# Raj's integration questions via Facebook

• MHB
• Prove It
In summary, the area enclosed between the two functions \displaystyle \begin{align*} y = 6 - x^2 \end{align*} and \displaystyle \begin{align*} y = 3 - 2\,x \end{align*} is found by first finding the points of intersection between the two functions, which are at \displaystyle \begin{align*} x = 3 \textrm{ and } x = -1 \end{align*}. The area is then calculated by taking the integral of the higher function minus the lower function from the point of intersection at $x = -1$ to the point of intersection at $x = 3$. This
Prove It
Gold Member
MHB
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*}

Prove It said:
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.

Greg Bernhardt

## 1. What is the purpose of Raj's integration questions via Facebook?

The purpose of Raj's integration questions via Facebook is to gather data and insights from a wider audience of people using the platform. By posting integration questions on Facebook, Raj is able to reach a larger and more diverse group of individuals and get a better understanding of their thoughts and opinions.

## 2. How often does Raj post integration questions on Facebook?

Raj typically posts integration questions on Facebook once a week. This allows for a consistent flow of data without overwhelming followers with too many questions at once.

## 3. Are the integration questions on Facebook related to a specific topic?

No, the integration questions are not limited to a specific topic. Raj may post questions related to various subjects, such as technology, social issues, or current events. This allows for a diverse range of responses and insights.

## 4. Can anyone participate in Raj's integration questions on Facebook?

Yes, anyone with a Facebook account can participate in Raj's integration questions. The questions are open to all individuals, regardless of their background or expertise. Raj encourages people from all walks of life to share their thoughts and opinions.

## 5. How does Raj use the data collected from integration questions on Facebook?

Raj uses the data collected from integration questions on Facebook to gain insights and understand the perspectives of a larger audience. This information can be used for research, product development, or to inform decision-making in various fields. Raj may also share the data with others to promote further discussion and analysis.

• General Math
Replies
4
Views
11K
• General Math
Replies
4
Views
10K
• General Math
Replies
1
Views
9K
• General Math
Replies
1
Views
9K
• General Math
Replies
1
Views
10K
• General Math
Replies
1
Views
10K
• General Math
Replies
2
Views
11K
• General Math
Replies
1
Views
10K
• General Math
Replies
1
Views
10K
• General Math
Replies
2
Views
10K