Answer to Question #114634 in Calculus for ANJU JAYACHANDRAN

Question

Let A denote the area of the graph of f(x)=rootx on [0, 1] and let B denote the area of the graph of f(x)=x^2 on [0, 1]. Explain geometrically why A+B=1

Answer 1

Let's consider the graph of $y = \sqrt{x}$ for $x \in [0,1].$(green on the pic.2 and pic.1) Since $y = \sqrt{x}$ , and $y \ge 0, x \ge 0$ , we obtain that:

$y^2 = x$.

Let's also notice, that there are two points of intersection of these functions, namely: $(0,0) , (1,1)$ . Thus, graphs of given functions $\in$ square with the side = 1. Its vertices are:$(0,0), (1,0), (1,1), (0,1)$

Thus if we consider $x$ as a function and $y$ as a variable, the area under the graph $x = y^2$(where y is a variable), which is equal to $S$, according to the pic.2, is also equal(since these graphs are for same functions on same intervals) to the area of graph $y = x^2$ (but with x as a variable)(NOT shaded part of not numbered picture).

So, according to the pic.2, A+S is the area of the square, mentioned above. But S is, correspondingly to the task, equal to B, and the area is obviously equal to $1 \cdot 1 = 1$ .

Hence, A+B = 1.