Problem 170 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{-1}^{2}\left(x^{2}-3 x\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understand the Problem

We need to evaluate the definite integral \( \int_{-1}^{2}(x^{2}-3x)\, dx \) using the Fundamental Theorem of Calculus, Part 2.

Step 2: Apply the Fundamental Theorem of Calculus, Part 2

According to the theorem, if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \). In this case, we need to find the antiderivative of \( x^2 - 3x \).

Step 3: Find the Antiderivative

The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \), and the antiderivative of \( -3x \) is \( -\frac{3x^2}{2} \). Thus, the antiderivative of \( x^2 - 3x \) is \( F(x) = \frac{x^3}{3} - \frac{3x^2}{2} + C \).

Step 4: Evaluate the Antiderivative at the Bounds

Calculate \( F(2) \) and \( F(-1) \). Substitute into the expression to find: \[ F(2) = \frac{2^3}{3} - \frac{3(2)^2}{2} = \frac{8}{3} - \frac{12}{2} = \frac{8}{3} - 6 = \frac{8}{3} - \frac{18}{3} = -\frac{10}{3} \] and \[ F(-1) = \frac{(-1)^3}{3} - \frac{3(-1)^2}{2} = \frac{-1}{3} - \frac{3}{2} = \frac{-1}{3} - \frac{9}{6} = \frac{-1}{3} - \frac{3}{2} = \frac{-1}{3} - \frac{9}{6} = \frac{-1}{3} - \frac{18}{6} = \frac{-1}{3} - 3 = \frac{-10}{3}. \]

Step 5: Subtract to Find the Integral

Using the Fundamental Theorem of Calculus, Part 2, we find the definite integral: \( \int_{-1}^{2}(x^2 - 3x)\, dx = F(2) - F(-1) = -\frac{10}{3} - \frac{-10}{3} = 0 \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical concept that connects differentiation with integration. This theorem is divided into two parts, and for definite integrals, we often use Part 2. This part states that if you have a continuous function over an interval, you can calculate its integral using an antiderivative. This theorem simplifies the process of finding the area under a curve by converting it into a question of evaluating a function.
In other words, Part 2 of the theorem says: If you already know one antiderivative of a function, you can find the exact integral by evaluating this antiderivative at the boundary values and subtracting. This process directly links the integral of a function to something much simpler involving basic algebra.

Antiderivative

An antiderivative is essentially the reverse of taking a derivative. When you find the antiderivative of a function, you are determining another function whose derivative is the original function.
For example, with the function in our exercise, \( x^2 - 3x \), the antiderivative is \( F(x) = \frac{x^3}{3} - \frac{3x^2}{2} + C \), where \( C \) represents the constant of integration. This constant is crucial when computing indefinite integrals, but in definite integrals, it cancels out during computation.
Understanding how to find antiderivatives is key in calculus because it allows us to solve integrals, predict motion in physics, calculate areas, and so forth. Learning to recognize patterns of basic functions and their antiderivatives speeds up this process significantly.

Calculus Problem Solving

Solving calculus problems often requires a blend of recognizing patterns, using known formulas, and applying understanding of concepts like derivatives and integrals. For our problem, the task is to evaluate a definite integral, which involves multiple steps.
- **Step 1:** Understand what you are being asked to do. Recognize that you need the definite integral, which calls for an antiderivative. - **Step 2:** Find the antiderivative of the function. - **Step 3:** Evaluate this antiderivative at the given upper and lower limits. - **Step 4:** Subtract the evaluations to find the solution. By breaking down complex problems into smaller steps and following logical processes, solving calculus problems becomes more manageable and systematic.

Evaluation of Integrals

The evaluation of integrals is about finding actual numerical values that represent the accumulated quantity of interest, such as area. With definite integrals, this means calculating the difference between the values of an antiderivative at two points.
For example, evaluate \( F(2) \) and \( F(-1) \) for our antiderivative \( F(x) = \frac{x^3}{3} - \frac{3x^2}{2} + C \). You then subtract these two results:

• Calculate \( F(2) = \frac{8}{3} - 6 = -\frac{10}{3} \)
• Calculate \( F(-1) = \frac{-1}{3} - 3 = -\frac{10}{3} \)
• Subtract: \(-\frac{10}{3} - ( -\frac{10}{3}) = 0 \)

Therefore, the definite integral evaluates to zero, depicting the net area between the curve and the x-axis over the interval is zero. This simple arithmetic step is how the Fundamental Theorem of Calculus applies practically to solve problems.