Find the area bounded by the parabola y=...

Find the area bounded by the parabola `y=x^2+1` and the straight line `x+y=3.`

Text Solution

Verified by Experts

The two curves meet at points where `3-x=x^(2)+1`,
`i.e.," "x^(2)+x-2=0`
`"or "(x+2)(x-1)=0 or x=-2,1`
`therefore" Required area "=int_(-2)^(1)[(3-x)-(x^(2)+1)]dx`
`=int_(-2)^(1)(2-x-x^(2))dx`
`=[2x-(x^(2))/(2)-(x^(3))/(3)]_(-2)^(1)`
`=(2-(1)/(2)-(1)/(3))-(-4-(4)/(2)+(8)/(3))`
`=(9)/(2)` sq. units

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The area bounded by the parabola y = x^(2) and the line y = 2x is

`(4)/( 3)`

`(2)/( 3)`

`( 51)/( 3)`

`( 20)/( 3)`

The area bounded by the parabola x^2=y and is latus rectum is :

`a) (8)/(3)`

b)`(2)/(3)`

`c) (4)/(3)`

d)`(1)/(6)`

The area bounded by the parabola y^(2) = x and its latus rectum is

`2 int_(0)^((1)/(4))y dx `

`2 int_(0)^((1)/(4))sqrt(x) dx `

`(1)/(6)`

`(2)/(3)`