# 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 ?

$$\int_1^4 e^y \, dy$$

## 4 Answers

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 $$\int_1^4 \ln(x)\,dx=\int_{\ln(1)}^{\ln(4)}ue^u\,du.$$

• For way 2, why did the $1$ And $4$ changed to $\ln 1$ and $\ ln 4$ ? – user307640 Aug 20 '17 at 13:52
• Way 2 is used what is commonly called $u$-substitution. In this case $x=e^u$. The limits have to be adjusted to be the $u$ values at those points. – Ross Millikan Aug 20 '17 at 14:10
• A useless substitution imho. You will need to perform integration by parts anyway. – user370967 Aug 20 '17 at 14:55
• @Math_QED : Not necessarily if you know that a primitive is of the form $(ax+b)e^x$. But yes, normally we solve this type of integral by a integration by part. – Surb Aug 20 '17 at 15:23
• The principal flaw in this answer is that it treats it as if the question is how to find the area referred to in the first line of the question. But BELOW the illustration in the posted question you see the actual question. – Michael Hardy Aug 22 '17 at 3:34

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: $$x=e^y, 0\le y\le \ln 4, 1\le x \le 4.$$ $$\int_{y=0}^{y=\ln 4} 4-e^y \, dy=4\ln 4-3.$$

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: $$\int_0^{\ln 4} 4 \,dy - \int_0^{\ln 4} e^y\, dy = 4\ln 4 - 3.$$

• $\displaystyle \int_0^{\ln 4} e^y\, dy=4-1=3\neq 4\ln(4)-3$ – Pedro Gomes Aug 22 '17 at 20:09
• @PedroGomes : I'm afraid your point escapes me. It is routine to see that $\displaystyle \int_0^{\ln4} e^y \, dy = 4-1=3.$ But how is it relevant to anything discussed here to point out that that is not $4\ln4 - 3\text{ ?}$ Did someone claim it is $4\ln4-3\text{ ?} \qquad$ – Michael Hardy Aug 22 '17 at 21:44
• OP wants to find the area under the curve that according to the solution of the integral $\int_1^4 \ln(x)\,dx$ is $4\ln (4) -3$. The my doubt arises when you come and give the impression the same solution is attainable through $\displaystyle \int_0^{\ln4} e^y \, dy$. I am intrigued. What did you mean on your answer? – Pedro Gomes Aug 23 '17 at 20:14
• oh . . . ok, I'll transform my answer into a complete solution. – Michael Hardy Aug 23 '17 at 20:37
• Now I understood what you were thinking. – Pedro Gomes Aug 23 '17 at 21:16

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.