Problem 47 - Calculus for Biology and Medicine

Question

In Problems \(39-48\), first make an appropriate substitution and then use integration by parts to evaluate each integral. $$ \int_{1}^{4} \ln (\sqrt{x}+1) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 3\ln(3) - \frac{5}{2} \).

Step 1: Choose a Substitution

Let's first simplify the integral by making a substitution. Let \( u = \sqrt{x} + 1 \). Then, \( \sqrt{x} = u - 1 \) and \( x = (u - 1)^2 \). This changes \( d\sqrt{x} = \frac{1}{2\sqrt{x}} \cdot dx \), so \( dx = 2(u - 1) du \). The limits change accordingly: when \( x = 1 \), \( u = 2 \), and when \( x = 4 \), \( u = 3 \).

Step 2: Transform the Integral

Substitute into the integral: \[ \int_{1}^{4} \ln(\sqrt{x}+1) \, dx = \int_{2}^{3} \ln(u) \, 2(u-1) \, du. \] This simplifies to \[ 2 \int_{2}^{3} u \ln(u) \, du - 2 \int_{2}^{3} \ln(u) \, du. \]

Step 3: Integration by Parts for First Integral

Use integration by parts on the first integral \( \int u \ln(u) \, du \) with \( v = \ln(u) \) and \( dw = u \, du \). Then, \( dv = \frac{1}{u} \, du \) and \( w = \frac{u^2}{2} \). Apply the formula \( \int v \, dw = vw - \int w \, dv \):\[ \int u \ln(u) \, du = \frac{u^2}{2} \ln(u) - \int \frac{u^2}{2} \cdot \frac{1}{u} \, du = \frac{u^2}{2} \ln(u) - \frac{1}{2} \int u \, du = \frac{u^2}{2} \ln(u) - \frac{u^2}{4}. \]

Step 4: Evaluate First Integral at Bounds

Evaluate the result from Step 3 from \( u = 2 \) to \( u = 3 \): \[ \left[ \frac{u^2}{2} \ln(u) - \frac{u^2}{4} \right]_2^3 = \left( \frac{9}{2} \ln(3) - \frac{9}{4} \right) - \left( 2 \ln(2) - 1 \right) = \frac{9}{2} \ln(3) - \frac{9}{4} - 2 \ln(2) + 1. \]

Step 5: Integration by Parts for Second Integral

Use integration by parts on the second integral \( \int \ln(u) \, du \) with \( v = \ln(u) \) and \( dw = du \). Then, \( dv = \frac{1}{u} \, du \) and \( w = u \). Apply \( \int v \, dw = vw - \int w \, dv \):\[ \int \ln(u) \, du = u \ln(u) - \int u \cdot \frac{1}{u} \, du = u \ln(u) - u. \]

Step 6: Evaluate Second Integral at Bounds

Evaluate the result from Step 5 from \( u = 2 \) to \( u = 3 \): \[ [u \ln(u) - u]_2^3 = (3 \ln(3) - 3) - (2 \ln(2) - 2) = 3 \ln(3) - 3 - 2 \ln(2) + 2. \]

Step 7: Combine Results

Substitute the evaluated integrals back into the transformed integral from Step 2:\[ 2 \left( \frac{9}{2} \ln(3) - \frac{9}{4} - 2 \ln(2) + 1 \right) - 2 \left( 3 \ln(3) - 3 - 2 \ln(2) + 2 \right) = 9 \ln(3) - \frac{9}{2} - 4 \ln(2) + 2 - 6 \ln(3) + 6 + 4 \ln(2) - 4. \]

Step 8: Simplify Final Answer

Combine like terms to get the final result: \[ 3 \ln(3) + 5 - \frac{9}{2}. \] Simplify further: \[ 3 \ln(3) - \frac{5}{2}. \] Thus, the integral evaluates to \( 3\ln(3) - \frac{5}{2} \).

Key Concepts

Substitution Method in Integration

The substitution method is a widely used technique to simplify integrals. It involves changing the variable of integration to make the integral easier to solve. In our problem, we started by letting \( u = \sqrt{x} + 1 \). The goal was to turn the complicated integral into a form that is more straightforward. By substituting \( u \) for \( \sqrt{x} + 1 \), we transformed the variable from \( x \) to \( u \), which also changed the form of \( dx \) in terms of \( du \). This substitution meant that we had to change the limits of the integral from \( x = 1 \) to \( 4 \) to \( u = 2 \) to \( 3 \).

• Substitute the expression: Choose an expression within the integral that simplifies nicely into a single variable.
• Transform \( dx \): Express \( dx \) in terms of this new variable \( du \).
• Change limits: Adjust the integration limits to correspond to the new variable.

By doing this, we broke down a complex expression into manageable parts and set the stage for integrating more comfortably.

Evaluating Definite Integrals

Definite integrals result in a number that represents the area under a curve within specified limits. After performing a substitution, the integral becomes easier to evaluate. Once you have simplified by substitution and integration, the next step is evaluating the definite integrals, which involves determining the area under the curve from the lower to the upper limits.In this exercise, after using the substitution method and integration by parts, we evaluated the integral of \( e^u \ln(u) \) between the new limits \( u = 2 \) and \( u = 3 \). This step involves plugging these limits into the antiderivative expression and computing the difference. Each piece of the simplified integral must be evaluated separately:

• Calculate the antiderivative.
• Substitute the upper and lower bounds to find the difference.
• Simplify the expression found by the difference of evaluations.

Using these steps ensures accuracy in determining the value of a definite integral.

Natural Logarithm Integration

Integrating expressions involving the natural logarithm \( \ln(u) \), especially when paired with other functions like polynomials or exponentials, often requires special techniques like integration by parts. Integration by parts is based on the product rule for differentiation. In our exercise, we used this technique twice.First, for integrating \( u \ln(u) \), we set \( v = \ln(u) \) and \( dw = u \, du \). The derivative \( dv \) became \( \frac{1}{u} \, du \), and \( w \) became \( \frac{u^2}{2} \). The integration by parts formula \( \int u \, dv = uv - \int v \, du \) was then applied. The method helps transform an integral of \( u \ln(u) \) into simpler functions. Next, to integrate \( \ln(u) \), similar steps were followed, choosing \( v = \ln(u) \) and \( dw = du \). This step reduced \( \ln(u) \) into a more manageable form of \( u \ln(u) - u \).Here's a quick guide to integrating logarithmic functions:

• Select parts \( v \) and \( dw \) to simplify the integral.
• Use the integration by parts formula: \( \int v \, dw = vw - \int w \, dv \).
• Evaluate the result and always check the bounds for definite integrals.

This method was crucial in breaking down the integral in our exercise into parts that could be individually integrated and then evaluated.