Problem 17 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \sqrt{3-2 s} d s$$

Step-by-Step Solution

Answer

\(-\frac{1}{3} (3 - 2s)^{3/2} + C\)

Step 1: Identify the Integral Type

The given integral is \( \int \sqrt{3-2s} \, ds \). This is an indefinite integral of a square root function, which suggests a potential substitution to simplify the expression inside the square root.

Step 2: Choose a Suitable Substitution

Let us choose a substitution to simplify the integral. Set \( u = 3 - 2s \). Then, differentiate to find \( du \):\[ du = -2 \, ds \then \, ds = -\frac{1}{2} du \].

Step 3: Substitute and Transform the Integral

Replace \( 3-2s \) with \( u \) in the integral and substitute \( ds \) with \( -\frac{1}{2} \, du \). This gives:\[ \int \sqrt{u} \left(-\frac{1}{2}\right) \, du = -\frac{1}{2} \int u^{1/2} \, du \].

Step 4: Integrate with Respect to 'u'

Integrate \( u^{1/2} \) using the power rule for integrals:\[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \].Therefore:\[ -\frac{1}{2} \int u^{1/2} \, du = -\frac{1}{2} \times \frac{2}{3} u^{3/2} = -\frac{1}{3} u^{3/2} + C \],where \( C \) is the constant of integration.

Step 5: Substitute Back to 's' Terms

Replace \( u \) back with \( 3-2s \) to express the result in terms of the original variable \( s \):\[ -\frac{1}{3} (3-2s)^{3/2} + C \].

Step 6: Write the Final Answer

The evaluated integral is:\[-\frac{1}{3} (3 - 2s)^{3/2} + C \].

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus for evaluating integrals, especially when the original integral is difficult to solve directly. In this method, a new variable is introduced to simplify the expression and make the integration process more manageable.

• The goal is to transform the integral into a simpler form by substituting a function or expression with a new variable, say, \( u \).
• Once the substitution is made, the integrand and the differential are expressed in terms of this new variable.
• It's crucial to differentiate the substitution expression to find an expression for \( du \), which allows you to replace the original differential \( ds \) in the integral with \( du \).

A clear example is substituting \( u = 3 - 2s \) for the integral \( \int \sqrt{3-2s} \, ds \). Here, differentiating yields \( du = -2 \, ds \), and thus \( ds = -\frac{1}{2} \, du \). The substitution converts the integral into terms of \( u \), making it far easier to integrate.

Power Rule

The power rule for integration is a direct extension of the power rule for differentiation, and is quite simple, yet very powerful.

• It states that the integral of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).
• This rule simplifies the integration of polynomial expressions and can easily be applied to any term raised to a power.

In our example, after substituting the original integral, we end up with \( \int u^{1/2} \, du \). Applying the power rule, the integral becomes \( \frac{u^{3/2}}{3/2} = \frac{2}{3}u^{3/2} \). The power rule allows for straightforward integration of terms with fractional exponents, making it an essential technique in solving indefinite integrals.

Square Root Function

The square root function often appears in integral calculus, particularly in expressions that need simplification. Integrating expressions involving square roots can be challenging, but with some techniques, they become manageable.

• When you see \( \sqrt{a - bx} \), your first thought should be whether substitution can simplify it. This function can often be rewritten using exponential notation, changed to \( (a-bx)^{1/2} \).
• This transformation allows us to use the power rule for integrating, since it's now in a polynomial form.

In this particular problem, the original integral \( \int \sqrt{3-2s} \, ds \) was converted to \( \int u^{1/2} \, du \), where \( u = 3 - 2s \). This conversion is crucial since it transforms the square root into a more standard form suitable for integration techniques like the power rule. By changing the function inside the integral to an exponent, it becomes much simpler to handle.