Problem 21 - Applied Calculus

Question

Use integration by parts twice to evaluate the integral. $$\int(\ln t)^{2} d t$$

Step-by-Step Solution

Answer

The integral evaluates to \( t (\ln t)^2 - 2t \ln t + 2t + C \).

Step 1: Identify Parts for Integration by Parts

Integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). For the integral \( \int (\ln t)^2 \, dt \), choose \( u = (\ln t)^2 \) and \( dv = dt \).

Step 2: Compute Derivatives and Integrals

Find \( du \) and \( v \). Since \( u = (\ln t)^2 \), differentiate to get \( du = 2(\ln t) \frac{1}{t} dt = \frac{2\ln t}{t} dt \). Integrate \( dv = dt \) to get \( v = t \).

Step 3: Apply Integration by Parts Formula

Substitute into the integration by parts formula: \[ \int (\ln t)^2 \, dt = t (\ln t)^2 - \int t \cdot \frac{2\ln t}{t} \, dt \] which simplifies to \[ t (\ln t)^2 - 2 \int \ln t \, dt \].

Step 4: Use Integration by Parts Again

For \( \int \ln t \, dt \), use integration by parts again. Set \( u = \ln t \) and \( dv = dt \). Thus, \( du = \frac{1}{t} dt \) and \( v = t \).

Step 5: Compute Second Integral by Parts

Using the integration by parts formula again, we get: \[ \int \ln t \, dt = t \ln t - \int 1 \, dt = t \ln t - t + C \].

Step 6: Substitute Back Into Original Expression

Substitute \( \int \ln t \, dt = t \ln t - t + C \) into the expression from Step 3: \[ t (\ln t)^2 - 2 (t \ln t - t) + C \].

Step 7: Simplify the Expression

Simplify the expression: \[ t (\ln t)^2 - 2t \ln t + 2t + C \]. This is the final result of the integral.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve a wide variety of integrals, especially when dealing with the product of functions. The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] where you identify two functions within the integral, labeling one as \( u \) and the other as \( dv \).

• First, you differentiate \( u \) to find \( du \).
• Second, you integrate \( dv \) to find \( v \).

Once you have \( u \), \( du \), \( v \), and \( dv \), you substitute these into the formula to transform the original integral into a simpler problem.
In the example provided, using integration by parts twice allowed us to evaluate the integral \( \int (\ln t)^2 \, dt \). First, we set \( u = (\ln t)^2 \) and \( dv = dt \), which then led us through a series of calculations, applying the integration by parts formula twice to finally simplify the expression.

Definite Integrals

Definite integrals compute the accumulation of a quantity, which is represented by the area under a curve over a specific interval. Though our example does not include limits, understanding definite integrals is key in calculus. To compute a definite integral:

• First, evaluate the antiderivative of the function.
• Next, apply the limits of integration to the antiderivative.
• Finally, subtract the lower limit evaluation from the upper limit evaluation.

The notation \( \int_{a}^{b} f(x) \, dx \) signifies the definite integral of \( f(x) \) from \( x = a \) to \( x = b \).
In our indefinite example, even though we did not evaluate specific limits, understanding definite integrals helps us grasp the complete range of calculus applications, particularly in real-world problems involving physical quantities like distance, area, or volume.
The key difference from indefinite integrals is that definite ones result in a numerical value, marking a specific quantity of accumulated change.

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm \( \ln(x) \), frequently appear in calculus problems. They are essential for dealing with products, powers, and exponential growth. The natural logarithm is the inverse of the exponential function and is defined as:

• \( \ln(e^x) = x \)
• \( e^{\ln(x)} = x \) for \( x > 0 \)

Logarithms are particularly useful in integration by parts.
In our exercise, the function \( (\ln t)^2 \) was chosen as \( u \) in the integration by parts formula, exploiting the derivative of \( \ln t \) to simplify the integral. Understanding the rules of logarithms can drastically reduce the difficulty of integration, simplifying products and powers present in a problem.
The logarithmic function’s properties make it a versatile tool, further expanding the methods to solve complex integrals, as demonstrated here, with the step-by-step breakdown of originally intricate expressions into more manageable components.