Let f( ) denote the area of the region in the first quadrant bounded by x=0,x=1,y^2=x and …

Question

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

Accepted Answer

Given Curves ${y}^{2}=x,y=|\alpha x-5|-|1-\alpha x|+\alpha {x}^{2}x=0$ and $x=1$

The area of region bounded by these curve is $f(\alpha )$.

To find $f(0)+f(1)$.

for $\alpha =0$

$\begin{gathered} y=|0	imes x-5|-|1-0	imes x|+0	imes {x}^{2} \ y=5-1=4 \end{gathered}$

so the region bounded by $x=0,x=1,y=\sqrt{x}$

$\begin{gathered} y=4 \ \begin{gathered} f(0)={\int }_{1}^{0}(4-\sqrt{x})dx \ f(0)={[4x-\frac{2}{3}{x}^{3/2}]}_{1}^{0}=4-\frac{2}{3}=\frac{10}{3} \end{gathered} \end{gathered}$

If, When $\alpha =1$, the second equation becomes:

$y=|x-5|-|1-x|+{x}^{2}$

For the interval $0\le x\le 1$ :

$\begin{gathered} |x-5|=-(x-5)=5-x \ |1-x|=1-x \ y=(5-x)-(1-x)+{x}^{2}=4+{x}^{2} \end{gathered}$

The region is bounded by $x=0,x=1,y=\sqrt{x}$, and $y=4+{x}^{2}$ as shown in graph is.

$\begin{gathered} f(1)={\int }_{1}^{0}(4+{x}^{2}-\sqrt{x})dx \ f(1)={[4x+\frac{{x}^{3}}{3}-\frac{2}{3}{x}^{3/2}]}_{1}^{0}=4+\frac{1}{3}-\frac{2}{3}=4-\frac{1}{3}=\frac{11}{3} \end{gathered}$

So, $(f(0)+f(1))=\frac{10}{3}+\frac{11}{3}=\frac{21}{3}=7$

Solutions · 1