Question

Let \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}^{2}}\) and \(\mathrm{F}(\mathrm{x})=-\frac{1}{\mathrm{x}} .\) Find \(\mathrm{F}(1)-\mathrm{F}(-1)\). Does \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{F}(1)-\mathrm{F}(-1) ?\) Explain

Short answer

The difference between F(1) and F(-1) is -2, and it is equal to the integral of f(x) from -1 to 1, which is also -2.

Step-by-step solution

Evaluate F(x) at x = 1 and x = -1

To find F(1) and F(-1) using the given expression, replace x with 1 and -1: F(1) = -\frac{1}{1} = -1 F(-1) = -\frac{1}{-1} = 1

Calculate F(1) - F(-1)

Now, compute the difference between F(1) and F(-1): F(1) - F(-1) = -1 - 1 = -2

Evaluate the integral of f(x) from -1 to 1

We are given f(x) = \frac{1}{x^2}, and we need to evaluate the integral of f(x) from -1 to 1: \int_{-1}^{1} \frac{1}{x^2} dx Since we are already given the antiderivative F(x) = -\frac{1}{x}, we can use the Fundamental Theorem of Calculus to evaluate this integral: F(1) - F(-1) = -(-\frac{1}{1} + \frac{1}{-1}) = -2

Compare the difference and the integral

Now that we have calculated both F(1) - F(-1) and the integral of f(x) from -1 to 1, we can compare their values: F(1) - F(-1) = -2 \int_{-1}^{1} \frac{1}{x^2} dx = -2 Since both values are equal, the answer to the question is: Yes, \int_{-1}^{1} f(x) dx = F(1) - F(-1).

Key concepts

Understanding the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, serving as one of the most important principles in calculus. In essence, this theorem tells us that integration and differentiation are inverse processes. It breaks down into two parts. The first part states that if you take the integral of a function and then differentiate the result, you'll get back to the original function. Meanwhile, the second part allows us to evaluate definite integrals when the antiderivative of the function is known.

The first part is crucial when we are dealing with indefinite integrals as it ensures that the differentiation of an antiderivative of a function gives us the original function. For example, if you have a function f(x) and you know its antiderivative F(x), then the derivative of F(x) will yield f(x).

The second part simplifies the evaluation of definite integrals, which calculate the net area under the curve of a function between two points. According to this part, all you need is the antiderivative of the function, and then you can simply calculate the difference between its evaluations at the upper and lower bounds of the integral. It's expressed as \[\int_{a}^{b} f(x) dx = F(b) - F(a)\] where \(a\) and \(b\) are the limits of integration, and F(x) is any antiderivative of f(x).

Getting back to our initial problem, since F(x) is given as the antiderivative of f(x), the Fundamental Theorem of Calculus directly applies, allowing us to compute the definite integral by evaluating F(x) at x = 1 and x = -1 and taking the difference.

Antiderivatives and their Role in Integral Calculus

Antiderivatives, also known as indefinite integrals, are the reverse operation of differentiation. If the derivative of a function F(x) gives us f(x), then we can say F(x) is an antiderivative of f(x). Symbolically, this is expressed as \(F'(x) = f(x)\), which means taking the derivative of F(x) results in f(x).

It is crucial in calculus to find antiderivatives for several reasons:

• They are used to solve differential equations which model many natural phenomena.
• They are essential for computing definite integrals using the Fundamental Theorem of Calculus. This is because, as mentioned, the evaluation of a definite integral from \(a\) to \(b\) involves finding \(F(b) - F(a)\), where F is an antiderivative of the function we're integrating.

In the particular problem we're scrutinizing, F(x) is provided as an antiderivative of f(x), thus streamlining our process significantly. Knowing that the integral of \(\frac{1}{x^2}\) is \( -\frac{1}{x} \), we can conclude F(1) and F(-1) for our bounds and subtract them to find the area under the curve between -1 and 1.

The Role of Integral Calculus in Mathematics

Integral calculus is a branch of mathematics focusing on the accumulation of quantities and the areas under and between curves. It allows us to find the total accumulation of a quantity, such as distance traveled over a time interval, from a rate of change (such as a velocity-time graph). The process of performing integral calculus involves two key components: antiderivatives and the evaluation of definite integrals.

When approaching problems involving integral calculus, understanding the nature of the function you're working with is crucial. One must consider factors such as continuity and whether the function yields a finite area between its curve and the x-axis within the specified bounds.

In our exercise, the continuity of f(x) = \(\frac{1}{x^2}\) over the interval [-1, 1] is called into question due to the discontinuity at x = 0, which would typically prevent the use of the Fundamental Theorem of Calculus. However, the exercise sidesteps this issue by offering F(x), the antiderivative of f(x), and directly asks to compare \(F(1) - F(-1)\) with the definite integral \(\int_{-1}^{1} \frac{1}{x^2} dx\). This is a subtle hint that we should be more careful when dealing with integrals over intervals that include discontinuities of the function, as the fundamental theorem assumes continuity over the interval.