Evaluating The Definite Integral Of (3 - 6/x^2) From 1 To 2

In the realm of calculus, evaluating definite integrals is a fundamental skill. Definite integrals allow us to calculate the area under a curve between two specified limits. This article delves into the step-by-step evaluation of the definite integral $\int_1^2\left(3-\frac{6}{x^2}\right) dx$. This exploration will not only provide the solution but also illuminate the underlying concepts and techniques involved in integral calculus. Understanding these concepts is crucial for students, engineers, and anyone working with mathematical modeling and analysis. The ability to accurately compute definite integrals opens doors to solving a myriad of real-world problems, from calculating the distance traveled by an object with varying velocity to determining the volume of complex shapes. This particular integral serves as a good example of how to apply basic integration rules and the fundamental theorem of calculus. By breaking down the problem into manageable steps, we'll demonstrate how to find the antiderivative of the integrand and then evaluate it at the limits of integration to obtain the final result. This process highlights the power and elegance of calculus in solving problems that would be difficult or impossible to tackle using purely algebraic methods. Moreover, mastering definite integrals is essential for progressing to more advanced topics in calculus and differential equations, making this a cornerstone skill in mathematical education and application.

Understanding the Integral

The integral$\int_1^2\left(3-\frac{6}{x^2}\right) dx$ represents the definite integral of the function $f(x) = 3 - \frac{6}{x^2}$ with respect to $x$, evaluated from the lower limit $x = 1$ to the upper limit $x = 2$. To solve this, we will utilize the fundamental theorem of calculus, which states that if $F(x)$ is an antiderivative of $f(x)$, then $\int_a^b f(x) dx = F(b) - F(a)$. Therefore, the first step is to find the antiderivative of the given function. The function $f(x) = 3 - \frac{6}{x^2}$ can be rewritten as $f(x) = 3 - 6x^{-2}$. This form is more amenable to applying the power rule for integration, which is a key technique in finding antiderivatives. The power rule states that the integral of $x^n$ with respect to $x$ is $\frac{x^{n+1}}{n+1}$, provided that $n$ is not equal to -1. By applying this rule, we can find the antiderivative of each term in the function. The antiderivative of 3 is simply $3x$, since the derivative of $3x$ with respect to $x$ is 3. For the term $-6x^{-2}$, we apply the power rule: adding 1 to the exponent -2 gives -1, and dividing by the new exponent -1 gives us $\frac{-6x^{-1}}{-1} = 6x^{-1}$, which can be rewritten as $\frac{6}{x}$. Therefore, the antiderivative of $-6x^{-2}$ is $\frac{6}{x}$. Combining these results, we find that the antiderivative of $f(x) = 3 - \frac{6}{x^2}$ is $F(x) = 3x + \frac{6}{x}$. This antiderivative is crucial for the next step in evaluating the definite integral, which involves plugging in the limits of integration and subtracting the values.

Finding the Antiderivative

To evaluate the definite integral, we first need to find the antiderivative of the function $f(x) = 3 - \frac{6}{x^2}$. We can rewrite the function as $f(x) = 3 - 6x^{-2}$. The antiderivative, denoted as $F(x)$, is a function whose derivative is $f(x)$. We find $F(x)$ by integrating $f(x)$ term by term. The integral of a constant, such as 3, is simply the constant times $x$. Thus, the integral of 3 is $3x$. For the second term, $-6x^{-2}$, we use the power rule for integration, which states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $C$ is the constant of integration. Applying this rule to $-6x^{-2}$, we add 1 to the exponent, giving us $-1$, and then divide by the new exponent: $\frac{-6x^{-1}}{-1} = 6x^{-1}$. This can be rewritten as $\frac{6}{x}$. Therefore, the antiderivative of $-6x^{-2}$ is $\frac{6}{x}$. Combining the antiderivatives of both terms, we get $F(x) = 3x + \frac{6}{x} + C$. Since we are evaluating a definite integral, the constant of integration $C$ will cancel out when we subtract $F(1)$ from $F(2)$, so we can ignore it for the purposes of this calculation. The antiderivative $F(x) = 3x + \frac{6}{x}$ is now ready to be used in the fundamental theorem of calculus to evaluate the definite integral. This theorem is the cornerstone of integral calculus and allows us to find the exact area under the curve of the function between the specified limits of integration. The next step involves evaluating this antiderivative at the upper and lower limits and then finding the difference.

Applying the Fundamental Theorem of Calculus

Now that we have the antiderivative $F(x) = 3x + \frac{6}{x}$, we apply the fundamental theorem of calculus to evaluate the definite integral $\int_1^2\left(3-\frac{6}{x^2}\right) dx$. The fundamental theorem of calculus states that $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$, and $a$ and $b$ are the lower and upper limits of integration, respectively. In our case, $a = 1$ and $b = 2$. First, we evaluate $F(2)$: $F(2) = 3(2) + \frac{6}{2} = 6 + 3 = 9$. Next, we evaluate $F(1)$: $F(1) = 3(1) + \frac{6}{1} = 3 + 6 = 9$. Finally, we subtract $F(1)$ from $F(2)$: $F(2) - F(1) = 9 - 9 = 0$. Therefore, the value of the definite integral $\int_1^2\left(3-\frac{6}{x^2}\right) dx$ is 0. This result indicates that the signed area under the curve of the function $f(x) = 3 - \frac{6}{x^2}$ from $x = 1$ to $x = 2$ is zero. This could mean that the area above the x-axis is equal to the area below the x-axis within the given interval. The fundamental theorem of calculus provides a powerful and efficient way to compute definite integrals, bypassing the need for more complex methods like Riemann sums in many cases. By finding the antiderivative and evaluating it at the limits of integration, we can precisely determine the area or net change represented by the integral.

Conclusion

In conclusion, the value of the definite integral $\int_1^2\left(3-\frac{6}{x^2}\right) dx$ is 0. We arrived at this solution by first finding the antiderivative of the integrand, which is $F(x) = 3x + \frac{6}{x}$, and then applying the fundamental theorem of calculus. This theorem allowed us to evaluate the integral by calculating the difference between the antiderivative at the upper and lower limits of integration. The process involved rewriting the function, applying the power rule for integration, and carefully evaluating the antiderivative at the given limits. The result of 0 signifies that the net signed area under the curve of the function $f(x) = 3 - \frac{6}{x^2}$ from $x = 1$ to $x = 2$ is zero. This example demonstrates the power and utility of integral calculus in determining areas and other quantities related to functions. Understanding and applying the fundamental theorem of calculus is a crucial skill for anyone working in mathematics, science, or engineering. This technique not only provides a method for solving definite integrals but also lays the foundation for more advanced topics in calculus and its applications. The ability to accurately compute definite integrals is essential for solving real-world problems and making meaningful predictions in various fields.