Mlopez14
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O(0,0) and P(1,4) are two points on the curve with the equation y = \frac{8x^{5}}{4-2x^{3}} . By using the substitution method u = 4 - 2x3 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 \frac{du}{dx} thus finding -\frac{1}{6x^{2}}.

\displaystyle\int^1_0 (\frac{8x^{5}}{u})-(\frac{1}{6x^{2}})du

After simplifying I get:

\frac{4}{3}\displaystyle\int^1_0 -\frac{x^{3}}{u}}du

By substituting -\sqrt[3]{\frac{u-4}{2}} into u , I get

\frac{4}{3}\displaystyle\int^1_0 \frac{u-4}{2u}du

Forgetting about the finite integration, just integrating that last part gives me 2/3 (4-2x3- 4In|4-2x3|+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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*****deadness
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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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mqb2766
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(Original post by Mlopez14)
O(0,0) and P(1,4) are two points on the curve with the equation y = \frac{8x^{5}}{4-2x^{3}} . By using the substitution method u = 3 - 2x3 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 \frac{du}{dx} thus finding -\frac{1}{6x^{2}}.

\displaystyle\int^1_0 (\frac{8x^{5}}{u})-(\frac{1}{6x^{2}})du

After simplifying I get:

\frac{4}{3}\displaystyle\int^1_0 \frac{8x^{3}}{6u}}

By substituting -\sqrt[3]{\frac{u-4}{2}} into u , I get

\frac{4}{3}\displaystyle\int^1_0 \frac{u-4}{2u}

Forgetting about the finite integration, just integrating that last part gives me 2/3 (4-2x3- 4In|4-2x3|+c) which I know is wrong, I know the real answer but I don't know how to get there.
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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Mlopez14
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(Original post by *****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
I edited the first two issues, those were some typo from my part. For the cube root, using u = 4-2x3 , 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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*****deadness
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(Original post by Mlopez14)
I edited the first two issues, those were some typo from my part. For the cube root, using u = 4-2x3 , 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
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?
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mqb2766
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(Original post by Mlopez14)
I edited the first two issues, those were some typo from my part. For the cube root, using u = 4-2x3 , 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
The limits not changed.
Last edited by mqb2766; 1 month ago
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Mlopez14
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(Original post by 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?
When I substituted the negative cube root, it cancelled itself out with the negative sign and the cubed numerator
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Mlopez14
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(Original post by mqb2766)
The x^3 sub is still the wrong sign and the limits not changed.
(Original post by *****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?
Got the same result, it was me typing the thing who messed up.
Also why would my limits change?
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mqb2766
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(Original post by Mlopez14)
Got the same result, it was me typing the thing who messed up.
Also why would my limits change?
If you evaluate in terms of u. Standard practice for substitutiin.

Edit for sign error, t he previous line is now negative so ok.
Last edited by mqb2766; 1 month ago
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Mlopez14
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#10
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(Original post by mqb2766)
Can you show this in steps?
y = -\frac{4}{3}\int^0_1 \frac{x^{3}}{u}
y = -\frac{4}{3}\int^0_1 -\frac{(\sqrt[3]{\frac{u-4}{2}})^{3}}{u}
y = \frac{4}{3}\int^0_1\frac{\frac{u-4}{2}}{u}
y = \frac{4}{3}\int^0_1\frac{u-4}{2u}
y = \frac{4}{3}*\frac{1}{2}\int^0_1 \frac{u-4}{u}
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mqb2766
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(Original post by Mlopez14)
y = -\frac{4}{3}\int^0_1 \frac{x^{3}}{u}
y = -\frac{4}{3}\int^0_1 -\frac{(\sqrt[3]{\frac{u-4}{2}})^{3}}{u}
y = \frac{4}{3}\int^0_1\frac{\frac{u-4}{2}}{u}
y = \frac{4}{3}\int^0_1\frac{u-4}{2u}
y = \frac{4}{3}*\frac{1}{2}\int^0_1 \frac{u-4}{u}
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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Mlopez14
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(Original post by mqb2766)
Sure. The previous line (starting point) changed sign when you edited it, and I didn't notice.
Are you ok for the limits?
Based on what I have picked up the limits being 0 and 1 should become 4 and 2 respectively right?
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mqb2766
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(Original post by Mlopez14)
Based on what I have picked up the limits being 0 and 1 should become 4 and 2 respectively right?
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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Mlopez14
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(Original post by mqb2766)
Yes. The limits have to match the integrating variable.
So as soon as you integrate wrt u, you must change the limits.
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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mqb2766
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(Original post by 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?
Integration gives the signed area between the curve and the x axis. When the function is negative, the signed area is negative.
I'll check the values now.
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mqb2766
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(Original post by 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?
The integral is negative in that region, but as you're going from 4 to 2, this makes the final result positive, as the original function is. What did you actually do?
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Mlopez14
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It seems I got my value reversed originally it should've been \int^1_0 hence becoming \int^2_4
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mqb2766
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(Original post by Mlopez14)
It seems I got my value reversed originally it should've been \int^1_0 hence becoming \int^2_4
That's correct.
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Mlopez14
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Well thank you for all of your help
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old_engineer
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(Original post by Mlopez14)
Well thank you for all of your help
Since this is an "or otherwise" question, it might just be worth noting that the given curve equation rearranges to y = (-4x^2) + 16x^2 / (4 - 2x^3), which can be integrated directly (without substitution) quite easily.
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