Problem 7
Question
In Exercises \(7-12,\) use differentiation to verify the antiderivative formula. $$\int \csc ^{2} u d u=-\cot u+C$$
Step-by-Step Solution
Verified Answer
The verification of the antiderivative formula, \( \int \csc ^{2} u du = -\cot u+C \), is achieved by differentiating \(-\cot u+C\) to receive the original function \(\csc ^{2} u\).
1Step 1: Write down the provided integral and its equivalent
The provided integral is \( \int \csc ^{2} u du \) and it is given that this is equivalent to \(-\cot u+C\).
2Step 2: Differentiating the obtained function
To verify the antiderivative formula, differentiate the obtained function which is \(-\cot u+C\). The derivative of \(-\cot u\) is \(\csc ^{2} u\) and the derivative of \(C\) is \(0\) (since \(C\) is a constant). Therefore, the derivative of \(-\cot u+C\) is \(\csc ^{2} u\).
3Step 3: Comparing the functions
As the derivative of \(-\cot u+C\) is \(\csc ^{2} u\) which is the original function. Thus, this verifies the antiderivative formula.
Key Concepts
AntiderivativeIntegrationCalculusMathematical Verification
Antiderivative
An antiderivative of a function is another function whose derivative is the original function we started with. Imagine it like the reverse process of differentiation. When we take the antiderivative, we are basically "undoing" the derivative. In the context of the exercise provided, we are given the antiderivative formula \(\int \csc^{2} u \, du = -\cot u + C\). Here, "C" represents the constant of integration that can arise because when we differentiate a constant, it vanishes.
To verify that \(-\cot u + C\) is indeed the antiderivative of \(\csc^2 u\), the procedure includes finding the derivative of \(-\cot u + C\) and checking if it returns the original integrand \(\csc^2 u\). This is crucial as it ensures that the transition between the function and its antiderivative is accurate.
To verify that \(-\cot u + C\) is indeed the antiderivative of \(\csc^2 u\), the procedure includes finding the derivative of \(-\cot u + C\) and checking if it returns the original integrand \(\csc^2 u\). This is crucial as it ensures that the transition between the function and its antiderivative is accurate.
Integration
Integration is like assembling the pieces of a puzzle where all the pieces are fragments of a derivative. This process is essential in calculus for determining antiderivatives and in calculating areas under curves, among other applications.
When we encounter an integral like \(\int \csc^{2} u \, du\), we are tasked with finding a function that, when differentiated, yields \(\csc^{2} u\). Integration helps us move from the rate of change back to the original quantity. This particular integral results in \(-\cot u + C\).
Always remember that:
When we encounter an integral like \(\int \csc^{2} u \, du\), we are tasked with finding a function that, when differentiated, yields \(\csc^{2} u\). Integration helps us move from the rate of change back to the original quantity. This particular integral results in \(-\cot u + C\).
Always remember that:
- Indefinite integrals result in a family of functions since constants vanish in differentiation;
- The constant of integration \(C\) plays a key role in defining the uniqueness of the antiderivative.
- The verification of the integration through differentiation ensures the process is correctly executed.
Calculus
Calculus is a branch of mathematics that explores both the rate of change and the accumulation of quantities. In this exercise, we delve into integral calculus, which primarily deals with finding antiderivatives and solving integrals. Calculus has two main branches:
We use integral calculus to determine the function before differentiation occurred. In this example, the antiderivative \(-\cot u + C\) represents an accumulation process, while its differentiation validates the earlier integrative step. This synergy between differentiation and integration is at the heart of calculus, allowing us to analyze and predict changes quantitatively.
- Differential calculus, which focuses on rates of change and slopes of curves;
- Integral calculus, which is concerned with accumulation of quantities and areas under and between curves.
We use integral calculus to determine the function before differentiation occurred. In this example, the antiderivative \(-\cot u + C\) represents an accumulation process, while its differentiation validates the earlier integrative step. This synergy between differentiation and integration is at the heart of calculus, allowing us to analyze and predict changes quantitatively.
Mathematical Verification
Verification is a crucial step in mathematical problem-solving that ensures our solution is correct and logical. In verifying an antiderivative, we need to differentiate to check if it matches with the integrand. Verification serves as a mathematical "double-check" to ensure truthfulness of the result.
In the current exercise:
In the current exercise:
- We started with the claimed antiderivative \(-\cot u + C\) for the integrand \(\csc^{2} u\).
- By differentiating \(-\cot u + C\), we obtained \(\csc^{2} u\), confirming our original function. This confirms coherence in the problem-solving process.
Other exercises in this chapter
Problem 6
In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=
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In Exercises \(1-10,\) find the general solution to the exact differential equation. $$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{x}}$$
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In Exercises \(1-10,\) find the indefinite integral. $$\int 3 x^{2} e^{2 x} d x$$
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In Exercises \(5-14,\) evaluate the integral. $$\int \frac{2 x^{3}}{x^{2}-4} d x$$
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