Finding Integrals: ∫ (5x^2 + sqrt(x) - 4/x^2) dx

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Homework Help Overview

The original poster presents several integrals to evaluate, including polynomial, trigonometric, and rational functions. The discussion revolves around the correctness of the integrals and the methods used to solve them.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the possibility of splitting integrals to simplify calculations. There are questions about the correctness of the original poster's solutions and the presence of factors in the antiderivatives. Some participants express confusion over the notation used for definite integrals.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's interpretations and calculations. Some guidance has been offered regarding the splitting of integrals and the need to consider factors in antiderivatives, but no consensus has been reached on the correctness of the solutions presented.

Contextual Notes

There are mentions of potential errors in the calculations, particularly regarding the factor of 1/2 in one of the integrals. Participants are also navigating the implications of definite versus indefinite integrals.

Mathsishard123
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I have these integrals to find:

∫ (5x^2 + sqrt(x) - 4/x^2) dx

∫ [cos(x/2) - sin(3x/2)] dx

∫ s/sqrt(s^2 + 4) ds (upper coordinate is 5 lower coordinate is 1)

I have worked it out as:

∫〖(5x^2+√x〗-4/x^2) dx=5x^3/(2+1)+x^(1/2+1)/(1+1/2)-4x^(-2+1)/(-2+1)+C=5/3 x^3+2/3x^(3/2)+4/x+C


∫〖(cos⁡(x/2)-sin⁡(3x/2) )dx=2 sin⁡(x/2)+2/3 cos⁡(3x/2)+C〗


∫_1^5〖s/√(s^2+4)〗ds=1/2 √(s^2+4) (1≤s≤5)=1/2 (5^2+1)^(1/2)-1/2(1^2+1)^(1/2)=√26/2-√2/2

Do these look right?
 
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The first one is kind of messy but you know you can split integrals like that up?
∫ (5x^2 + sqrt(x) - 4/x^2) dx = ∫5x^2 dx + ∫sqrt(x) dx - ∫4/x^2 dx
That should make it easier. The second integral is right.
By coordinate, do you mean bounds in a definite integral? so ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
sqrt(5^2+4)-sqrt(1^2+4), so sqrt(29) - sqrt(5) should be the last one.
 
Last edited:
Panphobia said:
The first one is kind of messy but you know you can split integrals like that up?
∫ (5x^2 + sqrt(x) - 4/x^2) dx = ∫5x^2 dx + ∫sqrt(x) dx - ∫4/x^2 dx
That should make it easier. The second integral is right.
By coordinate, do you mean bounds in a definite integral? so ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.
Panphobia said:
sqrt(5^2+4)-sqrt(1^2+4), so sqrt(29) - sqrt(5) should be the last one.
 
Mark44 said:
No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.
Are you talking about this? ∫ s/sqrt(s^2 + 4) ds = sqrt(s^2 + 4) + C
Look
u = s^2 + 4
du = 2s ds(1/2)∫du/sqrt(u)

(2/2) sqrt(u)

sqrt(s^2 + 4) + C
Where am I missing the factor of 1/2?
 
Mathsishard123 said:
∫_1^5〖s/√(s^2+4)〗ds=1/2 √(s^2+4) (1≤s≤5)=1/2 (5^2+1)^(1/2)-1/2(1^2+1)^(1/2)=√26/2-√2/2

Do these look right?

You missed a factor of 2.

ehild
 
Mark44 said:
No, this is incorrect. The OP had it right the first time. Your version is missing the factor of 1/2 in the antiderivative.

Panphobia is correct.
 
I shouldn't do integration in my head.:cry:
 
Mark44 said:
I shouldn't do integration in my head.:cry:

Why not? I did, and it was right. :-p

Just messin' with you. :biggrin:
 

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