Let f and g be defined over the interval [a, b] with g(x) ≤ f(x) for all x in [a, b] then the area A of the region bounded by there two curves and the lines x = a and x = b is given by
Sketch the graphs , shade the bounded region and find the area bounded by the given expressions.
Problem 1 :
y = x^{2} + 1, y = x, x = -1 and x = 2
Solution:
So, the required area is 4.5 square units.
Problem 2 :
y = √x and y = x/4
Solution:
Let us find the point of intersection,
y = √x ---> (1)
y = x/4 ---> (2)
(1) = (2)
So, the required area is 32/3 square units.
Problem 3 :
x = 1/y^{2}, y = x and y = 2
Solution:
Let us find the point of intersection,
x = 1/y^{2} ---> (1)
y = x ---> (2)
(1) = (2)
1/y^{2} = y
y^{3} = 1
y = 1
So, the required area is 1.5 square units.
Problem 4 :
x = 2 - y^{2}, y = -x
Solution:
Let us find the point of intersection,
x = 2 - y^{2} ---> (1)
y = -x ---> (2)
(1) = (2)
-y = 2 - y^{2}
y^{2} - y - 2 = 0
(y - 2) (y + 1) = 0
y = -1 and y = 2
So, the required area is 4.5 square units.
Problem 5 :
y = x^{2} and y = 4x
Use vertical rectangles (dx).
Solution:
Let us find the point of intersection,
y = x^{2} ---> (1)
y = 4x ---> (2)
(1) = (2)
4x = x^{2}
x^{2} - 4x = 0
x(x - 4) = 0
x = 0 and x = 4
So, the required area is 32/3 square units.
Problem 6 :
y = x^{2} and y = 4x
Use horizontal rectangles (dy).
Solution:
Let us find the point of intersection,
y = x^{2}
x = √y ---> (1)
y = 4x
x = 1/4 y ---> (2)
(1) = (2)
So, the required area is 32/3 square units.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM