Problem 40 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{1}^{2} \frac{x^{2}+1}{x} d x $$

Step-by-Step Solution

Answer

\( \frac{3}{2} + \ln 2 \)

Step 1: Simplify the Integrand

First, we need to simplify the integrand \( \frac{x^2 + 1}{x} \). We can do this by dividing each term in the numerator by \( x \): \( \frac{x^2}{x} + \frac{1}{x} = x + \frac{1}{x} \). So the integral becomes \( \int_{1}^{2} \left( x + \frac{1}{x} \right) dx \).

Step 2: Integrate Each Term

Next, find the antiderivative of each term separately. The antiderivative of \( x \) with respect to \( x \) is \( \frac{x^2}{2} \), and the antiderivative of \( \frac{1}{x} \) is \( \ln|x| \). Thus, the antiderivative of the entire expression is \( \frac{x^2}{2} + \ln|x| \).

Step 3: Evaluate the Definite Integral

Now, we need to evaluate the definite integral from 1 to 2. This means we substitute the upper limit and the lower limit into the antiderivative: \( \left[ \frac{x^2}{2} + \ln|x| \right]_{1}^{2} = \left( \frac{2^2}{2} + \ln|2| \right) - \left( \frac{1^2}{2} + \ln|1| \right) \).

Step 4: Calculate the Values

Calculate the expression by substituting the values: \( \frac{4}{2} + \ln 2 \) for the upper limit, and \( \frac{1}{2} + \ln 1 \) for the lower limit. So, we have: \( 2 + \ln 2 - \left( \frac{1}{2} + 0 \right) = 2 + \ln 2 - \frac{1}{2} \).

Step 5: Simplify the Result

Finally, simplify the result: \( 2 - \frac{1}{2} + \ln 2 = \frac{4}{2} - \frac{1}{2} + \ln 2 \). This simplifies to \( \frac{3}{2} + \ln 2 \).

Key Concepts

Antiderivative

An antiderivative, often called an indefinite integral, is the reverse process of differentiation. In simple terms, if you have a function and you want to find a function whose derivative is the original function, you're looking for an antiderivative.
To solve the definite integral, \[\int_{1}^{2} \left(x + \frac{1}{x}\right) dx,\]we need to find the antiderivative for each component of the expression:

• The antiderivative of \(x\) is \(\frac{x^2}{2}\).
• The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\), which is the natural logarithm of the absolute value of \(x\).

Thus, these produce the antiderivative \(\frac{x^2}{2} + \ln|x|\). This is crucial for solving definite integrals, which require evaluation at specific bounds.

Integrand Simplification

Simplifying the integrand is an essential step in evaluating integrals. It often makes the integration process more straightforward by breaking down complex expressions into simpler parts.
In this exercise, the integrand is \(\frac{x^2 + 1}{x}\). Using basic algebra, divide each term in the numerator by \(x\) to simplify:\[\frac{x^2}{x} + \frac{1}{x} = x + \frac{1}{x}.\]Breaking down the expression in this way reduces difficulty in integration since each component can be easily handled separately.
The skill of simplification is not only key in calculus but also a powerful tool in solving more complex mathematical problems.

Definite Integral

A definite integral, unlike an indefinite integral, calculates the net area under a curve between two specified limits. This area represents the accumulation of quantities over a given range.
Once you've found the antiderivative of the integrand, evaluate the definite integral over the interval \[\int_{1}^{2} \left(x + \frac{1}{x}\right) dx\] by substituting the bounds into the antiderivative:

• For the upper bound 2: \(\frac{2^2}{2} + \ln|2|\).
• For the lower bound 1: \(\frac{1^2}{2} + \ln|1|\).

Then find the difference between these values to determine the accumulated area between 1 and 2 under the function.

Calculus Problem Solving

Solving calculus problems, like the one presented here, involves a systematic approach to break down and tackle each step of the process.
Here is a structured way to solve such a problem:

• Firstly, simplify the integrand if necessary to make the integral easier to solve. This was achieved by separating \(\frac{x^2 + 1}{x}\) into simpler terms.
• Find the antiderivative, which serves as the foundation for evaluating the integral over a specific interval.
• Next, substitute the upper and lower bounds into the antiderivative to find the net value — this gives the area under the curve.
• Finally, simplify the result to reach the most accurate and simplest answer possible. In our scenario, that leaves us with \(\frac{3}{2} + \ln 2\).

This structured problem-solving method in calculus is applicable across a vast range of problems, assisting students in tackling challenging exercises with confidence.