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Using integration, find the area bounded...

Using integration, find the area bounded by the curve `x^2=4y` and the line `x=4y-2.`

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`x^(2)=4yimplies y=(x^(2))/(4) " " ` ...(1)
`x=4y-2impliesy=(x+2)/(4) " " ` ...(2)
The points of intersection of the parabola and straight line are C(2, 1) and `D(-1, (1)/(4))`.

Draw the graph of both curves.
Required area = area of ABCDA - area of ABCODA
`=int_(-1)^(2)(x+2)/(4)dx - int_(-1)^(2)(x^(2))/(4)dx`
`=(1)/(4)[(x^(2))/(2)+2x]_(-1)^(2)-(1)/(4)[(x^(3))/(3)]_(-1)^(2)`
`=(1)/(4)[(2+4)-((1)/(2)-2)]-(1)/(12)[8-(-1)]`
`=(1)/(4)(6+(3)/(2))-(1)/(2)xx9`
`=(15)/(8)-(3)/(4)=(9)/(8)xx9` sq. units.
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