Area Under Curves Question 10
Question 10 - 31 January - Shift 1
Let for $x \in R$
$f(x)=\frac{x+|x|}{2}$ and $g(x)={\begin{matrix} x, & x<0 \\ x^{2} & x \geq 0\end{matrix} .$.
Then area bounded by the curve $y=(f \circ g)(x)$ and the lines $y=0,2 y-x=15$ is equal to
Show Answer
Answer: 72
Solution:
Formula: Area between two curves - Area enclosed between two curves intersecting at one point and the $x$-axis.
$f(x)=\frac{x+|x|}{2} $
Or
$f(x)={\begin{matrix} x, & x \geq 0 \\ 0 & x < 0 \end{matrix} $
$g(x)={\begin{matrix} x, & x<0 \\ x^{2} & x \geq 0\end{matrix}$
$f o g(x)=f[g(x)] ={ \begin{matrix} g(x), & x \geq 0 \\ 0, & g(x)<0 \end{matrix}$
$g(x)={\begin{matrix} 0, & x<0 \\ x^{2}, & x \geq 0\end{matrix}$
$2 y-x=15$
$A=\int_0^{3}(\frac{x+15}{2}-x^{2}) dx+\frac{1}{2} \times \frac{15}{2} \times 15$
$\frac{x^{2}}{4}+\frac{15 x}{2}-.\frac{x^{3}}{3}|_0 ^{3}+\frac{225}{4}$
$=\frac{9}{4}+\frac{45}{2}-9+\frac{225}{4}=\frac{99-36+225}{4}$
$=\frac{288}{4}=72$
