What will be the area which is bounded by $x=0,y=\pi/4$ and $y=\arctan (x)$?

Question

What will be the area which is bounded by $x=0,y=\pi/4$ and $y=\arctan (x)$? I know that I have to compute $$\int_{0}^{\infty} \arctan x\ dx$$ but how does it go in this case?

Answer 1

The area of the figure you have described is:

$$ \int_0^{\pi/4}\tan y\, dy=\left[-\log|\cos y|\right]_0^{\pi/4}={\small\frac{1}{2}}\log2. $$

Answer 2

The integral in the question diverges. Notice how the integrand approaches $\pi/2$ and $x \to \infty$. However from the comments I'm guessing you mean,

$$\int_0^1 \arctan x \mathrm dx$$

(as $\tan \pi/4= 1$) From here you can use integration by parts. Remember,

$$\frac{\mathrm d(\arctan x)}{\mathrm dx} = \frac 1 {1+x^2}$$

So, with $\mathrm du=1$, $v = \arctan x$ $\implies$ $u=x$, $\mathrm dv = 1/(1+x^2)$,

$$\int_0^1 \arctan x \mathrm dx = [x\arctan x]_0^1 - \int_0^1 \frac x {1+x^2} \mathrm dx$$

I'm assuming you can take it from here.

Answer 3

Your essentially trying to find the area of the first picture with the arctan(x) graph, which is the same as the area of the second picture with the tan(x) graph, soon enough both are just themselves reflected across y=x.

The area of the second graph is $\int_0^\frac{π}{4} tan(x) dx $

$\int tan(x) dx = ln(sec(x))+C$

$[ln(sec(x))]_0^\frac{π}{4}$

$ln(sec(\frac{π}4))-ln(sec(0))$

$ln(\sqrt{2})-ln(1)$

$\frac{ln(2)}2$

I know this is easy and has already received enough answers, but I just couldn’t resist.