Let f() denote the area of the region bounded by x=0, x=1, y^2=x and y = | x - 5| - |1- x|…

Question

Let $f(\alpha)$ denote the area of the region bounded by $x=0$, $x=1$, $y^2=x$ and $y = |\alpha x - 5| - |1-\alpha x| + \alpha x^2$. Then $\bigl(f(0)+f(1)\bigr)$ is equal to:

Accepted Answer

At $\alpha = 0$: $y = |{-5}| - |1| + 0 = 4$. Region bounded by $x=0,x=1,y=\sqrt{x},y=4$:

$$f(0) = \int_0^1(4-\sqrt{x})\,dx = \left[4x-\frac{2}{3}x^{3/2}\right]_0^1 = \frac{10}{3}$$

At $\alpha=1$: for $0\le x\le1$, $|x-5|=5-x$, $|1-x|=1-x$, so $y=(5-x)-(1-x)+x^2 = 4+x^2$.

$$f(1) = \int_0^1(4+x^2-\sqrt{x})\,dx = \left[4x+\frac{x^3}{3}-\frac{2}{3}x^{3/2}\right]_0^1 = \frac{11}{3}$$

$$f(0)+f(1) = \frac{10}{3}+\frac{11}{3} = 7$$

Solutions · 1