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evaluate


∫(2/x-3√x+10)dx

values of u are -2 and 5

correct answer gets rep and respect

I would love to have a legitimate answer to this but I don't. So instead I'm going to post this so I can cover my stupidity...

I'm 12 and what is this?

Fuck math, 75% of the shit you learn you don't even use it in your future atleast with my career. But I love basic math that is needed for daily basis

Edit: People forget the basic math by doing so much complicated stuff that the math you need you forget.

you just wanted an answer, not a tutor...its right anyways

Yeah I was just doing some parametric equations the other day an forgot how to add. It was the weirdest thing.

solofgt i am not that smart m8 im only in algebra 1

wat the fuck is a ∫

2(lnx-x^1,5+5x)

2(ln5-5^1,5+25)-2(ln(-2)-(-2)^1,5-10)

2(ln5-5^1,5+35+(-2)^1,5-ln-2)-Answer

I'm a confident hangover guy, probably wrong

I am going to provide two answers since the given information isn't very clear, it feels like we are missing a part of the problem or a piece of information (The fact that values are given for u makes it seem like this is a u substitution problem but this problem does not require u substitution)

You can break the integral into three parts and then integrate each term separately.

First the integral of 10 is 10x.

Second the term -3 sqrt x is another simple integral, raise the power by 1 and divide by the new power. Dividing by 3/2 is equivalent to multiplying by 2/3. Which makes the term -2 x^(3/2).

Third the last integral is one that requires knowledge of some of the basic rules of special cases for integrals. When you have 1/x this becomes ln(abs(x)). You use the absolute value because ln of a number is only defined when the number is greater than 0 (the limit of ln(x) as x approaches 0 is negative infinity). Factor out the 2 from 2/x and then integrate giving you an answer for this part of 2*ln(abs(x)).

Then the final step is using the fact that it was an indefinite integral, which means there is a constant and since we are not given information that allows us to calculate the constant it will be represented with the variable 'c'.

Adding the terms together results in 10x-2x^(3/2)+2*ln(abs(x))+c

If the given information about u is meant to be u substitution (for some weird reason), then see the below.

If you are meant to use u substitution then u=x du=dx
The only possibility for u would be they are the range for the integral after you substitute u in for x.
Since u=x, the problem becomes ∫(2/x-3√x+10)dx evaluated for x=-2 to x=5. This is a definite integral and does not require a constant after you take the integral.

Using the solved version of the integral stated above, but without using c and making the adjustments for the problem to become a definite integral we have the following equation:

[10x-2x^(3/2)+2*ln(abs(x))] evaluated from x=-2 to x=5
Which is equivalent to
[10x-2x^(3/2)+2*ln(abs(x))] x=5 - [10x-2x^(3/2)+2*ln(abs(x))] x=-2
[10*5-2*(5)^(3/2)+2*ln(5)]-[10*(-2)-2*(-2)^(3/2)+2*ln(2)]
[50-2*(5)^(3/2)+2*ln(5)]-[(-20)-2(-2)^(3/2)+2*ln(2)]
50-2*(5)^(3/2))+2*ln(5)+20+2(-2)^(3/2)-2*ln(2)
49.47+2(-2)^(3/2)
49.47-5.656*i

Hopefully this answers your question. If not you need to clarify your original post and give more detailed information.



Edit:



First this is very basic calculus. Second this is high school math....

Got the same impression