Question

Integration by parts?

Answer 1

\[\int\limits_{4}^{9} \frac{ \ln y }{ y }\]

Answer 2

oops, forgot the dy at the end

Answer 3

Is IBP required for this question? Or is that just a suggestion?

Answer 4

A substitution is all you need.

Answer 5

The question requires it

Answer 6

Okay, when a logarithm is involved, the typical suggestion for IBP involves setting the differentiated part (\(u\)) to the logarithm:
\[\begin{matrix}
u=\ln y&&&dv=\dfrac{dy}{y}\\
du=\frac{dy}{y}&&&v=\ln y
\end{matrix}\]
(note: I' omitting the absolute values you wold usually see on the antiderivative of \(\dfrac{1}{y}\) because what I'm really doing is setting \(e^u=y\), which gives \(u=\ln y\)).

From the formula for IBP, you get
\[\int_4^9\frac{\ln y}{y}\,dy=(\ln y)^2-\int_4^9\frac{\ln y}{y}\,dy\]
See where to go from here?

Answer 7

Oops, that term after the equals sign should be \(\large\left[\big(\ln y\big)^2\right]_4^9\).

Answer 8

I think so... how would you take the integral of ln(y)/y? Doing it again?

Answer 9

@SithsAndGiggles I apologize for not being active, but thank you for your help! One quick question: I messed up when tying the problem, and it should be square root of y in the denominator. Would my u, du, v and dv values be the same (except changed for the square root)?

Answer 10

*typing

Answer 11

That would make the resulting integral a bit easier, actually.
\[\begin{matrix}u=\ln y&&&dv=\frac{dy}{\sqrt y}\\
du=\frac{dy}{y}&&&v=2\sqrt y\end{matrix}\]
so then
\[\int_4^9\frac{\ln y}{\sqrt y}\,dy=\left[2\sqrt y\ln y\right]_4^9-2\int_4^9\frac{dy}{\sqrt y}\]

Answer 12

I think the thing that's confusing me is how to choose the dv and v values for this particular problem..

Answer 13

Like I said before, the \(u\) term tends to be the logarithm if you're given an integral involving a logarithm. Powers of \(y\) are much easier to integrate than logarithms of \(y\).

Also, regarding the question you had earlier (I know it's after the fact, but I think it's an important enough detail to talk about):
\[\int_4^9\frac{\ln y}{y}\,dy=\left[(\ln y)^2\right]_4^9-\int_4^9\frac{\ln y}{y}\,dy\]
Notice that you have the same two terms on both sides of the equation. It's like solving the equation
\[x=1-x\]
for \(x\). In this case, you would have done the following:
\[\begin{align*}\int_4^9\frac{\ln y}{y}\,dy&=\left[(\ln y)^2\right]_4^9-\int_4^9\frac{\ln y}{y}\,dy\\\\
2\int_4^9\frac{\ln y}{y}\,dy&=\left[(\ln y)^2\right]_4^9\\\\
\int_4^9\frac{\ln y}{y}\,dy&=\frac{1}{2}\left[(\ln y)^2\right]_4^9
\end{align*}\]
Notice that you would get the same result had you used the substitution, \(t=\ln y\), which gives \(dt=\dfrac{dy}{y}\), and so
\[\int_4^9\frac{\ln y}{y}\,dy=\int_{\ln 4}^{\ln9} t\,dt=\frac{1}{2}\left[(\ln9)^2-(\ln4)^2\right]\]

Answer 14

Got it. Thank you for this explanation!
I'm unclear on another thing in your previous reply, where you got dy/sqrt(y). Why is that? When I did it and plugged in the v and du, I got sqrt(y)/y, so I'm not sure where I went wrong.

Answer 15

*after the integral

Answer 16

That's because \(\dfrac{\sqrt y}{y}\) can be reduced to \(\dfrac{1}{\sqrt y}\).

Recall that \(\dfrac{1}{\sqrt2}\times\dfrac{\sqrt 2}{\sqrt2}=\dfrac{\sqrt2}{2}\). The same principle is at play here.

Answer 17

Awesome. Thank you so much for everything :)

Answer 18

yw!