Problem 43 - Calculus: Early Transcendentals

Question

In Example 5.1.2 we showed that \( \displaystyle \int^1_0 x^2 \, dx = \frac{1}{3} \). Use this fact and the properties of integrals to evaluate \( \displaystyle \int^1_0 (5 - 6x^2) \, dx \).

Step-by-Step Solution

Answer

The integral \( \int_0^1 (5 - 6x^2) \, dx = 3 \).

Step 1: Understand the Integral Components

The integral we need to evaluate is \( \int_0^1 (5 - 6x^2) \, dx \). This can be split into separate integrals: \( \int_0^1 5 \, dx \) and \( - \int_0^1 6x^2 \, dx \).

Step 2: Evaluate the Constant Integral

The integral \( \int_0^1 5 \, dx \) is straightforward. With \( c = 5 \), the integral is \( 5 \int_0^1 1 \, dx = 5 \cdot (1 - 0) = 5 \).

Step 3: Use Given Integral Result

We already know from Example 5.1.2 that \( \int_0^1 x^2 \, dx = \frac{1}{3} \). We can use this to evaluate \( - \int_0^1 6x^2 \, dx \) as \( -6 \cdot \frac{1}{3} = -2 \).

Step 4: Combine the Results

Now combine the results of Step 2 and Step 3: \( 5 + (-2) = 3 \). This is the value of the integral \( \int_0^1 (5 - 6x^2) \, dx \).

Key Concepts

Definite Integrals

The concept of definite integrals is central in calculus, representing the accumulation of quantities. A definite integral is written in the form \( \int_a^b f(x) \, dx \), where \(a\) and \(b\) are the limits of integration. These limits help us define the interval over which we want to calculate the integral.
Unlike indefinite integrals, which include a constant of integration, definite integrals provide a specific numerical value. This value is essentially the total area between the curve \( f(x) \) and the x-axis, from \(x = a\) to \(x = b\). The process of calculating this is called integration. It gives a precise interpretation of how much a function accumulates between two points.
Understanding definite integrals involves recognizing their critical role in measuring everything from physical quantities to probabilities, where each application often benefits from seeing the integral as an area, volume, or other cumulative measurement.

Integration Properties

Understanding the properties of integrals can significantly simplify the computation. One key property is that the integral of a sum is the sum of the integrals. This is mathematically expressed as: \[ \int_a^b (f(x) + g(x)) \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx \]
In our exercise, we used this property to break down \( \int_0^1 (5 - 6x^2) \, dx \) into more manageable parts: \( \int_0^1 5 \, dx \) and \( -\int_0^1 6x^2 \, dx \).
Another useful property is the constant multiple rule, which states \( \int_a^b c \cdot f(x) \, dx = c \cdot \int_a^b f(x) \, dx \). This rule simplifies calculations when constants appear within the integrand.
These properties anchor our understanding that integrals, despite potentially complex functions, obey intuitive arithmetic-like rules that assist in evaluation.

Integral Evaluation

The process of integral evaluation involves computing the actual value from a given definite integral. Evaluation typically requires applying integration techniques, including recognizing specific functions whose antiderivatives are known.
In our example, once split, each integral is evaluated separately. For constants like \( \int_0^1 5 \, dx \), straightforward evaluation gives a multiplication of the constant by the length of the interval, producing \(5\). For \( \int_0^1 6x^2 \, dx \), it uses the previously calculated integral \( \int_0^1 x^2 \, dx = \frac{1}{3} \) to facilitate computation.
The final part, combining results, involves simple addition or subtraction of the evaluated parts. Here, \(5 - 2 = 3\) gives the total integral value, illustrating how breaking down the problem into parts leads to a straightforward solution.

Area Under a Curve

The integral’s value often gives us the area under a curve, specifically for functions above the x-axis within the limits of integration. For instance, the integral \( \int_0^1 x^2 \, dx \) calculates the area under \(x^2\) from \(x=0\) to \(x=1\). This area, \(\frac{1}{3}\), derives from the antiderivative of the function at the specified limits.
In some cases, like \( \int_0^1 (5 - 6x^2) \, dx \), the integral’s value results from balancing areas above and below the x-axis, which can cancel each other out partially. Negative areas (where the function lies below the x-axis) are subtracted from positive areas.
Understanding this concept uncovers why we are interested in these calculations: they encapsulate the function’s total effect over a segment of the x-axis, which can represent anything from land area to probability, making it incredibly versatile and useful in practical applications.