MHB Raj's integration questions via Facebook

[Prove It]

1. Find the area enclosed between $\displaystyle \begin{align*} y = 6 - x^2 \end{align*}$ and $\displaystyle \begin{align*} y = 3 - 2\,x \end{align*}$.

2. Find the area enclosed between $\displaystyle \begin{align*} y = \sqrt{4 - x^2} \end{align*}$ and the line $\displaystyle \begin{align*} x - y + 2 = 0 \end{align*}$.

1. The graphs intersect where the functions are equal, so

$\displaystyle \begin{align*} 6 - x^2 &= 3 - 2\,x \\ 0 &= x^2 - 2\,x - 3 \\ 0 &= \left( x - 3 \right) \left( x + 1 \right) \\ x &= 3 \textrm{ or } x = -1 \end{align*}$

The higher function is $\displaystyle \begin{align*} y = 6 - x^2 \end{align*}$ (check with a graph if you like), so the area is

$\displaystyle \begin{align*} A &= \int_{-1}^3{ \left[ \left( 6 - x^2 \right) - \left( 3 - 2\,x \right) \right] \,\mathrm{d}x } \\ &= \int_{-1}^3{ \left( 3 + 2\,x - x^2 \right) \,\mathrm{d}x } \\ &= \left[ 3\,x + x^2 - \frac{x^3}{3} \right] _{-1}^3 \\ &= \left[ 3 \left( 3 \right) + 3^2 - \frac{3^3}{3} \right] - \left[ 3\left( -1 \right) + \left( -1 \right) ^2 - \frac{ \left( -1 \right) ^3}{3} \right] \\ &= 9 - \left( -3 + 1 + \frac{1}{3} \right) \\ &= 9 - \left( -2 + \frac{1}{3} \right) \\ &= 9 - \left( -\frac{5}{3} \right) \\ &= \frac{27}{3} + \frac{5}{3} \\ &= \frac{32}{3} \,\textrm{units}^2 \end{align*}$2. The graphs intersect where the functions are equal, and the second can be rewritten as $\displaystyle \begin{align*} y = x + 2 \end{align*}$ so

$\displaystyle \begin{align*} \sqrt{4 - x^2} &= x + 2 \\ 4 - x^2 &= \left( x + 2 \right) ^2 \\ 4 - x^2 &= x^2 + 4\,x + 4 \\ 0 &= 2\,x^2 + 4\,x \\ 0 &= 2\,x \left( x + 2 \right) \\ x &= 0 \textrm{ or } x = -2 \end{align*}$

The top function is a semicircle centred at the origin of radius 2 units. The line cuts off the right angle triangle with base and height of 2 units. So the area we want is

$\displaystyle \begin{align*} A &= \frac{\pi \cdot 2^2}{4} - \frac{2 \cdot 2}{2} \\ &= \left( \pi - 2 \right) \,\textrm{units}^2 \end{align*}$

 

[chwala]

1. The graphs intersect where the functions are equal, so

$\displaystyle \begin{align*} 6 - x^2 &= 3 - 2\,x \\ 0 &= x^2 - 2\,x - 3 \\ 0 &= \left( x - 3 \right) \left( x + 1 \right) \\ x &= 3 \textrm{ or } x = -1 \end{align*}$

The higher function is $\displaystyle \begin{align*} y = 6 - x^2 \end{align*}$ (check with a graph if you like), so the area is

$\displaystyle \begin{align*} A &= \int_{-1}^3{ \left[ \left( 6 - x^2 \right) - \left( 3 - 2\,x \right) \right] \,\mathrm{d}x } \\ &= \int_{-1}^3{ \left( 3 + 2\,x - x^2 \right) \,\mathrm{d}x } \\ &= \left[ 3\,x + x^2 - \frac{x^3}{3} \right] _{-1}^3 \\ &= \left[ 3 \left( 3 \right) + 3^2 - \frac{3^3}{3} \right] - \left[ 3\left( -1 \right) + \left( -1 \right) ^2 - \frac{ \left( -1 \right) ^3}{3} \right] \\ &= 9 - \left( -3 + 1 + \frac{1}{3} \right) \\ &= 9 - \left( -2 + \frac{1}{3} \right) \\ &= 9 - \left( -\frac{5}{3} \right) \\ &= \frac{27}{3} + \frac{5}{3} \\ &= \frac{32}{3} \,\textrm{units}^2 \end{align*}$2. The graphs intersect where the functions are equal, and the second can be rewritten as $\displaystyle \begin{align*} y = x + 2 \end{align*}$ so

$\displaystyle \begin{align*} \sqrt{4 - x^2} &= x + 2 \\ 4 - x^2 &= \left( x + 2 \right) ^2 \\ 4 - x^2 &= x^2 + 4\,x + 4 \\ 0 &= 2\,x^2 + 4\,x \\ 0 &= 2\,x \left( x + 2 \right) \\ x &= 0 \textrm{ or } x = -2 \end{align*}$

The top function is a semicircle centred at the origin of radius 2 units. The line cuts off the right angle triangle with base and height of 2 units. So the area we want is

$\displaystyle \begin{align*} A &= \frac{\pi \cdot 2^2}{4} - \frac{2 \cdot 2}{2} \\ &= \left( \pi - 2 \right) \,\textrm{units}^2 \end{align*}$

Your steps are correct! In addition to this i would always require my students to show step-step working to solution...in general they ought to start with formula for finding Area bound by given functions i.e $A=\int_a^b f(x) dx$ ...before plugging in the values.