# Integration

#### skeeter

##### Elite Member
the coordinates of P and Q are located at two of the three values of x that make y = 0. What values of x make y = 0?

Surely you are capable of expanding the factored form of the function?

Try to set up two integral expressions that determine the shaded area ... show us what you think is the way to go.

#### Honeypuffmonster

##### New member
the coordinates of P and Q are located at two of the three values of x that make y = 0. What values of x make y = 0?

Surely you are capable of expanding the factored form of the function?

Try to set up two integral expressions that determine the shaded area ... show us what you think is the way to go.

Ok, I know the coordinates for P are (-3,0) and Q (4,0) and when I expand and simplify the expression I get x^3-x^2-12x. I'm just having difficulty finding the area of both curves. I'm ok when doing integration with both points but after that I got a bit confused. It would be great if you could explain how I can achieve that.

#### Subhotosh Khan

##### Super Moderator
Staff member
Ok, I know the coordinates for P are (-3,0) and Q (4,0) and when I expand and simplify the expression I get x^3-x^2-12x. I'm just having difficulty finding the area of both curves. I'm ok when doing integration with both points but after that I got a bit confused. It would be great if you could explain how I can achieve that.

#### Honeypuffmonster

##### New member
I currently do not have a phone with me to take a picture of my work but I know how to find the coordinates by using the the curve y=x(x+3)(x-4). So when y=0, 3+x=0 and 4-x=0 so x can either be -3 or 4. Expanding and simplifying the equation of the curve, you multiply the brackets out and after you collect and gather the like terms and get =x^3-x^2-12x. Is that ok?

#### skeeter

##### Elite Member
1. post the definite integral that evaluates to the area of A
don't evaluate it

2. do the same for area B

let's see what you got ...

#### Honeypuffmonster

##### New member
1. post the definite integral that evaluates to the area of A
don't evaluate it

2. do the same for area B

let's see what you got ...
So you mean for area A, I would make -3 the lower bound and 0 the upper bound while integrating (x^3-x^2-12x)dx. If so the answer I got is 99/4 and for area B, the lower bound being 4 and upper bound 0, I got 160/3. Is this correct?

#### Subhotosh Khan

##### Super Moderator
Staff member
So you mean for area A, I would make -3 the lower bound and 0 the upper bound while integrating (x^3-x^2-12x)dx. If so the answer I got is 99/4 and for area B, the lower bound being 4 and upper bound 0, I got 160/3. Is this correct?
Skeeter said:
1. post the definite integral that evaluates to the area of A
don't evaluate it

2. do the same for area B

#### skeeter

##### Elite Member
• Honeypuffmonster

#### skeeter

##### Elite Member
total area = [MATH]\int_{-3}^0 f(x) \, dx - \int_0^4 f(x) \, dx = \frac{99}{4} + \frac{160}{3} = \frac{937}{12}[/MATH]
or

total area = [MATH]\int_{-3}^4 |f(x)| \, dx = \frac{937}{12}[/MATH]

• Honeypuffmonster

#### HallsofIvy

##### Elite Member
Notice that region B is below the x-axis. That means the integral will be negative while are is always positive. That is why skeeter subtracts that integral instead of adding it.

• Honeypuffmonster

#### Honeypuffmonster

##### New member
total area = [MATH]\int_{-3}^0 f(x) \, dx - \int_0^4 f(x) \, dx = \frac{99}{4} + \frac{160}{3} = \frac{937}{12}[/MATH]
or

total area = [MATH]\int_{-3}^4 |f(x)| \, dx = \frac{937}{12}[/MATH]
Notice that region B is below the x-axis. That means the integral will be negative while are is always positive. That is why skeeter subtracts that integral instead of adding it.

Thank you for helping me, I finally understand 