Evaluate the following integrals as a limit of sums:



Formula used:



where,



Here, a = 1 and b = 3


Therefore,




Let,



Here, f(x) = 3x2 + 1 and a = 1



Now, by putting x = 1 in f(x) we get,


f(1) = 3(12) + 1 = 3(1) + 1 = 3 + 1 = 4


f(1 + h)


= 3(1 + h)2 + 1


= 3{h2 + 12 + 2(h)(1)} + 1


= 3(h)2 + 3 + 3(2h) + 1


= 3(h)2 + 4 + 6h


Similarly, f(1 + 2h)


= 3(1 + 2h)2 + 1


= 3{2(2h)2 + 12 + 2(2h)(1)} + 1


= 3(2h)2 + 3 + 3(4h) + 1


= 3(2h)2 + 4 + 12h


{∵ (x + y)2 = x2 + y2 + 2xy}




In this series, 4 is getting added n times



Now take 3h2 and 6h common in remaining series





Put,



Since,















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