Problem 9 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{1}^{3} 6 x^{2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 52.

Step 1: Identify the function to integrate

Our function is \( f(x) = 6x^2 \). We need to find its antiderivative, which is the first step in applying the Fundamental Theorem of Calculus.

Step 2: Find the antiderivative

The antiderivative of \( f(x) = 6x^2 \) is found by increasing the power of \( x \) by 1 and dividing by the new power. Therefore, the antiderivative \( F(x) \) is \( F(x) = 2x^3 \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, to find \( \int_{a}^{b} f(x) \, dx \), we evaluate the antiderivative at the upper limit \( b \) and at the lower limit \( a \), and find the difference: \( F(b) - F(a) \).

Step 4: Evaluate at the bounds

Our integral bounds are from \( x = 1 \) to \( x = 3 \). So we calculate \( F(3) = 2(3)^3 = 54 \) and \( F(1) = 2(1)^3 = 2 \).

Step 5: Find the value of the definite integral

Subtract the value of the antiderivative at the lower bound from the value at the upper bound: \( F(3) - F(1) = 54 - 2 = 52 \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in mathematical analysis. It bridges the concept of differentiation with integration, two crucial aspects of calculus. There are two parts to this theorem that you need to know:

• The first part: It guarantees that every continuous function has an antiderivative. This antiderivative can be used to evaluate definite integrals.
• The second part: It states that if you have an antiderivative of a function, you can evaluate the integral by finding the difference between the values of this antiderivative at two points.

In simpler terms, this theorem tells us that to evaluate a definite integral like \( \int_{a}^{b} f(x) \, dx \), find the antiderivative (let's call it \( F(x) \)), then compute \( F(b) - F(a) \). This process effectively gives us the net area under the curve \( f(x) \) from \( x = a \) to \( x = b \). Remembering this connection between differentiation and integration is incredibly helpful when tackling integrals in calculus.

Antiderivative

An antiderivative is essentially the "reverse" of a derivative. If you have a function \( f(x) \), its antiderivative \( F(x) \) is a function such that the derivative of \( F(x) \) is \( f(x) \). For example, if \( f(x) = 6x^2 \), the antiderivative could be \( F(x) = 2x^3 + C \), where \( C \) is the constant of integration, elucidating the fact that there are infinitely many such antiderivatives differing by a constant.
However, when calculating definite integrals, the constant \( C \) goes away when you take the difference \( F(b) - F(a) \) because \( C - C = 0 \). Finding the antiderivative involves recognizing which function, when differentiated, leads you back to \( f(x) \). In practice, for power functions like \( f(x) = x^n \), raise the power by one and divide by the new power, making the antiderivative \( \frac{x^{n+1}}{n+1} \). This simple technique is fundamental in solving many calculus problems.

Integration Bounds

Integration bounds are the limits between which you evaluate your definite integral. These are noted at the top and bottom of the integral sign—for example, the bounds in \( \int_{1}^{3} 6x^2 \, dx \) are 1 and 3, respectively.

• Lower bound \( a \): The place where integration starts.
• Upper bound \( b \): The place where integration ends.

These bounds are critical because they define the section of the function that you're interested in, which in turn determines the "net" area that you are calculating. By evaluating the antiderivative at these bounds and subtracting \( F(a) \) from \( F(b) \), you find the cumulative value of the function over that interval. In this way, the bounds essentially "box in" the part of the function that the definite integral is measuring.