Problem 19 - Applied Calculus

Question

Both exactly [e.g. \(\ln (3 \pi)]\) and numerically [e.g. \(\ln (3 \pi) \approx 2.243]\). $$ \int_{1}^{3} t \ln t d t $$

Step-by-Step Solution

Answer

The exact solution is \( \frac{9}{2} \ln 3 - 2 \) and the numerical solution is approximately 2.944.

Step 1: Identify the Integration Technique

To integrate the function \( \int t \ln t \, dt \), we use integration by parts, which is suitable for products of functions. Recall the formula: \[ \int u \, dv = uv - \int v \, du \]where \( u \) and \( dv \) are parts of the original function.

Step 2: Choose the Functions for Integration by Parts

For integration by parts, let \( u = \ln t \) and \( dv = t \, dt \). Then find \( du \) and \( v \):- \( du = \frac{1}{t} \, dt \)- \( v = \int t \, dt = \frac{t^2}{2} \)

Step 3: Apply Integration by Parts Formula

Substitute \( u \, (dv, du, v) \) into the integration by parts formula:\[\int t \ln t \, dt = \frac{t^2}{2} \cdot \ln t - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt = \frac{t^2}{2} \ln t - \int \frac{t}{2} \, dt\]Simplify the second integral.

Step 4: Simplify and Integrate the Remaining Term

Compute the remaining integral:\[\int \frac{t}{2} \, dt = \frac{1}{2} \int t \, dt = \frac{1}{2} \cdot \frac{t^2}{2} = \frac{t^2}{4}\] Thus, we have: \[ \frac{t^2}{2} \, \ln t - \frac{t^2}{4} \] This is the indefinite integral.

Step 5: Evaluate the Definite Integral

Substitute the limits 1 and 3 into the expression:\[\left[ \frac{t^2}{2} \ln t - \frac{t^2}{4} \right]_1^3 = \left( \frac{9}{2} \ln 3 - \frac{9}{4} \right) - \left( \frac{1}{2} \ln 1 - \frac{1}{4} \right)\]Recall that \( \ln 1 = 0 \). Simplify this to get:\[ \left( \frac{9}{2} \ln 3 - \frac{9}{4} \right) - \left( -\frac{1}{4} \right) = \frac{9}{2} \ln 3 - 2\]

Step 6: Numerical Evaluation

Calculate \( \ln 3 \approx 1.0986 \). Substitute to get a numerical answer:\[\int_{1}^{3} t \ln t \, dt \approx \frac{9}{2} \times 1.0986 - 2 = 4.9437 - 2 = 2.9437\]

Key Concepts

Indefinite Integrals

Indefinite integrals represent a family of functions that includes all antiderivatives of a given function. Also known as the antiderivative, it is usually expressed with an additive constant, denoted as +C, which accounts for any constant term that might have been differentiated away when finding the original function.
For instance, given a function \( f(x) \), its indefinite integral is \( \int f(x)\, dx \), which includes all functions whose derivative is \( f(x) \).
Here's an example: Given \( f(x) = x^2 \), the indefinite integral is \( \int x^2 \, dx = \frac{x^3}{3} + C \).
In the problem exercise, after applying integration by parts, the indefinite integral of the function \( t \ln t \) was computed as \( \frac{t^2}{2} \ln t - \frac{t^2}{4} \). This result gives the antiderivative but does not specify a constant of integration since it might be evaluated further in definite integrals.

Definite Integrals

Definite integrals are used to calculate the total accumulation of quantities and are expressed with upper and lower limits of integration. Unlike indefinite integrals, definite integrals provide a specific numerical value representing the net area under a curve over an interval.
To compute a definite integral, use the Fundamental Theorem of Calculus: Find the antiderivative and then evaluate it at the upper limit and subtract the value of the antiderivative evaluated at the lower limit.
Mathematically expressed as \[\int_{a}^{b} f(x) \, dx = F(b) - F(a),\]where \( F \) is any antiderivative of \( f \).
In the exercise example, evaluating the definite integral \( \int_{1}^{3} t \ln t \, dt \) was performed by inserting the limits into the expression obtained after integration by parts. The final result provides the total change from 1 to 3, yielding \( \frac{9}{2} \ln 3 - 2 \), which gives us a numerical value for the area.

Numerical Evaluation of Integrals

Numerical evaluation involves approximating the value of an integral, which can be especially useful when dealing with complex functions or when an explicit antiderivative does not exist. Methods like Simpson's rule, midpoint rule, or trapezoidal rule are some common approaches.
In real examples, even when analytical solutions are available (as seen in the step-by-step solution), numerical approximations provide a practical means to verify results by calculating specific numbers. For instance, if we know that \( \ln 3 \approx 1.0986 \), we can use this to approximate the value of the definite integral over specified limits.

• This gives a result of \( \int_{1}^{3}t \ln t \, dt \approx 2.9437 \).
• Such calculations can provide insights into the relative scale of the integral's value.

Moreover, for functions difficult to integrate symbolically, numerical methods deliver a useful tool for exploring mathematical models and performing physical computations where precision is key.