# Finite Integration by substiution

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*O(0,0) and P(1,4) are two points on the curve with the equation . By using the substitution method u = 4 - 2x*

^{3}

*or otherwise, find the area bounded by the arc OP of the curve, the x - axis and the line x = 1.*

What I first did was differentiating u and then finding thus finding .

After simplifying I get:

By substituting into u , I get

Forgetting about the finite integration, just integrating that last part gives me 2/3 (4-2x

^{3}- 4In|4-2x

^{3}|+c) which I know is wrong, I know the real answer but I don't know how to get there.

Last edited by Mlopez14; 1 month ago

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#2

Why did you get just u at the denominator? On the second line, why did you still have 8/6 in the integral after taking out 4/3? How did you get to the third line, the cube root of (u-4)/2 doesn't seem to be what you need to me

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#3

(Original post by

What I first did was differentiating u and then finding thus finding .

After simplifying I get:

By substituting into u , I get

Forgetting about the finite integration, just integrating that last part gives me 2/3 (4-2x

**Mlopez14**)*O(0,0) and P(1,4) are two points on the curve with the equation . By using the substitution method u = 3 - 2x*^{3}*or otherwise, find the area bounded by the arc OP of the curve, the x - axis and the line x = 1.*What I first did was differentiating u and then finding thus finding .

After simplifying I get:

By substituting into u , I get

Forgetting about the finite integration, just integrating that last part gives me 2/3 (4-2x

^{3}- 4In|4-2x^{3}|+c) which I know is wrong, I know the real answer but I don't know how to get there.
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(Original post by

Why did you get just u at the denominator? On the second line, why did you still have 8/6 in the integral after taking out 4/3? How did you get to the third line, the cube root of (u-4)/2 doesn't seem to be what you need to me

*******deadness**)Why did you get just u at the denominator? On the second line, why did you still have 8/6 in the integral after taking out 4/3? How did you get to the third line, the cube root of (u-4)/2 doesn't seem to be what you need to me

^{3}, I made x the subject and hence got the cube root of (u - 4)/2, and then changed it in the equation to get the equation in term of u

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#5

(Original post by

I edited the first two issues, those were some typo from my part. For the cube root, using u = 4-2x

**Mlopez14**)I edited the first two issues, those were some typo from my part. For the cube root, using u = 4-2x

^{3}, I made x the subject and hence got the cube root of (u - 4)/2, and then changed it in the equation to get the equation in term of u
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#6

**Mlopez14**)

I edited the first two issues, those were some typo from my part. For the cube root, using u = 4-2x

^{3}, I made x the subject and hence got the cube root of (u - 4)/2, and then changed it in the equation to get the equation in term of u

Last edited by mqb2766; 1 month ago

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(Original post by

You've a few typo errors. Not sure if they're from typing up the post of in your actual working, but go through it carefully. You need to remember to change your limits. Also the numerator is x^3, there is no need to cube root then cube again. Things like that make errors more likely. In this case a sign error?

**mqb2766**)You've a few typo errors. Not sure if they're from typing up the post of in your actual working, but go through it carefully. You need to remember to change your limits. Also the numerator is x^3, there is no need to cube root then cube again. Things like that make errors more likely. In this case a sign error?

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(Original post by

The x^3 sub is still the wrong sign and the limits not changed.

**mqb2766**)The x^3 sub is still the wrong sign and the limits not changed.

(Original post by

As mqb2766 pointed out, leaving the cube would be safer, but it's alright as long you don't make mistakes. What did you get then?

*******deadness**)As mqb2766 pointed out, leaving the cube would be safer, but it's alright as long you don't make mistakes. What did you get then?

Also why would my limits change?

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#9

(Original post by

Got the same result, it was me typing the thing who messed up.

Also why would my limits change?

**Mlopez14**)Got the same result, it was me typing the thing who messed up.

Also why would my limits change?

Edit for sign error, t he previous line is now negative so ok.

Last edited by mqb2766; 1 month ago

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#11

Are you ok for the limits?

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(Original post by

Sure. The previous line (starting point) changed sign when you edited it, and I didn't notice.

Are you ok for the limits?

**mqb2766**)Sure. The previous line (starting point) changed sign when you edited it, and I didn't notice.

Are you ok for the limits?

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#13

(Original post by

Based on what I have picked up the limits being 0 and 1 should become 4 and 2 respectively right?

**Mlopez14**)Based on what I have picked up the limits being 0 and 1 should become 4 and 2 respectively right?

So as soon as you integrate wrt u, you must change the limits.

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(Original post by

Yes. The limits have to match the integrating variable.

So as soon as you integrate wrt u, you must change the limits.

**mqb2766**)Yes. The limits have to match the integrating variable.

So as soon as you integrate wrt u, you must change the limits.

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#15

(Original post by

After integrating, I get 2/3(2-4In(2)) or - 0.5151, how is possible, If I am going to find an area, it should be positive, unless I got my limits reversed?

**Mlopez14**)After integrating, I get 2/3(2-4In(2)) or - 0.5151, how is possible, If I am going to find an area, it should be positive, unless I got my limits reversed?

I'll check the values now.

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#16

**Mlopez14**)

After integrating, I get 2/3(2-4In(2)) or - 0.5151, how is possible, If I am going to find an area, it should be positive, unless I got my limits reversed?

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#18

(Original post by

It seems I got my value reversed originally it should've been hence becoming

**Mlopez14**)It seems I got my value reversed originally it should've been hence becoming

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#20

(Original post by

Well thank you for all of your help

**Mlopez14**)Well thank you for all of your help

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