Question

Why is it wrong to say that

Answer 1

\[\int\limits_{-1}^{1} \sqrt{1+ x ^{2}} dx = \int\limits_{2}^{2} \sqrt{u} du\]
using substitution of \[u = 1 + x^{2}\] ?

Answer 2

Well first off don't forget areas are not the same for \(u\) $$du/dx=2x\,dx$$

Answer 3

the problem is the \[\int\limits_{2}^{2}\] the rest is not the problem

Answer 4

because \[\int\limits_{2}^{2}\] of anything is 0

Answer 5

The problem is that \(u=1+x^2\) implicitly defines \(x(u)\) i.e. \(x\) as a function of \(u\), but on our interval of integration there is no single function of \(x\) that works. You can't substitute with both \(x(u)=\pm\sqrt{u-1}\)

Answer 6

thx a lot :D solved ^^

Answer 7

The correct method is to split in two parts first:$$\int_{-1}^0\sqrt{1+x^2}\,dx+\int_0^1\sqrt{1+x^2}\,dx$$On our first, let \(x(u)=-\sqrt{u-1}\) so our boundaries become:$$x=0\implies u=1\\x=-1\implies u=2$$Note that the factor by which our areas are now 'distorted' since we've changed our coordinate system to \(u\) is just the derivative:$$x'(u)=-\frac1{2\sqrt{u-1}}$$We fix up our first integral:$$\int_2^1\sqrt{1+(-\sqrt{u-1})^2}\cdot-\frac1{2\sqrt{u-1}}du=\frac12\int_1^2\sqrt{\frac{u}{u-1}}\ du$$For our second, let \(x(u)=\sqrt{u-1}\) and thus $$x=0\implies u=1\\x=1\implies u=2$$Our derivative is almost identical to the prior, \(x'(u)=1/(2\sqrt{u-1})\):$$\frac12\int_1^2\sqrt{\frac{u}{u-1}}\,du$$Summing them we get just:$$\int_1^2\sqrt{\frac{u}{u-1}}\,du$$

Answer 8

note this is not an easy integral to do without further change of variables!

Answer 9

the easier way would be to call x=tan y and x^2+1 = sec and integrated from there >.<

Answer 10

sec^2 there

Answer 11

right I was just showing you the proper way to do a substitution here

Answer 12

i understood thx a lot ^^