Chapter 8.2, Problem 12E

To determine

To find the area enclosed by the graphs of the two curves.

Answer to Problem 12E

The area enclosed by the graphs of the two curves is $\frac{343}{48}$ squared units.

Explanation of Solution

Concept used: Area between curves.

If f and g are continuous with $f\left(x\right)\ge g\left(x\right)$ throughout [a,b], then the area between

The curves $y=f\left(x\right)\text{ and }y=g\left(x\right)$ from a to b is the integral of $\left[f-g\right]$ from ato b.

  $\begin{array}{l}A={\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]}dx\\ \end{array}$

Given information:

The given function is, $y^2=x+3\text{ , }y=2x.$

Calculation: The area enclosed by the graphs of the two curves is obtained as,

  

Set up each equation in term of y,

  $\begin{array}{l}{y}^2=x+3y=2x\\ x=y^2-3x=\frac{y}{2}\end{array}$

Set both the equations equal to each other,

  $\begin{array}{l}{y}^2-3=\frac{y}{2}\left(\text{solve for the limits}\right)\\ 2y^2-y-6=0\\ \left(2y+3\right)\left(y-2\right)=0\\ y=-\frac{3}{2},2\end{array}$

The curves $y^2=x+3$ and $y=2x$ intersect at $y=-\frac{3}{2},2.$

Now, area bounded by these region is,

  $\begin{array}{l}\text{ A}={\displaystyle {\int }_a^b\left[f\left(t\right)-g\left(t\right)\right]}dt\\ ={\displaystyle {\int }_{-3/2}^2\left[\frac{y}{2}-\left(y^2-3\right)\right]}dy\\ ={\left[\frac{y^2}{4}-\frac{1}{3}{y}^3+3y\right]}_{-3/2}^2\\ =\left[\frac{{\left(2\right)}^2}{4}-\frac{1}{3}{\left(2\right)}^3+3\left(2\right)\right]-\left[\frac{{\left(-3/2\right)}^2}{4}-\frac{1}{3}{\left(-3/2\right)}^3+3\left(-3/2\right)\right]\\ =1-\frac{8}{3}+6-\left(\frac{9}{16}+\frac{9}{8}-\frac{9}{2}\right)\\ =4\sqrt{2}+4\sqrt{2}-\frac{4\sqrt{2}}{3}-\frac{4\sqrt{2}}{3}\\ =\frac{343}{48}\end{array}$

Hence, the area enclosed by the graphs of the two curves is $\frac{343}{48}$ squared units.