# Finding exact area intergration?

##
Evaluate

#int_-1^4 x(3 − x) dx# .

b) Find the exact area of the shaded region in the figure.

Evaluate

b) Find the exact area of the shaded region in the figure.

##### 3 Answers

#### Explanation:

We have,

Let's Evaluate

So,

And,

So, The Actual Integral :-

so, The Area Under the Curve =

Hope this helps.

The area is

#### Explanation:

The area is

The total "actual area" is

#### Explanation:

We seek that "actual" area bounded by the curve

We will split this into

First, for convenience, consider the general case:

# I(a,b) = int_a^b \ x(3 − x) \ dx #

# \ \ \ \ \ \ \ \ \ \ = int_a^b \ 3x-x^2 \ dx #

# \ \ \ \ \ \ \ \ \ \ = [3/2x^2-1/3x^3]_a^b \ #

# \ \ \ \ \ \ \ \ \ \ = 3/2(b^2-a^2)-1/3(b^3-a^3)#

Then we consider the three regions separately:

# I(-1,0) = 3/2(0-1)-1/3(0+1) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -3/2-1/3 = -11/6#

# I(0,3) = 3/2(9-0)-1/3(27-0) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 27/2-9 = 9/2#

# I(3,4) = 3/2(16-9)-1/3(64-27) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 21/2-37/3 = -11/6 # (expected via symmetry)

Then the total "actual area" is:

# A = |I(-1,0)| + |I(0,3)| + |I(3,4)|#

# \ \ \ = |-11/6| + |9/2| + |-11/6|#

# \ \ \ = 11/6 + 9/2 + 11/6#

# \ \ \ = 49/6#

Note this is not the same as the "net" area, where area below the axis is counted negatively, which results in the incorrect answer:

# I(-1,4) = int_(-1)^4 \ x(3 − x) \ dx = 5/6#