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@ganeshie8
\[\int\limits_0^4 f'(x) dx = f(4) - f(0)\]
you can work the left side integral using geometry look at the given diagram, whats the area under graph between x=0 and x=4 ?
the area under graph between x=0 & x=4 is 2
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Yes!
\[\int\limits_0^4 f'(x) dx = f(4) - f(0)\] so the value of ntegral on left hand side equals \(4\) : \[4= f(4) - f(0)\]
and you're given the value of \(f(4)\) plug that in and solve \(f(0)\)
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2
f(0) = 2 is right !
thanks!!!!!!
yw
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