Find the integral of:

  • Thread starter LDC1972
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  • #1
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Homework Statement



Find the integral of

Homework Equations



∫(5x^2 + √x - 4/x^2 dx

The Attempt at a Solution



∫(5x^2 + √x - 4/x^2 dx

= 5 ∫x^2 dx + ∫√x dx ∫- 4/x^2
= 5 [x^3/3] + ∫√x dx ∫- 4/x^2
= 5 [x^3/3] + [x^1/2] dx ∫- 4/x^2
= 5 [x^3/3] + [x^1/2] - [-4x^-2]
= 5[x^3/3] + [x^1/2] + [4/x] + C
= 5x^3/3 + x^1/2 + 4/x + C

Does my final answer look right for this? I'm really struggling with integrals, and am hoping I'm getting to grip with it now, but if someone can verify that would be great.

Thanks so much,

Lloyd
 

Answers and Replies

  • #2
CAF123
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$$\sqrt{x} = x^{\frac{1}{2}},$$ You must then integrate this. The rest is fine.
 
  • #3
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$$\sqrt{x} = x^{\frac{1}{2}},$$ You must then integrate this. The rest is fine.
Thanks for fast reply.

So do I add 1 to exponent = x^2/3
Divide by same so = x^2/3 / 2/3

From there can I simplify this to just x?

Thanks

Lloyd
 
  • #4
CAF123
Gold Member
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88
So do I add 1 to exponent
Yes, but 1/2 + 1 ≠ 2/3.

Also xa/a ≠ x, for example x2/2 = 1/2 when x = 1.
 
  • #5
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Thanks for fast reply.

So do I add 1 to exponent = x^2/3
Divide by same so = x^2/3 / 2/3

From there can I simplify this to just x?

Thanks

Lloyd
Really sorry I was being dumb there...

I wrote in pen 3/2 then typed 2/3.

So to recap:

x^1/2
Add 1 to exp' = X^3/2
Divide by same = X^3/2 / 3/2

From here there must be some simplification, as the fraction is improper?

Or is that the conclusion?

Thank you,

Lloyd
 
  • #6
128
0

Homework Statement



Find the integral of

Homework Equations



∫(5x^2 + √x - 4/x^2 dx

The Attempt at a Solution



∫(5x^2 + √x - 4/x^2 dx

= 5 ∫x^2 dx + ∫√x dx ∫- 4/x^2
= 5 [x^3/3] + ∫√x dx ∫- 4/x^2
= 5 [x^3/3] + [x^1/2] dx ∫- 4/x^2
= 5 [x^3/3] + [x^1/2] - [-4x^-2]
= 5[x^3/3] + [x^1/2] + [4/x] + C
= 5x^3/3 + x^1/2 + 4/x + C

Does my final answer look right for this? I'm really struggling with integrals, and am hoping I'm getting to grip with it now, but if someone can verify that would be great.

Thanks so much,

Lloyd
[itex]\int \sqrt{x} dx[/itex] = [itex]\int \ {x^{\frac{1}{2}}} dx [/itex]

= [itex]\frac{2}{3} x^{\frac{3}{2}}[/itex]
 
  • #7
CAF123
Gold Member
2,914
88
From here there must be some simplification, as the fraction is improper?
Yes, you can write $$\frac{x^{\frac{3}{2}}}{3/2} = x^{\frac{3}{2}} \frac{1}{3/2} = \frac{2}{3} x^{\frac{3}{2}}$$
 
  • #8
65
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Yes, you can write $$\frac{x^{\frac{3}{2}}}{3/2} = x^{\frac{3}{2}} \frac{1}{3/2} = \frac{2}{3} x^{\frac{3}{2}}$$
Thank you both.

I'm still not 100% on these, but have come a long way (from knowing nothing to where I am now) in the last 7 hours.

Thankfully calculus is only a mall proportion of the study I am doing.

Thanks again,

Lloyd
 
  • #9
128
0
Thank you both.

I'm still not 100% on these, but have come a long way (from knowing nothing to where I am now) in the last 7 hours.

Thankfully calculus is only a mall proportion of the study I am doing.

Thanks again,

Lloyd
All you need to do now is simplify :))

[itex]\frac{5}{3}x ^{3} + \frac{2}{3} x^{\frac{3}{2}} + \frac{4}{x}[/itex]
 

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