Integrating Definite Integral: 3x^5 + 4x^4 + x^2

In summary, the conversation discusses a question about integrating a given function with limits and the concern about the term x^-1. It is clarified that the integral of x^-1 is ln(x) and that it is not infinite when taken at the given limits. The person also thanks for the explanation and acknowledges that x^-1 is a term that needs to be memorized.
  • #1
noobie!
58
0
there's a que which baffles me today,so i hope any1 of you could help me to clear my doubt,here is the que:

integrate with limit 3 on top n 1 below (3x^5 + 4x^4 + x^2)/x^3 dx ..
so here goes my sol please rectify me..
SOL: 3x^2 + 4x + x^-1
then integrate it
x^3 + 2x^2..<<<<then what about x^-1 ?if we integrate it the denominator will become infinite so any value + infinity it will become infinity?am i wright?pl help me,thanks a lot...
 
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  • #2
The integral of x^-1 is ln(x)... taken at the limits given, that's not infinity. You're method for solving is sound, but x^-1 is just one that you have to memorize.
 
  • #3
Dr. Lady said:
The integral of x^-1 is ln(x)... taken at the limits given, that's not infinity. You're method for solving is sound, but x^-1 is just one that you have to memorize.

thanks a lot,i didnt thought of that..thanks...:smile:
 

FAQ: Integrating Definite Integral: 3x^5 + 4x^4 + x^2

What is the purpose of integrating a function?

Integrating a function is used to find the area under the curve of the function. It is also used to find the total change or accumulation of a quantity over a given interval.

How do you integrate a polynomial function?

To integrate a polynomial function, you follow the power rule, which states that you add 1 to the exponent and then divide the coefficient by the new exponent. For example, for the function 3x^5, you would add 1 to the exponent to get 6, and then divide the coefficient 3 by 6 to get 0.5. Therefore, the integrated function would be 0.5x^6.

What is the difference between indefinite and definite integration?

Indefinite integration results in a general solution with a constant term, while definite integration gives a specific numerical value for the area under the curve. Indefinite integration involves adding a constant, while definite integration involves evaluating the function at two given points and subtracting the results.

Can definite integration be used to find the average value of a function?

Yes, definite integration can be used to find the average value of a function over a given interval. The average value is equal to the definite integral of the function divided by the length of the interval.

What are some real-world applications of integrating a function?

Integrating a function has various real-world applications, such as calculating the distance traveled by an object with a given velocity function, finding the total cost of a product with a changing price function, and determining the amount of medication in a patient's body over time with a given dosage function.

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