# What are all the functions that satisfy (Calculus II)

• mnech
In summary: So what is the final answer?In summary, the problem is to solve for f(x) in the equation (\int f(x)dx)(\int \frac{1}{f(x)} dx) = -1. By dividing both sides by integral(f(x)dx), taking derivatives, and applying L'Hopital's incorrectly, the solutions f(x) = e^x and -e^x were found. However, this method is incorrect and the correct approach is to reduce the problem to f''=f, which has solutions of the form f(x) = ke^x and -ke^x.

#### mnech

The problem:
$$(\int f(x)dx)(\int \frac{1}{f(x)} dx) = -1$$
solve for f(x)

The attempt at a solution

divide both sides by integral(f(x)dx):
$$\int \frac{1}{f(x)} dx = \frac{-1}{\int f(x)dx}$$

took derivative of both sides:
$$\frac{1}{f(x)} = \frac{f(x)}{(\int f(x) dx)^2}$$

multiplied both sides by f(x):
$$1 = \frac{f(x)^2}{(\int f(x) dx)^2}$$

square root:
$$1 = \frac{f(x)}{\int f(x) dx}$$

Now, here I applied L'Hopital's incorrectly (this isn't even a limit I know!) by taking derivative of top and bottom on the right side; it seemed to work later that's why I tried to stick with it:
$$1 = \frac{f'(x)}{f(x)}$$

took antiderivative of both sides...
$$x = ln|f(x)|$$

$$e^x = |f(x)|$$

$$f(x) = e^x and -e^x$$

when plugging those two answers into the original equation it seems to work...
but something isn't right and anyway there is no way that only those two functions satisfy that. help?

(first time trying to use the tex tags so i have my fingers crossed that I did not make a mistake in formatting..)

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When you took an antiderivative there is a constant of integration possible. So ln(|f(x)|)=x+C. Now what? Also when you took the square root, the square root of 1 could also be -1. That gives you more solutions.

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And actually the use of l'Hopital is completely incorrect. It's not even a limit. Do it this way. When you get to integral(f(x))^2=f(x)^2, just differentiate both sides and cancel the f(x). Then differentiate again. Much simpler. Now you just have the differential equation d^2(f(x))/dx^2=f(x).

However, I am unsure how a -1 will give me more answers. when taking the square root of both sides, it would become:

$$\frac{|f(x)|}{|\int f(x) dx|} = (negative?)1$$

sorry, but can you elaborate, by including -e^x in the answer (from the absolute value another step or two down), how is that excluding anything?

As far as the constants go, we could bring them both to the right side, bringing us $$e^{ln|f(x)| + c} = e^c * |f(x)|$$ we could just call that constant $$e^c = k$$ bring it to the left side thus making the answer $$ke^x and -ke^x$$

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Um, how about e^(-x)?? You are missing stuff related to those. Like I said, if you reduce the problem to f''=f there are ODE type techniques to handle that.

sorry, but can you elaborate?
ODE type techniques?

Ordinary Differential Equations. More systematic ways of doing the same thing you've been doing.

## 1. What is Calculus II?

Calculus II is the second part of the study of calculus, which is a branch of mathematics that deals with the rates of change of functions. It builds upon the concepts and techniques learned in Calculus I, including derivatives and integrals, and introduces new topics such as sequences, series, and differential equations.

## 2. What are some common functions that are studied in Calculus II?

Some common functions that are studied in Calculus II include polynomial functions, exponential and logarithmic functions, trigonometric functions, and rational functions. These functions are often used to model real-world situations and their derivatives and integrals are calculated to analyze their behavior and solve problems.

## 3. What are the main concepts and techniques learned in Calculus II?

The main concepts and techniques learned in Calculus II include techniques of integration, convergence and divergence of sequences and series, parametric and polar equations, and applications of derivatives and integrals in physics and engineering. These concepts and techniques are essential for further studies in mathematics, science, and engineering.

## 4. How is Calculus II different from Calculus I?

Calculus II builds upon the concepts and techniques learned in Calculus I, but it introduces new topics such as sequences, series, and differential equations. It also focuses more on applications of derivatives and integrals, rather than just their theoretical aspects. Additionally, Calculus II involves more challenging and complex problems that require a deeper understanding of the concepts.

## 5. What are some tips for success in Calculus II?

Some tips for success in Calculus II include practicing regularly, seeking help from professors or tutors when needed, and working through challenging problems step-by-step. It is also important to understand the underlying concepts and connections between different topics, rather than just memorizing formulas. Additionally, actively participating in class and seeking additional resources, such as textbooks or online tutorials, can also aid in understanding and mastering the material.