Calculate the area

[canvas]

Calculate the area cut by the given lines:

[math]x=-1,\,\,\,y=0,\,\,\,y=\arctan(x),\,\,\,x=\sqrt{3}\\\\A=\int_{-1}^0[0-\arctan(x)]dx+\int_0^{\sqrt{3}}[\arctan(x)-0]dx=\frac{\pi}{4}-\frac{3}{2}\ln(2)+\frac{\sqrt{3}\pi}{3}\\\\\text{but looking at the graph, we see that:}\\\\\int_{-1}^0[0-\arctan(x)]dx+\int_{0}^1[\arctan(x)-0]dx=0\\\text{because}\,\,\arctan(x)\,\,\text{is an odd function}\\\text{so the area should be:}\\\\A=\int_{1}^{\sqrt{3}}[\arctan(x)-0]dx=\frac{\sqrt{3}\pi}{3}-\frac{1}{2}\ln(2)-\frac{\pi}{4}\\\\\text{Finally we got two different aswers, can you guys point my mistake?}[/math]

 

[BigBeachBanana]

You're integrating [imath]-\arctan(x)[/imath] for [imath]P_1[/imath] and [imath]+\arctan(x)[/imath] for [imath]P_2[/imath].
[math]-P_1=P_2\implies -(P_1)+P_2=2*P_2[/math] over the interval [imath](-1,1)[/imath].

 

[canvas]

Yes I do, we know that integral over the interval [-a, a] from an odd function equals 0, change my mind

 

[BigBeachBanana]

Yes I do, we know that integral over the interval [-a, a] from an odd function equals 0, change my mind

[math]\int_{-1}^{0} \arctan(x)\,dx + \int_{0}^{1} \arctan(x)\,dx =-P+P=0\\ \int_{-1}^{0} \red{-}\arctan(x)\,dx + \int_{0}^{1} \arctan(x)\,dx =\red{-}(-P)+P=2P\\[/math]

 

[canvas]

I'm an idiot seriously or maybe it's too late, 1 a.m. my time, I'm sorry for asking that trivial questions...

 

[BigBeachBanana]

I'm an idiot seriously or maybe it's too late, 1 a.m. my time, I'm sorry for asking that trivial questions...

Proof that even intelligent people sometimes make mistakes, or we shouldn't do math at 1 am.