Question

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. Sketch the graph of the integrand and shade the region whose net area you have found. $$\int_{1 / 2}^{2}\left(1-\frac{1}{x^{2}}\right) d x$$

Short answer

Question: Evaluate the definite integral of the function $1 - \frac{1}{x^2}$ from $1/2$ to $2$ and sketch the graph of the integrand. Answer: The definite integral of the function is 3. The graph of the integrand will have a horizontal asymptote at y = 1 and a vertical asymptote at x = 0. It will be decreasing when x is between 0 and 1, and increasing when x is greater than 1. Shade the region under the curve between x = 1/2 and x = 2, which represents a net area of 3 units.

Step-by-step solution

Evaluate the antiderivative of the function

First, we need to find the antiderivative of the given function. We can rewrite the function as follows: $$\int\left(1-\frac{1}{x^{2}}\right)~d x$$ Now, let's find its antiderivative: $$\int\left(1-\frac{1}{x^{2}}\right)~d x = \int1~dx - \int\frac{1}{x^2}~dx$$ We can now integrate the two terms separately: $$\int1~dx = x + C_1$$ $$\int\frac{1}{x^2}~dx = -\frac{1}{x} + C_2$$ So, the antiderivative of the given function is: $$x - \frac{1}{x} + C$$

Apply the Fundamental Theorem of Calculus

Now, we can use the Fundamental Theorem of Calculus to evaluate the definite integral. The theorem states: $$ \int_{a}^{b}f(x)~dx = F(b) - F(a), $$ where F(x) is the antiderivative of f(x). Here, our given limits are a = 1/2 and b = 2. Substitute these values into the theorem. $$\int_{1/2}^{2}\left(1-\frac{1}{x^2}\right)~dx = \left[x - \frac{1}{x}\right]_{1/2}^2$$

Evaluate the antiderivative at the given limits

Now, let's evaluate the antiderivative at the given limits: $$ \begin{aligned} \left[x - \frac{1}{x}\right]_{1/2}^2 &= \left[(2) - \frac{1}{(2)}\right] - \left[(\frac{1}{2}) - \frac{1}{(\frac{1}{2})}\right] \\ &= (2 - \frac{1}{2}) - (\frac{1}{2} - 2) \\ &= \frac{3}{2} - (-\frac{3}{2}) \\ &= \frac{3}{2} + \frac{3}{2} \\ \end{aligned} $$ So, the definite integral is: $$\int_{1/2}^{2}\left(1-\frac{1}{x^2}\right)~dx = 3$$

Sketch the graph of the integrand

The integrand function is f(x) = 1 - (1/x^2). Plot this function on the interval [1/2, 2]. Shade the region under the curve between these points, which represents the net area of 3 units. To plot the function, observe the following properties: 1. As x approaches infinity, f(x) approaches 1 (The horizontal asymptote is y = 1). 2. When x = 1, f(x) = 0. 3. The function is decreasing when x is between 0 and 1, and increasing when x is greater than 1. With these observations, sketch the graph of the function on the interval [1/2, 2] and shade the region whose net area we have found to be 3. The function will resemble a horizontal asymptote at y=1 and a vertical asymptote at x=0.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, acting as a central pillar in calculus. It states that if you have a continuous function, then integrating that function over a closed interval gives you the net change of its antiderivative. In simpler terms, it helps us evaluate definite integrals by using antiderivatives.

This theorem has two parts: the first asserts that the derivative of an integral of a function is the function itself; the second part allows us to compute definite integrals. It's expressed mathematically as:\[ \int_{a}^{b} f(x)\,dx = F(b) - F(a) \]where \( F\) is any antiderivative of \( f\). For this exercise, it means that after finding the antiderivative of \(1 - \frac{1}{x^2}\), we simply evaluate at the upper limit \( b = 2 \) and subtract the value at the lower limit \( a = \frac{1}{2} \).

This ends with simplifying to \([F(b) - F(a)] = [2 - \frac{1}{2}] - [\frac{1}{2} - 2]\), resulting in 3 units of net area.

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. It is foundational in finding the exact value of definite integrals. For the function \( f(x) = 1 - \frac{1}{x^2} \), we begin by breaking it down into simpler components.To find the antiderivative:

• The integral of \(1\) with respect to \(x\) is \(x\).
• For \(\frac{1}{x^2}\), rewrite it as \(x^{-2}\), and integrate to get \(-\frac{1}{x}\).

Combining these results leads us to an antiderivative \(F(x) = x - \frac{1}{x} + C\), where \(C\) is a constant. This function helps us compute the definite integral by evaluating its value at the specific bounds.

Net Area Under Curve

The concept of net area under a curve involves finding the total area between the curve of a function and the x-axis, accounting for areas above and below the axis. In the definite integral \[ \int_{1/2}^{2} \left(1 - \frac{1}{x^2}\right) \, dx \] we find that the net area totals to \(3\), calculated using the antiderivative and applying the Fundamental Theorem of Calculus.

When dealing with definite integrals, areas above the x-axis are positive, and areas below are negative. The sum gives the net area, offering insight into the balance between these regions. In this case, the net area also represents the "total effect" of the function from \(x=\frac{1}{2}\) to \(x=2\).

The key steps include:

• Finding the antiderivative \( F(x) = x - \frac{1}{x} \).
• Evaluating \( F(x) \) at the upper and lower bounds of the interval.
• Subtracting these values to find the net result.

By following these steps, you compute the exact value of the integral, confirming the net area of 3 units.

Integrand Function Graphing

Graphing the integrand function, \( f(x) = 1 - \frac{1}{x^2} \), provides a visual insight into the function's behavior and the area we're evaluating. Here's what to look for when graphing:

• Note the vertical asymptote at \(x=0\). It implies that the function becomes unbounded as \(x\) approaches zero.
• The horizontal asymptote at \(y=1\) reflects the function's limit as \(x\) becomes very large.
• The function crosses the x-axis when \(x=1\), separating the positive and negative areas.

In the specific range \([\frac{1}{2}, 2]\), the shape helps guide us to understand how the positive and negative areas sum to net area \(3\). By sketching the graph:

• Plot points at intervals, such as \(x=\frac{1}{2}\), \(x=1\), and \(x=2\).
• Connect these major points with a smooth curve.
• Shade the area under the curve between the given limits.

This visual representation complements the numerical evaluation, solidifying the learning of calculating definite integrals.