find d/dx of the integral from 1 to 2x of the (sqrt t^2+1)dt at x=sqrt2

Question

the answer is 6. but i am not sure how to find d/dx.

i tried taking the integral then doing that from 1 to 2x and then taking d/dx then plugging in sqrt2 for x but it was wrong

Answer 1

Better take a look at

http://mathmistakes.info/facts/CalculusFacts/learn/doi/doif.html

So, what you have is

√(1+(2x)^2)(2) = 2√(1+4x^2)
at x=√2, that is 2√(1+8) = 2*3 = 6

Answer 2

DRAW A PICTURE

The change of an integral at the right hand limit is the area defined by the height , f(x), times dx
In other words the answer is f(2sqrt2)
2sqrt [2sqrt 2)^2+1] = 2sqrt(8+1) =2* 3