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,