Endless Embedded Integrals

Calculus Level 2

a = ∫ a 1 x d x \large a=\int_{a}^{1}x\ dx

Which of the following is a solution of a a ?

2 − 1 \sqrt{2}-1 No solution 1 1 2 \sqrt{2} 2 2

Nicholas Walters by Nicholas Walters , 21, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Jordan Cahn
Mar 19, 2019

a = ∫ a 1 x d x a = [ x 2 2 ] x = a 1 a = 1 2 ( 1 − a 2 ) a 2 + 2 a − 1 = 0 a = − 2 ± 8 2 a = − 1 ± 2 \begin{aligned} a &= \int_a^1 x\mathrm{d}x \\ a &= \left[\frac{x^2}{2}\right]_{x=a}^1 \\ a &= \frac{1}{2}(1-a^2) \\ a^2+2a-1 &= 0 \\ a &= \frac{-2\pm\sqrt{8}}{2} \\ a &= -1\pm\sqrt{2} \end{aligned}

Although there are two possible values for a a , only one, − 1 + 2 = 2 − 1 -1+\sqrt{2} = \boxed{\sqrt{2}-1} appears as an answer choice.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

But they are both valid solutions or not?

Peter van der Linden - 2 years, 2 months ago

Log in to reply

Both are valid, but the answer options only list one of them.

Jordan Cahn - 2 years, 2 months ago

Thanks. I've updated the problem statement to reflect this.

In the future, if you have concerns about a problem's wording/clarity/etc., you can report the problem. See how here .

Brilliant Mathematics Staff - 2 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...