Problem 4 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{1}^{3} x \sqrt{3 x^{2}-2} d x$$

Step-by-Step Solution

Answer

The definite integral \(\int_{1}^{3} x \sqrt{3 x^{2}-2} d x\) evaluates to \(\boxed{12}\) by performing a substitution, finding the antiderivative and applying the Fundamental Theorem of Calculus.

Step 1: Perform substitution

Let's substitute $$u = 3x^2 - 2$$ The derivative of \(u\) with respect to \(x\) is: $$\frac{d u}{d x}=6 x$$ Now, we must express everything in terms of \(u\). To do that, we need to solve for \(x\) in terms of \(u\): $$x = \sqrt{\frac{u + 2}{3}}$$ Further, we need to find the limits of integration for \(u\). Substitute the original limits of integration for \(x\) in the expression for \(u\): - For the lower limit, where \(x = 1\): $$u = 3(1)^2 - 2 = 1$$ - For the upper limit, where \(x = 3\): $$u = 3(3)^2 - 2 = 25$$ Now, substitution gives $$\int_{1}^{3} x \sqrt{3 x^{2}-2} d x = \int_{1}^{25} \sqrt{u} \frac{d u}{6 \sqrt{\frac{u + 2}{3}}}$$

Step 2: Simplify the integrand and find the antiderivative

$$\begin{aligned}\int_{1}^{25} \sqrt{u} \frac{d u}{6 \sqrt{\frac{u + 2}{3}}} &= \int_{1}^{25} \frac{\sqrt{u}}{6 \sqrt{\frac{u}{3}}} \cdot \frac{\sqrt{u}}{\sqrt{u}} d u \\\\ &= \int_{1}^{25} \frac{\sqrt{u}}{2\sqrt{u}} d u\end{aligned}$$ We can simplify the integrand further: $$\int_{1}^{25} \frac{1}{2} du$$ Now we find the antiderivative: $$\frac{1}{2} \int_{1}^{25} d u = \frac{1}{2}(u)\Bigg|_1^{25}$$

Step 3: Apply the Fundamental Theorem of Calculus

Now, we need to substitute the limits of integration for \(u\) and subtract the resulting terms: $$\frac{1}{2}(u)\Bigg|_1^{25} = \frac{1}{2}(25)- \frac{1}{2}(1) = \boxed{12}$$ So, the value of the definite integral is 12.

Key Concepts

Substitution Method

The substitution method is a technique used in calculus to simplify the process of finding an integral. This method involves changing variables to make the integration straightforward. In this exercise, we substitute a new variable, denoted as \( u \), to replace a more complex expression within the integral.

To effectively use the substitution method:

• Identify a portion of the integrand that can be substituted with a single variable \( u \). Here, we choose \( u = 3x^2 - 2 \).
• Determine the differential \( du \), which involves differentiating \( u \) with respect to \( x \). In our case, \( \frac{du}{dx} = 6x \).
• Express \( dx \) in terms of \( du \). This enables the rewriting of the entire integral in terms of \( u \), removing the \( x \) variable.

This method simplifies our original complex integrand \( x \sqrt{3x^2 - 2} \) into a simpler form, making it easier to find the antiderivative in the next step.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, providing a powerful tool to evaluate definite integrals. This theorem has two main components:

• The first part states that if a function is continuous over an interval, then it has an antiderivative on that interval.
• The second part allows us to evaluate this antiderivative by taking the difference of its values at the upper and lower bounds of the integral.

In this exercise, once we have found the antiderivative, the Fundamental Theorem of Calculus helps us to evaluate the definite integral efficiently. It tells us that:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]where \( F(x) \) is the antiderivative of \( f(x) \).

Applying this, after we resolved the integral in terms of \( u \), we calculate: \[\frac{1}{2}(u)\Bigg|_1^{25} = \frac{1}{2}(25)- \frac{1}{2}(1)\]Thus, simplifying to find the final result of 12.

Antiderivative

An antiderivative is the reverse process of differentiation; it usually refers to finding a function whose derivative is a given function. The concept of the antiderivative is central to solving integrals. Specifically, when you compute an indefinite integral, you are finding its antiderivative.

In our worked example, once we applied the substitution to simplify the integrand, we identified that the function \( \int \frac{1}{2} \quad du \) needed an antiderivative. The antiderivative of \( \frac{1}{2} \) with respect to \( u \) is simply \( \frac{1}{2}u \).

Finding antiderivatives regularly involves recognizing standard integral forms or applying techniques like substitution to manipulate the expression into a standard form. The simplification to \( \int \frac{1}{2} \, du \) made it trivial to find the antiderivative, highlighting the importance of substitution in complex integral problems.