Modulus + Integral

Calculus Level 3

$\ Let \ S = \int^{+\sqrt{2}}_{-\sqrt{2}} (1- |1-x^2|) dx$ S can be written in the form $\frac{a \sqrt{2} +b}{c}$ , where a,b, c are integers and the fraction is in its simplest form. What is the value of a+b+c?

Assumption and details:

1) |x| represents the modulus (distance from 0) of x

2) An example of a fraction in its simplest form: $\frac{17 \sqrt{2} +10}{9}$ as 17 and 9 are coprime as well as 10 and 9.

The answer is 3.

1 solution

Curtis Clement
Jul 29, 2015

For $\ |x| \leq\ 1$ the curve behaves like $\ x^2$ and for $\ |x| \geq\ 1$ the curve behaves like $\ 2-x^2$ . Separating the integral will give: $\ S = 2 \int_{0}^{\sqrt{2}} (1-|1-x^2|)dx$ $\ = 2[ \int_{0}^{1} x^2 dx + \int_{1}^{\sqrt{2}} (2-x^2)dx ]$ $\ = 2 [ \frac{x^3}{3} ]_0^1 +2[ 2x-\frac{x^3}{3}]_1^{\sqrt{2}}$ $\ = \frac{2}{3} + \frac{8 \sqrt{2} -10}{3} = \boxed{ \frac{8 \sqrt{2} -8}{3}}$ $\therefore\ a+b+c = 8-8+3 = 3$

Please elaborate the whole calculation, would you? Because I was using Wolfram to solve it. I was out of idea. Thank you

Math Mediocre - 5 years, 9 months ago

Modulus + Integral

The answer is 3.

1 solution

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