Question

Let $f:\left[–3,1\right]\to R$ be given as

$f\left(x\right)=\left\{\begin{array}{ll}\text{min}\left[\left(x+6,x^2\right)\right],& -3\le x\le 0\\ \text{max}\left[\sqrt{x},x^2\right]& 0\le x\le 1\end{array}\right.$

If the area bounded by $y=f\left(x\right)$ and $x-\text{axis}$ is $A$, then the value of $6A$ is equal to

Answer 1

Finding the value of $6A$using the given function:

Consider the given equation as,

$f\left(x\right)=\left\{\begin{array}{ll}\text{min}\left[\left(x+6,x^2\right)\right],& -3\le x\le 0\\ \text{max}\left[\sqrt{x},x^2\right]& 0\le x\le 1\end{array}\right.$

From the above given data, we can construct the $f\left(x\right)$ as in the image given.

If the area bounded by $y=f\left(x\right)$ and $x-\text{axis}$ is $A$

The area of bonded is

_-$$\begin{gathered} A={\int }_{-3}^{-2}\left(x+6\right)dx+{\int }_{-2}^0{x}^{2}dx+{\int }_0^1\sqrt{x}dx \\ A={\left[\left(\frac{x^2}{2}+6\right)\right]}_{-3}^{-2}+{\left[\frac{x^3}{3}\right]}_{-2}^0+{\left[\left(\frac{2}{3}\right)x^{\frac{3}{2}}\right]}_0^1\left[\because \frac{d{x}^n}{dx}=n{x}^{n-1}\right] \\ A=\left(\frac{{\left(-2\right)}^2}{2}-\frac{{\left(-3\right)}^2}{2}+6\right)+\left(\frac{{\left(2\right)}^3}{3}\right)+\left(\frac{2}{3}\right)1^{\frac{3}{2}} \\ A=\left(\frac{4-9+12}{2}\right)+\left(\frac{8}{3}\right)+\left(\frac{2}{3}\right) \\ A=\left(\frac{7}{2}\right)+\left(\frac{8}{3}\right)+\left(\frac{2}{3}\right) \\ A=\left(\frac{21+16+4}{6}\right) \\ A=\frac{41}{6} \\ 6A=41 \end{gathered}$$

Hence, the value of $6A$ is equal to $41$.