Question

Find the intersection points when the line y = 12-4x meets the curve y =12-x^3. Hence, sketch in the same diagram, the line y = 12 - 4x and the curve y=12-x^3 Calculate the area of the region bounded by the curve and the line

Answer 1

The equations of the line and the curve are

`\begin{gathered} y=12-4x \\ y=12-x^3 \end{gathered}`

We will equate their right sides to find the values of x

`12-4x=12-x^3`

Subtract 12 from both sides

`\begin{gathered} 12-12-4x=12-12-x^3 \\ -4x=-x^3 \end{gathered}`

Divide both sides by -1

`\begin{gathered} (-4x)/(-1)=(-x^3)/(-1) \\ 4x=x^3 \end{gathered}`

Subtract 4x from both sides

`\begin{gathered} 4x-4x=x^3-4x \\ 0=x^3-4x \end{gathered}`

Switch the two sides

`x^3-4x=0`

Take x as a common factor

`\begin{gathered} x((x^3)/(x)-(4x)/(x))=0 \\ x(x^2-4)=0 \end{gathered}`

Factor the bracket using the rule of the difference between two squares

`(a^2-b^2)=(a+b)(a-b)`
`\begin{gathered} (x^2-4)=(x+2)(x-2) \\ x(x+2)(x-2)=0 \end{gathered}`

Now, equate each factor by 0

`\begin{gathered} x=0 \\ x+2=0,x-2=0 \\ x=-2,x=2 \end{gathered}`

Substitute the values of x in the equation of the line to find their corresponding values of y

`\begin{gathered} y=12-4(0)=12 \\ y=12-4(-2)=12+8=20 \\ y=12-4(2)=12-8=4 \end{gathered}`

The points of intersection between the line and the curve are

(0, 12), (-2, 20), (2, 4)

Let us draw the graph

The red line represents the equation of the line

The blue curve represents the equation of the curve

To find the area bounded between the line and the curve we will use the integration

Since the line is above the curve at the point (-2, 20), then

We will subtract the curve from the line and use the values of x -2 and 2 as the limits of integration

`A=\int_(-2)^2[(12-4x)-(12-x^3)]dx`

Simplify the terms inside the brackets first.

`\begin{gathered} A=\int_(-2)^2[12-4x+12+x^3]dx \\ A=\int_(-2)^2[-4x+x^3]dx \end{gathered}`

In integration, we will add the power by 1 and divide the term by the new power

`A=[(-4x^(1+1))/(1+1)+(x^(3+1))/(3+1)]_(-2)^2`

Simplify it

`\begin{gathered} A=[(-4x^2)/(2)+(x^4)/(4)]_(-2)^2 \\ A=[-2x^2+(x^4)/(4)]_(-2)^2 \end{gathered}`

Substitute x by 2 and -2, then subtract the answer

`A=[-2(2)^2+(2^4)/(4)]-[-2(-2)^2+((-2)^4)/(4)]`

Solve each bracket

`A=8`

The area of the region bounded between the line and the curve is 8 square unit