Problem 25 - Thomas Calculus

Question

Evaluate the integrals by using a substitution prior to integration by parts. \(\int e^{\sqrt{3 s+9}} d s\)

Step-by-Step Solution

Answer

\(\frac{2}{3} e^{\sqrt{3s+9}} \left( \sqrt{3s+9} - 1 \right) + C\).

Step 1: Choose a substitution

Let's start by making a substitution to simplify the integral. We notice that the expression under the exponent, \(\sqrt{3s+9}\), is a bit complex. A logical substitution is \(u = \sqrt{3s+9}\). We need to express \(s\) and \(ds\) in terms of \(u\). Square both sides to get \(u^2 = 3s+9\).

Step 2: Express \(s\) and \(ds\)

From \(u^2 = 3s + 9\), solve for \(s\): \[s = \frac{u^2-9}{3}\] Now, differentiate with respect to \(u\): \[ds = \frac{2u}{3}\, du\].

Step 3: Substitute variables and simplify the integral

Substitute \(u\) into the integral: \[\int e^{\sqrt{3s+9}} ds = \int e^u \frac{2u}{3} du\]. Simplify the expression to become: \[\frac{2}{3} \int u e^u \, du\]. Now, we will use integration by parts to solve this integral.

Step 4: Integration by parts

Recall the integration by parts formula: \(\int v \, du = uv - \int u \, dv\). Let \(v = u\) and hence \(dv = du\). Let \(du = e^u du\) and so \(u = e^u\). Applying integration by parts: \(uv - \int u \,dv = u e^u - \int e^u du = u e^u - e^u + C\).

Step 5: Simplify the result

After applying integration by parts to the integral \(\frac{2}{3} \int u e^u \, du\), we get: \(\frac{2}{3} [u e^u - e^u] + C\). Simplify to find: \(\frac{2}{3} (u e^u - e^u) + C\).

Step 6: Back-substitution to the original variable

Replace \(u\) with \(\sqrt{3s+9}\) to express the result in terms of \(s\): \[= \frac{2}{3} \left( \sqrt{3s+9} e^{\sqrt{3s+9}} - e^{\sqrt{3s+9}} \right) + C\]. Simplifying further gives: \[= \frac{2}{3} e^{\sqrt{3s+9}} \left( \sqrt{3s+9} - 1 \right) + C\].

Key Concepts

Definite integrals

Definite integrals are crucial in calculus, offering a way to calculate the area under a curve between two points on a graph. They are represented as \[ \int_{a}^{b} f(x) \, dx \] where \(a\) and \(b\) are the limits of integration. This expression produces a numerical value rather than a function, showing how much area lies beneath the curve of the function \(f(x)\) from \(x = a\) to \(x = b\).
Definite integrals are widely utilized in physics, engineering, and statistics to solve real-world problems. Here, our focus is on indefinite integrals, since we are evaluating an integral without specified limits.

Integration techniques

Integration techniques are methods and strategies used to solve integrals, especially when simple antiderivatives are challenging to find. Two essential techniques in this context are:

• Substitution Method: A technique that simplifies integrals by changing variables, making them easier to evaluate. It's especially effective when dealing with nested functions or complexities within the integrand.
• Integration by Parts: Based on the product rule for differentiation, this technique is ideal for integrals involving the product of functions, such as polynomials and exponentials.

In the exercise we tackled, both substitution and integration by parts were needed to evaluate the integral. By combining these techniques, we managed to simplify the expression and successfully integrate it. Understanding these methods enhances the skill set for tackling a wide array of integral problems.

Substitution method

The substitution method is a powerful technique for simplifying integrals. It involves substituting part of the integrand with a single variable, known as "u-substitution." By changing variables, the complex part of the integral can become significantly easier to solve.
In this exercise, we identified \(u = \sqrt{3s+9}\) as our substitution, acknowledging the complexity of the expression under the square root. After determining \(u\), we express \(s\) and \(ds\) in terms of \(u\).

• We solve for \(s\) giving \(s = \frac{u^2-9}{3}\).
• By differentiating, we find \(ds = \frac{2u}{3} \, du\).

This substitution converts the integral \[ \int e^{\sqrt{3s+9}} ds \] into a simpler form, \[ \int e^u \frac{2u}{3} \, du \]. Applying the substitution method sets a foundation for using other techniques such as integration by parts.

Integration by parts

Integration by parts is a method derived from the product rule of differentiation. It is particularly effective for integrating products of functions, such as a polynomial multiplied by an exponential. This technique is governed by the formula: \[ \int u \, dv = uv - \int v \, du \].
In our exercise, once the substitution was made, we applied integration by parts to \[ \int u e^u \, du \].We identified the parts as:

• \(u = u\), hence \(du = du\)
• \(dv = e^u du\), leading to \(v = e^u\)

Using the integration by parts formula, \[ u e^u - \int e^u \, du \], the solution proceeds to \[ u e^u - e^u + C \].Returning to the original variable \(s\), we replace \(u\) with \(\sqrt{3s+9}\). This method, in conjunction with substitution, elegantly breaks down the integral into manageable parts, allowing for efficient problem-solving.