Problem 300 - Calculus Volume 1

Question

In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using \(u\) -substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of \(C\) that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)=\int_{a}^{x} f(t) d t,\) with \(a\) the left endpoint of the given interval. 298\. [T] \(\int(2 x+1) e^{x^{2}+x-6} d x\) over [-3,2] 299\. [T] \(\int \frac{\cos (\ln (2 x))}{x} d x\) on [0,2] 300\. [T] \(\int \frac{3 x^{2}+2 x+1}{\sqrt{x^{3}+x^{2}+x+4}} d x\) over [-1,2] 301\. [T] \(\int \frac{\sin x}{\cos ^{3} x} d x\) over \(\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]\) 302\. [T] \(\int(x+2) e^{-x^{2}-4 x+3} d x\) over [-5,1] 303\. [T] \(\int 3 x^{2} \sqrt{2 x^{3}+1} d x\) over [0,1]

Step-by-Step Solution

Answer

C = 0, as integrating from 0 to x matches antiderivative with C = 0.

Step 1: Choose a function to solve

Let's focus on solving the integral from exercise 299: \[ \int \frac{\cos (\ln (2x))}{x} \, dx \] over the interval \([0, 2]\).

Step 2: Identify the substitution

In this case, we can use \( u \)-substitution. Let's set \( u = \ln(2x) \). Thus, \( du = \frac{1}{x} \, dx \). Hence, \( \frac{1}{x} \, dx = du \).

Step 3: Perform the substitution

Substitute all the \( x \)-dependent expressions with \( u \):\[ \int \cos(u) \, du \]

Step 4: Integrate

The integral of \( \cos(u) \) is \( \sin(u) + C \). So we have:\[ \sin(u) + C \]

Step 5: Substitute back to original variable

Replace \( u \) with \( \ln(2x) \) to revert to the original variable:\[ \sin(\ln(2x)) + C \]

Step 6: Evaluate graph features

The graph of the function \( f(x) = \frac{\cos(\ln(2x))}{x} \) and its antiderivative \( F(x) = \sin(\ln(2x)) + C \) can be plotted over the interval \([0, 2]\).

Step 7: Estimate the constant C

To estimate the constant \( C \), compare the definite integral form of the area calculation:\[ F(x) = \int_{0}^{x} \frac{\cos(\ln(2t))}{t} \, dt \]and match with the antiderivative. The definite integral from 0 to 0 yields 0, hence \( F(0) = 0 = \sin(\ln(2 \cdot 0)) + C \), solving for \( C \) gives \( C = 0 \).

Key Concepts

Indefinite Integral

Indefinite integrals are a key component of calculus used to find the antiderivative of a given function. The notation \( \int f(x) \, dx \) is used to represent the indefinite integral of the function \( f(x) \). When solving for indefinite integrals through the process of integration, remember that it is crucial to include a constant, \( C \), because the derivative of any constant is zero. This constant represents the family of all possible antiderivatives of the function.
In the exercise mentioned, we use \( u \)-substitution, a technique that simplifies the integral by substituting a part of the integral with a new variable \( u \). This technique especially helps when the integral contains a composite function where an inner function is present. Here, we first assign \( u = \ln(2x) \), simplifying the integral into one involving \( u \) instead of \( x \).

• Identify the substitution: Choose \( u \) such that the derivative \( du \) is present in the integral.
• Substitute \( x \) terms: Replace \( x \)-terms using \( u \) and \( du \).
• Integrate with respect to \( u \): Solve the integral in terms of \( u \), giving you the antiderivative.

After performing these steps, do not forget to back-substitute the original terms to express the function in terms of \( x \). In this exercise, we derived an antiderivative \( \sin(\ln(2x)) + C \).

Definite Integral

Definite integrals calculate the area under a curve between specific limits. In mathematical notation, the definite integral of \( f(x) \) from \( a \) to \( b \) is represented as \( \int_{a}^{b} f(x) \, dx \). The process involves finding the antiderivative, evaluating it at the upper limit \( b \), and subtracting the evaluation result at the lower limit \( a \).
In step 7 of the solution, evaluating the definite integral helps confirm the antiderivative. By considering the function \( F(x) = \int_{0}^{x} \frac{\cos(\ln(2t))}{t} \, dt \), we try to match this with the indefinite integral result, \( \sin(\ln(2x)) + C \). This involves ensuring that the values from the definite integral function at \( a = 0 \) match.

• Set up the integral with limits: Choose \( a \) and \( b \) as the interval edges.
• Evaluate the antiderivative at the boundaries: Find \( F(b) \) and \( F(a) \).
• Calculate the area: Use \( F(b) - F(a) \) to find the exact area under the curve.

In the given problem, we find that \( F(0) = 0 \) aligns with \( \sin(\ln(0)) + C \), confirming that \( C = 0 \).

Graphing Functions

Graphing the function and its antiderivative helps visualize the behavior of both. The original function we are integrating, \( f(x) = \frac{\cos(\ln(2x))}{x} \), possesses distinct features like oscillations or peaks determined by the trigonometric cosine function applied to a logarithmic term.
When graphing:

• Start with the function's base features: Analyze the logarithmic and trigonometric components for patterns.
• Plot the graph over the specific interval: For our function, this is done over \([0, 2]\).
• Use antiderivative relations: The function \( F(x) = \sin(\ln(2x)) + C \) mirrors the integral's trends, demonstrating how accumulation changes over x.

Graphing supports a deeper understanding of how these mathematical expressions behave visually, letting you see, for instance, how changes in \( x \) gradually affect the output of \( F(x) \). It allows you to intuitively grasp how an integral signifying accumulation impacts graph shape in comparison to the rate-of-change perspective used in derivatives.