Area under the graph $y=\ln x$
Find the Area of the shaded region
$y= \ln x \rightarrow x = e^y $
I found that the curve cuts the $x$ axis at $1$ through substituting $0$ to $y$
Why I can't find the area under the graph through this and how should I find it ?
$$ \int_1^4 e^y \, dy $$
$\endgroup$4 Answers
$\begingroup$Way 1
By part $$\int_1^4 \ln(x)\,dx=\int_1^4 \underbrace{1}_{u'(x)}\underbrace{\ln(x)}_{v(x)} \, dx= \left[ x \ln(x) \vphantom{\frac 11} \right]_1^4-\int_1^4dx=\cdots$$
Way 2
If you make the substitution $x=e^u$, then you get $$\int_1^4 \ln(x)\,dx=\int_{\ln(1)}^{\ln(4)}ue^u\,du.$$
$\endgroup$ 6 $\begingroup$Note:
1-method: $$y=\ln x, 1\le x \le 4, 0\le y\le \ln 4.$$ $$\int_{x=1}^{x=4} \ln x \, dx =4\ln 4 -3.$$
2-method: $$x=e^y, 0\le y\le \ln 4, 1\le x \le 4.$$ $$\int_{y=0}^{y=\ln 4} 4-e^y \, dy=4\ln 4-3.$$
$\endgroup$ $\begingroup$The problem with $\displaystyle \int_1^4 e^y\,dy$ is that the bounds are wrong. It should be $\displaystyle \int_0^{\ln 4} e^y\, dy,$ because, as $x$ goes from $1$ to $4,$ then $y$ goes from $0$ to $\ln 4.$
PS: It has been said in comments that this may misunderstand the question, so I will add this:
When integrating with respect to $y,$ and putting the $x$-and $y$-axes in the conventional positions, then "upward" is to the right in the graph, so what you're looking for is the area "below" (i.e. to the left of) the line $x=4$ and "above" the curve (i.e. to its right). Thus ultimately you need this: $$ \int_0^{\ln 4} 4 \,dy - \int_0^{\ln 4} e^y\, dy = 4\ln 4 - 3. $$
$\endgroup$ 5 $\begingroup$We first get an integral by parts:
$$\int_1^4 \ln(x)\,dx=\int_1^4 \ln(x)\times 1\, dx$$
Then applying integral by parts formula we get
$$\int_1^4 \ln(x)\times 1\,dx=\ln(x)x|_1^4-\int_1^4 \frac{1}{x} x \, dx=4\ln(4)-\ln(1)-4+1=\\=4\ln(4)-\ln(1)-3=4\ln(4)-3$$
Then you know as you pointed out:
$$\int_1^4 \ln(x)\,dx=\int_{\ln(1)}^{\ln(4)}ue^u\, du$$
Your mistake was on the bounds of the integral that are not 4 and 1 for the integral but $\ln(1)$ and $\ln(4)$. So $\int_\limits{1}^{4}ue^u \, du$ is wrong it is instead $\int_\limits{\ln(1)}^{\ln(4)}ue^u \, du$.
$u=\ln(x)\implies x=e^u$, we know that $4=e^u$ and $1=e^u\implies$ $\ln(4)=u$ and $\ln(1)=u$
That is the reason why the bounds change.
There is another mistake $\ln(x)=u\implies x=e^u$, then $\frac{dx}{du}=\implies dx=e^u \, du$, then we have finally $\int_\limits{\ln(1)}^{ln(4)}ue^u \, du$, that is easily solved with polar coordinates.
However we already know:
$$\int_1^4 \ln(x)\,dx=4\ln(4)-3 \text{ Therefore: } \int_{\ln(1)}^{\ln(4)}ue^u \, du=4\ln(4)-3 $$ which is easier.
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