# What will be the area which is bounded by x = 0 , y = &#x03C0;<!-- π --> <m

What will be the area which is bounded by $x=0,y=\pi /4$ and $y=\mathrm{arctan}\left(x\right)$?
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pomastitxz27r
The integral in the question diverges. Notice how the integrand approaches $\pi /2$ and $x\to \mathrm{\infty }$. However from the comments I'm guessing you mean,
${\int }_{0}^{1}\mathrm{arctan}x\mathrm{d}x$
(as $\mathrm{tan}\pi /4=1$) From here you can use integration by parts. Remember,
$\frac{\mathrm{d}\left(\mathrm{arctan}x\right)}{\mathrm{d}x}=\frac{1}{1+{x}^{2}}$
So, with du=1, $v=\mathrm{arctan}x$ $\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}$ u=x, $\mathrm{d}v=1/\left(1+{x}^{2}\right)$,
${\int }_{0}^{1}\mathrm{arctan}x\mathrm{d}x=\left[x\mathrm{arctan}x{\right]}_{0}^{1}-{\int }_{0}^{1}\frac{x}{1+{x}^{2}}\mathrm{d}x$
I'm assuming you can take it from here.
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Edith Mayer
The area of the figure you have described is:
${\int }_{0}^{\pi /4}\mathrm{tan}y\phantom{\rule{thinmathspace}{0ex}}dy={\left[-\mathrm{log}|\mathrm{cos}y|\right]}_{0}^{\pi /4}=\frac{1}{2}\mathrm{log}2.$