Integral problem / mathematical induction

In summary, the speaker has completed step (a) and found that the integral of d(x^0)*e^(-ax)dx from 0 to infinity is equal to 1/a. However, they are unsure of how to proceed with the next steps and have not encountered a similar problem before. The advice given is to use integration by parts directly for part (b) and to use the result of part (b) for part (c) along with integration by parts.
  • #1
Samme013
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View attachment 3411
Ok so i got step (a) and found that $\int_{0}^{\infty} \,d (x^0)*e^(-ax)dx=1/a$
But i do not get how i should go about starting the next steps using the info from the first step(have not done a similar problem before so i need to get a grasp on the method)
 

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  • #2
Samme013 said:
View attachment 3411
Ok so i got step (a) and found that $\int_{0}^{\infty} \,d (x^0)*e^(-ax)dx=1/a$
But i do not get how i should go about starting the next steps using the info from the first step(have not done a similar problem before so i need to get a grasp on the method)

Hi Samme013,

To solve (b), use integration by parts directly. Do not use the result in (a). To solve (c), use the result of part (b) and integration by parts.
 

FAQ: Integral problem / mathematical induction

What is an integral problem?

An integral problem is a mathematical problem that involves finding the area under a curve or the accumulation of a quantity over a given interval. It is commonly used in calculus to solve problems related to motion, optimization, and physics.

How is an integral problem solved?

There are several methods for solving an integral problem, including using the fundamental theorem of calculus, substitution, and integration by parts. The specific method used depends on the complexity of the problem and the techniques that are most appropriate.

What is mathematical induction?

Mathematical induction is a proof technique used to show that a statement is true for all natural numbers. It involves two steps: the base case, which shows that the statement is true for the first natural number, and the inductive step, which shows that if the statement is true for one natural number, it is also true for the next natural number.

How is mathematical induction used in integral problems?

In integral problems, mathematical induction is used to prove that a statement about the integrals of a function is true for all natural numbers. This is done by first proving the statement for the base case, typically n=1, and then using the inductive step to show that if the statement is true for n, it is also true for n+1.

Are there any limitations to using mathematical induction in integral problems?

Yes, there are some limitations to using mathematical induction in integral problems. It can only be used to prove statements that are true for all natural numbers, and it may not work for more complex functions or integrals. Additionally, it is important to check the validity of the inductive step to ensure that the statement is true for all natural numbers, not just a subset of them.

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