Problem 22 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{1}^{4} \sqrt{x} \ln x d x $$

Step-by-Step Solution

Answer

\( 8 \ln 2 - \frac{28}{9} \)

Step 1: Identify the Parts for Integration by Parts

Integration by parts is given by the formula \[ \int u \, dv = uv - \int v \, du \]For the integral \( \int \sqrt{x} \ln x \, dx \), choose \( u = \ln x \) and \( dv = \sqrt{x} \, dx \) so that \( du = \frac{1}{x} \, dx \) and \( v = \frac{2}{3}x^{3/2} \).

Step 2: Apply the Integration by Parts Formula

Substitute \( u = \ln x \), \( v = \frac{2}{3}x^{3/2} \), \( du = \frac{1}{x} \, dx \), and original integral limits (1 to 4) into the formula. Calculate:\[ \int \sqrt{x} \ln x \, dx = \left[ \ln x \cdot \frac{2}{3}x^{3/2} \right]_1^4 - \int_1^4 \left( \frac{2}{3} x^{3/2} \cdot \frac{1}{x} \right) dx \]

Step 3: Evaluate the Boundary Term

Calculate the boundary term \( \left[ \ln x \cdot \frac{2}{3}x^{3/2} \right]_1^4 \):\[ \left[ \ln x \cdot \frac{2}{3}x^{3/2} \right]_1^4 = \left( \ln 4 \cdot \frac{2}{3} \cdot 4^{3/2} \right) - \left( \ln 1 \cdot \frac{2}{3} \cdot 1^{3/2} \right) \]Given \( \ln 1 = 0 \), it becomes \( \ln 4 \cdot \frac{16}{3} \).

Step 4: Simplify the Second Integral

The second integral is simplified as:\[ \int_1^4 \left( \frac{2}{3} x^{1/2} \right) dx = \frac{2}{3} \int_1^4 x^{1/2} \, dx \]Use the power rule for integration. This evaluates to:\[ \frac{2}{3} \cdot \left[ \frac{2}{3}x^{3/2} \right]_1^4 = \frac{4}{9} \right[ x^{3/2} \right]_1^4\]

Step 5: Finish the Calculations

Evaluate the integral from step 4 using the power rule, calculate:\[ \frac{4}{9} \left[ x^{3/2} \right]_1^4 = \frac{4}{9} (4^{3/2} - 1^{3/2}) \]And simplify the expression:\[ = \frac{4}{9} (8 - 1) = \frac{28}{9} \]

Step 6: Combine All Results

Combine the evaluated results into:\[ \ln 4 \cdot \frac{16}{3} - \frac{28}{9} \]Calculate numerically for the final result:\( \approx \frac{32}{3} \ln 2 - \frac{28}{9} \)

Key Concepts

Definite Integration

Definite integration is a process of calculating the integral of a function over a specified interval. It provides not only the antiderivative but also the net area under the curve of the function from one point to another. In our exercise, we are dealing with the definite integral \( \int_{1}^{4} \sqrt{x} \ln x \, dx \), meaning we want to find the total area accumulated from \( x=1 \) to \( x=4 \).

When performing definite integrals, after solving the indefinite integral, it's crucial to evaluate the result at the upper and lower limits, and then subtract the lower result from the upper. The definite nature of the integral confines its solution to these boundaries, which is shown in the formula:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

This ensures we calculate the signed area between the curve and the horizontal axis over the specified interval. When we apply integration by parts to evaluate a definite integral, we also use these limits in our final calculation to ensure accuracy.

Logarithmic Integration

Logarithmic integration involves evaluating integrals that contain the natural logarithm function \( \ln(x) \). This often requires specific techniques, as logarithmic functions can complicate direct integration. In our exercise, we use the natural logarithm as one of our parts within the integration by parts formula.

The strategy we used is to let \( u = \ln(x) \), making it easier to differentiate, because the derivative of \( \ln(x) \) is \( 1/x \). Recognizing \( \ln(x) \) typically pairs well with algebraic functions in integration by parts due to this simple derivative.

• Identifying \( u \) as \( \ln(x) \) implies \( du = \frac{1}{x} \, dx \)
• This choice simplifies the differentiation process and reduces the complexity of the integration steps.

Logarithmic integration, particularly paired with the integration by parts method, proves helpful because of how naturally logarithms simplify upon differentiation.

Power Rule in Integration

The power rule in integration allows us to easily integrate expressions where the base is \( x \) with a constant exponent. This is essential in simplifying terms like \( \sqrt{x} \), which can be rewritten as \( x^{1/2} \). This expression is common in our exercise, specifically when simplifying and evaluating the resulting integral from the integration by parts process.

The power rule formula for integration is:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for any real number \( n eq -1 \)

In this scenario, we see \( dv = \sqrt{x} \, dx \), simplified as \( x^{1/2} \, dx \). Utilizing the power rule, the integration becomes straightforward:

• \( v = \frac{2}{3} x^{3/2} \)
• This integral becomes simple arithmetic, enabling us to evaluate and substitute back into the integration by parts formula.

Applying the power rule simplifies many integrals and is a cornerstone technique when working through integrals involving polynomial powers.