Problem 61 - Applied Calculus

Question

Find the exact area. Under \(f(x)=1 /(x+1)\) between \(x=0\) and \(x=2\).

Step-by-Step Solution

Answer

The area is \( \ln(3) \).

Step 1: Identify the Integral Bounds and Function

The problem requires finding the area under the curve of the function \(f(x) = \frac{1}{x+1}\) from \(x=0\) to \(x=2\). This means we need to compute the definite integral of \(f(x)\) from \(x=0\) to \(x=2\).

Step 2: Write the Integral Expression

The area under the curve between \(x=0\) and \(x=2\) is given by the integral: \[\int_{0}^{2} \frac{1}{x+1} \, dx.\]

Step 3: Find the Antiderivative

The antiderivative of \( \frac{1}{x+1} \) is \( \ln|x+1|\). Thus, we have:\[F(x) = \ln|x+1|.\]

Step 4: Evaluate the Definite Integral

Apply the Fundamental Theorem of Calculus to evaluate the definite integral:\[\int_{0}^{2} \frac{1}{x+1} \, dx = \left[ \ln|x+1| \right]_{0}^{2} = \ln|2+1| - \ln|0+1|.\]Simplify it to:\[\ln(3) - \ln(1) = \ln(3) - 0 = \ln(3).\]

Step 5: Conclude the Result

The exact area under the curve \(f(x) = \frac{1}{x+1}\) from \(x=0\) to \(x=2\) is \(\ln(3)\).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration. It's hugely important for understanding calculus. When solving definite integrals, this theorem helps tremendously. It tells us that if we have a continuous function over an interval, the definite integral of that function can be found using its antiderivative. It's like finding the total "weight" that the curve adds up to over that interval.

Use the Fundamental Theorem of Calculus in these steps:

• Find the antiderivative of the function.
• Evaluate this antiderivative at the upper bound and the lower bound of the interval.
• Subtract the evaluated lower bound from the upper bound to get the result.

In our exercise, we used the fundamental theorem to evaluate the integral of \(\frac{1}{x+1}\) on the interval from 0 to 2. This gave us the area under the curve as \(\ln(3)\). Understanding this theorem is key to finding areas, solving integrals, and tying together different calculus concepts.

Antiderivative

The concept of an antiderivative is crucial when working with integrals. An antiderivative of a function is another function whose derivative gives back the original function. For example, consider the function \(f(x) = \frac{1}{x+1}\). Its antiderivative, as found in the solution, is \(F(x) = \ln|x+1|\). That means when you differentiate \(F(x)\), you get back \(f(x)\).

Antiderivatives are used when evaluating definite integrals:

• First, identify the antiderivative of the given function.
• Plug in your bounds into this antiderivative when using the Fundamental Theorem of Calculus.
• Substitute and solve to find the exact numerical value of the integral.

In our exercise, understanding that \(\ln|x+1|\) is the antiderivative of \(\frac{1}{x+1}\), allowed us to calculate the area precisely. Realizing this connection makes solving integrals how clear and systematic.

Area under Curve

When you think about the "area under a curve," imagine the space between the function and the x-axis in a given interval. Calculating this area precisely often requires calculus. That's because many curves aren't simple rectangles, making basic geometry insufficient. We use definite integrals to calculate these areas accurately.

Here's how you can determine the area under a curve:

• Define the interval over which you want to calculate the area.
• Set up the definite integral of the function representing the curve over that interval.
• Solve the integral using its antiderivative and evaluate at specified bounds.

In our example, finding the area between \(f(x) = \frac{1}{x+1}\) and the x-axis from \(x = 0\) to \(x = 2\) involves calculating the integral \(\int_{0}^{2} \frac{1}{x+1} \, dx\). The solution yields \(\ln(3)\), which is the total area these two bounds capture. Understanding areas under curves is essential for many real-world applications like physics and economics, where accurate calculations of quantities are crucial.