F 2 3 F 5 7 AND INTEGRATION OF F X FROM LIMIT 2 TO 5 IS 17 THEN WATS

solution is=12. as d inverse exist as a function the F need to be strictly increasing function.ans we change 'x' and 'y' axis.the area under curve F is fiven in the limit 2 to 5 is=17.so the area under limit 3 to 7 will be =5*7-(2*3+17)=35-(6+17)=12.
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-15
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12
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12
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12
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the ans is 5-2 =3
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ans =14
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Ans is 12
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since F(inverse) exists then F must be one-one and onto function.since nothing is mentioned about continuity of the function we can assume(without loss of generality) it to be continuous.then if we draw the graph of this function then it will be an increasing function.also if we draw graph of its inverse then the area between the curve of F(x) and x-axis will tranform to area between curve of Finverse(x) and y-axis.if we join the AB,CD,EB,FD where A=(3,0),B=(3,2),C=(7,0),D=(7,5),E=(0,2),F=(0,5)THEN from above arguments area(EBDF)=17,area(EBACDF)=29,THUS REQUIRED AREA ie area(ABDC)=29-17=12
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F(2)=3,F(5)=7 AND INTEGRATION OF F(X) FROM LIMIT 2 TO 5 IS 17 ........THEN WATS THE INTEGRATION VALUE OF F(inverse)X LIMIT FROM 3 TO 7 IS????