Flakqfqbq
2022-05-18
Answered

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

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pomastitxz27r

Answered 2022-05-19
Author has **16** answers

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,

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

So, with du=1, $v=\mathrm{arctan}x$ $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$ u=x, $\mathrm{d}v=1/(1+{x}^{2})$,

${\int}_{0}^{1}\mathrm{arctan}x\mathrm{d}x=[x\mathrm{arctan}x{]}_{0}^{1}-{\int}_{0}^{1}\frac{x}{1+{x}^{2}}\mathrm{d}x$

I'm assuming you can take it from here.

${\int}_{0}^{1}\mathrm{arctan}x\mathrm{d}x$

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

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

So, with du=1, $v=\mathrm{arctan}x$ $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$ u=x, $\mathrm{d}v=1/(1+{x}^{2})$,

${\int}_{0}^{1}\mathrm{arctan}x\mathrm{d}x=[x\mathrm{arctan}x{]}_{0}^{1}-{\int}_{0}^{1}\frac{x}{1+{x}^{2}}\mathrm{d}x$

I'm assuming you can take it from here.

Edith Mayer

Answered 2022-05-20
Author has **4** answers

The area of the figure you have described is:

${\int}_{0}^{\pi /4}\mathrm{tan}y\phantom{\rule{thinmathspace}{0ex}}dy={[-\mathrm{log}|\mathrm{cos}y|]}_{0}^{\pi /4}=\frac{1}{2}\mathrm{log}2.$

${\int}_{0}^{\pi /4}\mathrm{tan}y\phantom{\rule{thinmathspace}{0ex}}dy={[-\mathrm{log}|\mathrm{cos}y|]}_{0}^{\pi /4}=\frac{1}{2}\mathrm{log}2.$

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