A calculus problem by Dharani Chinta

Calculus Level 4

Let $y=g(x)$ be the image of $f(x)=x+\sin x$ about the line $x+y=0$ .

If the area bounded by $y=g(x)$ , $x$ -axis, $x=0$ and $x=2 \pi$ is $A$ , find $\lfloor A \rfloor$ .

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

1 solution

Tom Engelsman
Apr 26, 2020

The image of $f(x) = x + \sin(x)$ about $y = -x$ is just $g(x) = -f(x) = -x - \sin(x)$ . The required area then computes to:

$A = \int_{0}^{2\pi} 0 - g(x) dx = \int_{0}^{2\pi} f(x) dx = \int_{0}^{2\pi} x + \sin(x) dx = \frac{1}{2}x^2 - \cos(x)|_{0}^{2\pi} = 2\pi^{2}.$

Hence $\lfloor A \rfloor = \lfloor 2\pi^{2} \rfloor = \boxed{19}.$