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Suppose that the graph of y=f(x), contains the points (0, 4) and (2, 7). If f' is continuous then 02f(x)dx is equal to

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a
2
b
-2
c
3
d
none of these

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detailed solution

Correct option is C

f(0)=4,f(2)=7 and ∫02 f′(x)dx=f(2)−f(0)= 7−4=3


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