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Question
The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is _______.
Options
28 sq. units
3 sq. units
`(28)/(3)` sq. units
`(3)/(28)` sq. units
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Solution
`bb((28)/(3))` sq. units
Explanation:
The right-handed parabola in this example, y2 = 4x, has its vertex at the origin, and the lines parallel to the y-axis at x = 1 to x = 4 units distance are x = 1, x = 4.
Similarly y2 = 4x contains even power of y and is symmetrical about the x-axis.
So the required area = Area of ABCD
Area of ABCD = `int_1^4ydx=int_1^4sqrt(4x)dx`
It can be written as
= `2int_1^4sqrtxdx`
= `2[(x3/2)/(3/2)]_1^4`
= `2[(2x3/2)/3]_1^4`
Substituting the values we get
= `2((2(4)3/2)/3-(2(1)3/2)/3)`
= `4(8/3-1/3)`
= `4(7/3)`
= `28/3` sq. units
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