Problem 25 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 21 through 30 , evaluate the indicated definite integral. \(\int_{-1}^{2} 30(5 x-2)^{2} d x\)

Step-by-Step Solution

Answer

1710

Step 1 - Identify the substitution

Let’s simplify the integral using substitution. Set \(u = 5x - 2\), so that we can change the variable of integration.

Step 2 - Find the derivative

Calculate the derivative of \(u = 5x - 2\) with respect to \(x\). Thus, \(du/dx = 5\), or \(du = 5dx\).

Step 3 - Substitute into the integral

Rewrite \(dx\) in terms of \(du\): \(dx = du/5\). Substitute \(u\) and \(dx\) into the integral: \(\int_{-1}^{2} 30(5x-2)^{2} dx = \int_{-1}^{2} 30u^{2} (du/5)\).

Step 4 - Change limits of integration

Transform the limits of integration from \(x\) to \(u\): when \(x = -1\), \(u = 5(-1) - 2 = -7\), and when \(x = 2\), \(u = 5(2) - 2 = 8\). So the integral now is: \(\int_{-7}^{8} 30u^{2} (du/5)\).

Step 5 - Simplify the integral

Factor out constants to simplify: \(30/5 = 6\). Combing the constants, we get: \(6 \int_{-7}^{8} u^{2} du\).

Step 6 - Integrate

Integrate \(u^2\) to get \(\frac{u^3}{3}\): \(6 \int_{-7}^{8} u^{2} du = 6[\frac{u^3}{3}]_{-7}^{8}\).

Step 7 - Apply limits

Evaluate the definite integral by applying the upper and lower limits: \(6\left( \frac{8^3}{3} - \frac{(-7)^3}{3} \right) = 6 \left( \frac{512}{3} - \frac{-343}{3} \right) = 6 \left( \frac{512 + 343}{3} \right) = 6 \left( \frac{855}{3} \right) = 6 \times 285 = 1710\).

Key Concepts

Substitution Method

The substitution method is a technique used in integral calculus to simplify complex integrals. By substituting a part of the integral with a new variable, called 'u', we can transform the integral into a simpler form. In this exercise, we set \( u = 5x - 2 \). This helps us to change the variable of integration from \( x \) to \( u \). The goal is to make the integral easier to solve.

• Choose a substitution: Identify a function within the integral that you can replace with a simpler variable.
• Find the derivative: Calculate the derivative of the chosen function with respect to the original variable.
• Rewrite the integral: Substitute the simpler variable and its derivative into the integral.

This method is effective for integrals where a particular part of the function's structure stands out as suitable for substitution.

Integration Limits

When performing definite integrals with substitution, it's essential to change the integration limits according to the new variable. The limits \([-1, 2]\) for \( x \) were transformed to \([-7, 8]\) for \( u \). To achieve this, substitute the original limits into the substitution equation:

• Convert the lower limit: Substitute \( x = -1 \) into \( u = 5x - 2 \). This gives \( u = -7 \).
• Convert the upper limit: Substitute \( x = 2 \) into \( u = 5x - 2 \). This gives \( u = 8 \).

By adjusting the limits, we ensure the integral is evaluated correctly under the new variable.

Integral Calculus

Integral calculus is a branch of mathematics that deals with integration and its applications. Integration is the process of finding the integral of a function, which can be thought of as the reverse operation of differentiation. There are two main types of integrals: definite and indefinite.

• Definite integrals: Represent the area under a curve between two points and include limits of integration.
• Indefinite integrals: Represent the general form of the antiderivative and do not include limits of integration.

Definite integrals calculate a specific numerical value, like in our exercise where we found the integral of \(30(5x-2)^{2}\) from \(x = -1\) to \(x = 2\).

Derivative

Derivatives are central in calculus and represent the rate of change of a function with respect to a variable. In the substitution method, we need to find the derivative to replace the variable of integration correctly. In our exercise, we computed the derivative of \( u = 5x - 2 \) as \( du/dx = 5 \). This helped us express \( dx \) in terms of \( du \):

• Calculate the derivative: For \( u = 5x - 2 \), we found \( du/dx = 5 \).
• Rearrange the derivative: Express \( dx \) as \( dx = du/5 \).

Using derivatives in this way allows us to simplify and solve the integral more easily.