Question 73

# Doubt from area under curves

0.67 ?

Don't know the answer will come to know tomorrow. But upload your solution.

I m getting 1/3

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bro drag the graph of (y-1)=(x-1)^2+1 and draw the its image wrt the line y=x,

then we get our desired area as,

2\int^{{2}}
_{1}[(x^2-2x+2)-x]dx

we get,our desired area as \frac{1}{3}.

EDIT: please neglect the above post, i did a mistake previously,i.e i did 2(\frac{1}{3}) instead of \frac{1}{3}

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