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#