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If the area bounded by the curve y=x^2+1...

If the area bounded by the curve `y=x^2+1` and the tangents to it drawn from the origin is `A ,` then the value of `3A` is__-

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The correct Answer is:
2

Let the point of the curve be `(x,x^(2)+1).`
Now, the slope of tangent at this point is 2x, which is equal to the slope of the line joining `(x,x^(2)+1) and (0,0).`
`"Hence, "2x=(x^(2)+1)//x or 2x^(2)=x^(2)+1`
`"or "x^(2)+1 or x=pm1`

Hence, equation of tangent is `y=pm2x`
`"Now Area "2overset(1)underset(0)int(x^(2)+1-2x)dx=2overset(1)underset(0)int(x-1)^(2)dx=2[(x-1)^(3)/(3)]_(0)^(1)=2//3`
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