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The area between the curves y = x2 and y = x3 is:
- a)1/12 sq.units
- b)1/8 sq.units
- c)1/10 sq.unit
- d)1/6 sq.units
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The area between the curves y = x2and y = x3is:a)1/12 sq.unitsb)1/8 s...
y = x2 and y = x3
To find point of intersections :
x2 = x3
x2(x-1) = 0
So, x = 0,1
POI are (0,0) & (1,1)
= [x3/3 – x4/4]x=1 - [x3/3 – x4/4]x=0
= (1/3 – 1/4) – (0)
= 1/12 sq units
To find point of intersections :
x2 = x3
x2(x-1) = 0
So, x = 0,1
POI are (0,0) & (1,1)
= [x3/3 – x4/4]x=1 - [x3/3 – x4/4]x=0
= (1/3 – 1/4) – (0)
= 1/12 sq units
Most Upvoted Answer
The area between the curves y = x2and y = x3is:a)1/12 sq.unitsb)1/8 s...
First we note that the curves intersect at the points (0,0) and (1,1). Then we see that
x^3 < x^2
in this interval. Hence the area is given by
= 1/3 - 1/4 = 1/12.
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Community Answer
The area between the curves y = x2and y = x3is:a)1/12 sq.unitsb)1/8 s...
Area between the curves y = x^2 and y = x^3:
The area between two curves can be calculated by finding the points of intersection and then integrating the difference of the two functions within those bounds. In this case, the curves are y = x^2 and y = x^3.
Finding Points of Intersection:
To find the points of intersection, we set the two equations equal to each other:
x^2 = x^3
x^3 - x^2 = 0
x^2(x - 1) = 0
x = 0 or x = 1
Integrating the Difference:
To find the area between the two curves, we integrate the difference of the two functions within the bounds of 0 and 1:
∫(x^3 - x^2) dx
= [(1/4)x^4 - (1/3)x^3] evaluated from 0 to 1
= [(1/4) - (1/3)] - [0]
= 1/4 - 1/3
= 1/12 sq. units
Therefore, the area between the curves y = x^2 and y = x^3 is 1/12 sq. units.
The area between two curves can be calculated by finding the points of intersection and then integrating the difference of the two functions within those bounds. In this case, the curves are y = x^2 and y = x^3.
Finding Points of Intersection:
To find the points of intersection, we set the two equations equal to each other:
x^2 = x^3
x^3 - x^2 = 0
x^2(x - 1) = 0
x = 0 or x = 1
Integrating the Difference:
To find the area between the two curves, we integrate the difference of the two functions within the bounds of 0 and 1:
∫(x^3 - x^2) dx
= [(1/4)x^4 - (1/3)x^3] evaluated from 0 to 1
= [(1/4) - (1/3)] - [0]
= 1/4 - 1/3
= 1/12 sq. units
Therefore, the area between the curves y = x^2 and y = x^3 is 1/12 sq. units.
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