqrt{y}\) and \(v=(\sqrt{a}-u)\): $$\int \left(\frac{1}{u} + \frac{1}{\sqrt{a}-u}\right) d u = \ln |\sqrt{y}| - \ln |\sqrt{a}-\sqrt{y}| + C$$ We dropped \(C_1\) and \(C_2\) and wrote them as one single constant \(C\). This is the final result: $$\int \frac{d y}{y(\sqrt{a}-\sqrt{y})} = \ln \left|\frac{\sqrt{y}}{\sqrt{a}-\sqrt{y}}\right| + C$$

Key Concepts

Integration Techniques

When it comes to calculus, integrating complex expressions often requires a series of specialized techniques. Integration techniques are methods used to simplify integrals so they can be resolved, either into a basic form that we can integrate directly or into a form for which we already know the antiderivative. Techniques such as substitution (change of variables), integration by parts, trigonometric substitution, and partial fractions are part of this powerful toolkit.

The primary goal is to transform the original integral into an equivalent one that is easier to evaluate. Each technique has its own scenarios where it is most useful. For instance, a change of variables is particularly helpful when the integrand includes some function and its derivative. On the other hand, partial fraction decomposition is most useful when dealing with rational functions.

Change of Variables

The change of variables, also known as substitution, is a strategy where we replace a difficult-to-integrate expression with a new variable that simplifies the integral. This process involves two key steps: identifying a new variable to substitute and finding the differential of the new variable with respect to the original variable.

After substitution, we can convert the integral into a form with respect to the new variable, which is hopefully more straightforward to integrate. Changing variables can sometimes turn a challenging integral into a basic one or allow the use of other integration techniques like partial fractions.

Partial Fraction Decomposition

In calculus, partial fraction decomposition is a method used to break down rational functions into simpler fractions that can be integrated more easily. This technique is extremely useful when the integrand is a fraction with a polynomial in the numerator and a factorizable polynomial in the denominator.

Finding the Constants

Partial fraction decomposition relies on expressing the integrand as a sum of fractions with unknown constants, and then determining these constants by exploiting values that simplify the equation. Once the constants are found, the integral becomes a sum of simpler integrals, generally involving natural logarithms or arctangent functions when integrating with respect to these basic functions.

Definite Integrals

The concept of a definite integral is a central part of calculus, representing the accumulation of quantities and often interpreted as the area under a curve over an interval. In contrast to indefinite integrals which provide a family of functions (antiderivatives) plus a constant of integration, a definite integral has limits of integration indicating the bounds of the region we are interested in, and it results in a specific numerical value.

In practical terms, after applying integration techniques to find the antiderivative, we can evaluate a definite integral by calculating the difference between the values of the antiderivative at the upper and lower limits of integration.