AREA BETWEEN PARABOLA AND LINE USING INTEGRATION

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

area-between-two-curves

Sketch the graphs , shade the bounded region and find the area bounded by the given expressions.

Problem 1 :

y = x2 + 1, y = x, x = -1 and x = 2

Solution:

area-using-integration-q1
=2-1x2+1-(x) dx=13x3+x-12x22-1=13(2)3+2-12(2)2-13(-1)3+(-1)-12(-1)2=83+2-2--13-1-12=83+116=16+116=276=4.5

So, the required area is 4.5 square units.

Problem 2 :

y = √x and y = x/4

Solution:

area-using-integration-q2.png

Let us find the point of intersection,

y = √x ---> (1)

y = x/4 ---> (2)

(1) = (2)

x=x44x=x4=xx4=xx=0 and x=16
area-using-integration-q2-1.png
=160x-x4 dx=23x32-18x2160=23(16)32-18(16)2-[0]=1283-32=128-963=323

So, the required area is 32/3 square units.

Problem 3 :

x = 1/y2, y = x and y = 2

Solution:

Let us find the point of intersection,

x = 1/y2 ---> (1)

y = x ---> (2)

(1) = (2)

1/y2 = y

y3 = 1

y = 1

area-using-integration-q3.png
=21y-1y2dy=12y2+y-121=12(2)2+(2)-1-12(1)2+(1)-1=2+12-12+1=52-1=32=1.5

So, the required area is 1.5 square units.

Problem 4 :

x = 2 - y2, y = -x

Solution:

Let us find the point of intersection,

x = 2 - y2 ---> (1)

y = -x ---> (2)

(1) = (2)

-y = 2 - y2

y2 - y - 2 = 0

(y - 2) (y + 1) = 0

y = -1 and y = 2

area-using-integration-q4.png

So, the required area is 4.5 square units.

Problem 5 :

y = x2 and y = 4x

Use vertical rectangles (dx).

Solution:

Let us find the point of intersection,

y = x2 ---> (1)

y = 4x ---> (2)

(1) = (2)

4x = x2

x2 - 4x = 0

x(x - 4) = 0

x = 0 and x = 4

area-using-integration-q5.png
=404x-x2 dx=4x22-x3340=4(4)22-(4)33-[0]=32-643=323

So, the required area is 32/3 square units.

Problem 6 :

y = x2 and y = 4x

Use horizontal rectangles (dy).

Solution:

Let us find the point of intersection,

y = x2

x = √y ---> (1)

y = 4x 

x = 1/4 y ---> (2)

(1) = (2)

y=14y4y=y4=yyy=4y12=4y=16 and y=0
area-using-integration-q6.png
=160y-14y dy=23y32-18y2160=23(16)32-18(16)2-[0]=323

So, the required area is 32/3 square units.

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