Problem 1 - Calculus: One and Several Variables

Question

Calculate. $$\int \frac{d x}{1-\sqrt{x}}$$

Step-by-Step Solution

Answer

The short answer for the given integral is: \(\int \frac{d x}{1-\sqrt{x}} = -2 \sqrt{x} + 2 \ln |\sqrt{x} - 1| + C\)

Step 1: Substitution

To simplify the integrand and make the integration easier, we need to make a substitution. Let's take: \(u = \sqrt{x}\) Now, Differentiating both sides with respect to x, we get: \(d u = \frac{1}{2\sqrt{x}} d x\) Rearranging to find dx: \(d x = 2\sqrt{x} d u = 2u d u\) Now we substitute x and dx in the given integral: $$\int \frac{d x}{1-\sqrt{x}} = \int \frac{2 u d u}{1-u}$$

Step 2: Integration

Now we can integrate the new function with respect to u: $$\int \frac{2 u d u}{1-u} = 2\int \frac{u}{1-u}d u$$ To integrate the above expression, we can apply the natural logarithm property: $$2\int \frac{u}{1-u}d u = -2\int \frac{1-u+1}{1-u}d u = -2\int (1-\frac{1}{1-u}) d u $$ Now, integrate each term separately: $$-2\int (1-\frac{1}{1-u}) d u = -2\int d u + 2 \int \frac{1}{1-u}d u $$ Integrating each term, we get: $$-2 u + 2(- \ln |1-u|) + C = -2 u + 2 \ln |u - 1| + C$$

Step 3: Undo the substitution

Now, substitute back the original variable x using \(u = \sqrt{x}\): $$-2 \sqrt{x} + 2 \ln |\sqrt{x} - 1| + C$$ So, the final answer to the given integral is: $$\int \frac{d x}{1-\sqrt{x}} = -2 \sqrt{x} + 2 \ln |\sqrt{x} - 1| + C$$

Key Concepts

Definite integrals

Integrals are fundamental in mathematics, especially when determining the accumulation of quantities. A definite integral, specifically, involves limits and represents the net area under a curve, typically from point \(a\) to point \(b\) on the x-axis. It is written as \(\int_{a}^{b} f(x) \, dx\). This operation provides a precise value, which serves as the total accumulation of the function \(f(x)\) over the interval \([a,b]\).

Unlike indefinite integrals that have a constant \(C\), definite integrals result in a specific value and usually apply the Fundamental Theorem of Calculus. This theorem connects differentiation with integration, ensuring that if we know the derivative of a function, we can find its integral and vice versa.

• The first part of the theorem tells us that integration can be used to determine antiderivatives, the inverse process of differentiation.
• The second part asserts that if a function is continuous, the integral of its derivative is the function itself within the given limits \(a\) and \(b\).

For those learning calculus, grasping definite integrals is crucial for solving real-world problems involving rates and accumulations.

Logarithmic integration

In calculus, logarithmic integration is a method to solve integrals involving fractions. This typically occurs when the integrand is a rational function or involves expressions in the denominator, similar to \(\int \frac{u}{1-u} \, du\). Here, we apply the property of logarithms to separate terms.

This method relies on the identity \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), which helps rearrange complex fractions into forms suitable for integration. When integrating, expressions like \(\frac{1}{x}\) convert to natural logarithms because the derivative of \(\ln|x|\) is \(\frac{1}{x}\).

• The substitution method is often used before applying logarithmic integration. Here, the solution involved a substitution \(u = \sqrt{x}\) to simplify the integrand, making it feasible to integrate by parts.
• This approach is particularly useful because it transforms integrals into more manageable parts, streamlining the process of finding solutions.

Logarithmic integration is pivotal for integrals involving terms that don't simplify easily with basic rules.

Differentiation

Differentiation is the counterpart of integration, focusing on the instantaneous rate of change of a function. It involves calculating the derivative, which represents how a function changes as its input changes. In the context of the exercise, we differentiated \(u = \sqrt{x}\), finding \(\frac{du}{dx} = \frac{1}{2\sqrt{x}}\).

Differentiation rules, such as the power rule, product rule, and chain rule, help break down complex expressions for simpler calculation. These rules are useful for understanding how various functions behave and for solving real-life problems involving rates of change, such as velocity or growth rates.

• The power rule says that \(\frac{d}{dx}x^n = nx^{n-1}\) for any real number \(n\).
• The chain rule is particularly important when dealing with compositions like \(f(g(x))\) and states \(f'(g(x))g'(x)\). It was applied implicitly when we differentiated \(u = \sqrt{x}\).

Understanding differentiation is essential for calculus students, as it provides the foundation for more advanced concepts in mathematical analysis and helps solve various engineering and physics problems.