Author Topic: Integration by parts question (Read 911 times) Tweet Share

Kanye East · « **on:** October 21, 2013, 12:10:23 pm »

Hey everyone,
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.
$Using integration by parts, find:$
$\int\ y\sqrt{y+1}\ dy$

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

b^3 · « **Reply #1 on:** October 21, 2013, 12:40:03 pm »

Are you sure you need to do it by int by parts? It's easier to do it by a easy substitution and then expand it, rather than substitution and then int by parts, which is kinda overkill in this situation.
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 $\int(uv')dx=uv'-\int(u'v)dx$ .
So to start off with, we need to pick our $u$ and $v$ , we pick a $v$ that we know how to integrate and such that the $u$ we pick will make the term we need to integrate later something simpler. If we differentiate $y$ with respect to $y$ then we just get $1$ (which makes it simpler to integrate the last term), choosing this as $u$ means we get $\sqrt{y+1}$ as our $v'$ , which if we integrate that gives us $\frac{2}{3}(y+1)^\frac{3}{2}$ as $v$ .
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 $u$ and $v$ .
$\begin{alignedat}{1}\int y\sqrt{y+1}dy & =\int\left(\frac{d}{dy}\left(\frac{2}{3}\left(y+1\right)^{\frac{3}{2}}\right)y\right)dy \\ & =\left[\frac{2}{3}y\left(y+1\right)^{\frac{3}{2}}\right]-\int\frac{2}{3}\left(y+1\right)^{\frac{3}{2}}\left(1\right)dy \\ & =\frac{2}{3}y\left(y+1\right)^{\frac{3}{2}}-\frac{2}{3}\times\frac{2}{5}\left(y+1\right)^{\frac{5}{2}}+C,\: C\in\mathbb{R} \\ & =\frac{2}{3}\left(y+1\right)^{\frac{3}{2}}\left(y-\frac{2}{5}\left(y+1\right)\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{3}\left(y+1\right)^{\frac{3}{2}}\left(\frac{3}{5}y-\frac{2}{5}\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y-2\right)+C,\: C\in\mathbb{R} \\ \end{alignedat}$

Substitution method, not needed now

Anyways, the ~~easy~~ substitution way:
$\begin{alignedat}{1}\int y\sqrt{y+1}dy \\ \text{Let }u=y+1 \\ y=u-1 \\ \frac{dy}{du}=1 \\ dy=du \\ \int y\sqrt{y+1}dy & =\int\left(u-1\right)u^{\frac{1}{2}}du \\ & =\int\left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right)du \\ & =\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}+C,\: C\in\mathbb{R} \\ & =\frac{2}{5}\left(y+1\right)^{\frac{5}{2}}-\frac{2}{3}\left(y+1\right)^{\frac{3}{2}}+C,\: C\in\mathbb{R} \\ & =2\left(y+1\right)^{\frac{3}{2}}\left(\frac{1}{5}\left(y+1\right)-\frac{1}{3}\right)+C,\: C\in\mathbb{R} \\ & =2\left(y+1\right)^{\frac{3}{2}}\left(\frac{3}{15}y+\frac{3}{15}-\frac{5}{15}\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y-2\right)+C,\: C\in\mathbb{R} \end{alignedat}$

Longer redundant int by parts method

With int by parts, we are going to use $\int(uv')dx=uv'-\int(u'v)dx$ .
So to start off with, we need to pick our $u$ and $v$ , we pick a $v$ that we know how to integrate and such that the $u$ we pick will make the term we need to integrate later something simpler. If we differentiate $a-1$ with respect to $a$ then we just get $1$ (which makes it simpler to integrate the last term), choosing this as $u$ means we get $a^{\frac{1}{2}}$ as our $v'$ , which if we integrate that gives us $\frac{2}{3}a^\frac{3}{2}$ as $v$ .
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 $u$ and $v$ .
$\begin{alignedat}{1}\int y\sqrt{y+1}dy \\ \text{Let }a=y+1 \\ y=a-1 \\ \frac{dy}{da}=1 \\ dy=da \\ \int y\sqrt{y+1}dy & =\int\left(a-1\right)a^{\frac{1}{2}}da \\ & =\int\frac{d}{da}\left(\frac{2}{3}a^{\frac{3}{2}}\right)\left(a-1\right)da \\ & =\left[\frac{2}{3}a^{\frac{3}{2}}\left(a-1\right)\right]-\int\frac{2}{3}a^{\frac{3}{2}}\times\frac{d}{da}\left(a-1\right)da \\ & =\left[\frac{2}{3}a^{\frac{3}{2}}\left(a-1\right)\right]-\frac{2}{3}\int a^{\frac{3}{2}}da \\ & =\frac{2}{3}a^{\frac{3}{2}}\left(a-1\right)-\frac{2}{3}\times\frac{2}{5}a^{\frac{5}{2}}+C,\: C\in\mathbb{R} \\ & =\frac{2}{3}a^{\frac{3}{2}}\left(a-1-\frac{2}{5}a\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{3}a^{\frac{3}{2}}\left(\frac{3}{5}a-1\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{15}a^{\frac{3}{2}}\left(3a-5\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y+3-5\right)+C,\: C\in\mathbb{R} \\ & =\frac{2}{15}\left(y+1\right)^{\frac{3}{2}}\left(3y-2\right)+C,\: C\in\mathbb{R} \end{alignedat}$

EDIT: I've changed the substitution variable to $a$ instead of $u$ so that it's less confusing.

EDIT2: Finally got all the $u$ and $a$ changes right.

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.

Kanye East · « **Reply #2 on:** October 21, 2013, 03:06:14 pm »

Thanks for this!!

I think I will make a seperate thread in the maths board for ENG1091 questions

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Author Topic: Integration by parts question (Read 911 times) Tweet Share

Kanye East

Integration by parts question

b^3

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Kanye East

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