Post by: Kanye East on October 21, 2013, 12:10:23 pm
so I know that this isn't in the spesh course, but I know from previous posts, some students would know math beyond the spesh course.
MOD EDIT: While one of these methods is in the spesh course, the other isn't. So that people who read it don't get freaked out that they should have learnt something that they haven't been taught, moved it to just the Mathematics Board :)
Post by: b^3 on October 21, 2013, 12:40:03 pm
Now that I look at it in the right way it's not too bad, but anyways, just look at the Int by parts working.
Only just realised I did a susbsitution with the int by parts for no reason. So yeah that's in the spoiler now. The same idea works though.
With int by parts, we are going to use
So to start off with, we need to pick our
Then we split it up using our formula earlier, and work out way through. Really, the hard part with integration by parts is working out what to choose for
Substitution method, not needed now
Anyways, the easy substitution way:
u^{\frac{1}{2}}du<br />\\ & =\int\left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right)du<br />\\ & =\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{5}\left(y+1\right)^{\frac{5}{2}}-\frac{2}{3}\left(y+1\right)^{\frac{3}{2}}+C,\: C\in\mathbb{R}<br />\\ & =2\left(y+1\right)^{\frac{3}{2}}\left(\frac{1}{5}\left(y+1\right)-\frac{1}{3}\right)+C,\: C\in\mathbb{R}<br />\\ & =2\left(y+1\right)^{\frac{3}{2}}\left(\frac{3}{15}y+\frac{3}{15}-\frac{5}{15}\right)+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y-2\right)+C,\: C\in\mathbb{R}<br />\end{alignedat})
Longer redundant int by parts method
With int by parts, we are going to use dx=uv'-\int(u'v)dx)
.
So to start off with, we need to pick our
and 
, we pick a that we know how to integrate and such that the we pick will make the term we need to integrate later something simpler. If we differentiate 
with respect to 
then we just get 
(which makes it simpler to integrate the last term), choosing this as means we get 
as our 
, which if we integrate that gives us 
as .
Then we split it up using our formula earlier, and work out way through. Really, the hard part with integration by parts is working out what to choose for and .
a^{\frac{1}{2}}da<br />\\ & =\int\frac{d}{da}\left(\frac{2}{3}a^{\frac{3}{2}}\right)\left(a-1\right)da<br />\\ & =\left[\frac{2}{3}a^{\frac{3}{2}}\left(a-1\right)\right]-\int\frac{2}{3}a^{\frac{3}{2}}\times\frac{d}{da}\left(a-1\right)da<br />\\ & =\left[\frac{2}{3}a^{\frac{3}{2}}\left(a-1\right)\right]-\frac{2}{3}\int a^{\frac{3}{2}}da<br />\\ & =\frac{2}{3}a^{\frac{3}{2}}\left(a-1\right)-\frac{2}{3}\times\frac{2}{5}a^{\frac{5}{2}}+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{3}a^{\frac{3}{2}}\left(a-1-\frac{2}{5}a\right)+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{3}a^{\frac{3}{2}}\left(\frac{3}{5}a-1\right)+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{15}a^{\frac{3}{2}}\left(3a-5\right)+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y+3-5\right)+C,\: C\in\mathbb{R}<br />\\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y-2\right)+C,\: C\in\mathbb{R}<br />\end{alignedat})
So to start off with, we need to pick our
Then we split it up using our formula earlier, and work out way through. Really, the hard part with integration by parts is working out what to choose for
EDIT: I've changed the substitution variable to
EDIT2: Finally got all the
EDIT3: Don't know why I used a substitution with int by parts at first, fixed it and now that old redundant working is in the spoiler.
Post by: Kanye East on October 21, 2013, 03:06:14 pm
I think I will make a seperate thread in the maths board for ENG1091 questions :)