find d/dx of the integral from 1 to 2x of the (sqrt t^2+1)dt at x=sqrt2
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
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
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
To find the derivative of the given integral, we can use the Fundamental Theorem of Calculus. According to this theorem, if we have an integral that depends on a parameter, such as x in this case, then the derivative of that integral can be found by differentiating the integrand with respect to x and then evaluating it at the limits of integration.
Let's break down the process step by step:
1. Rewrite the integral in terms of x:
∫[1 to 2x] √(t^2 + 1) dt
2. Differentiate the integrand with respect to x:
d/dx (√(t^2 + 1)) = (1/2) * (t^2 + 1)^(-1/2) * 2t
3. Substitute 2x for t in our result:
(1/2) * ((2x)^2 + 1)^(-1/2) * 2 * (2x)
Simplifying this expression gives:
(1/2) * (4x^2 + 1)^(-1/2) * 4x
4. Evaluate the differentiated integrand at the upper limit of integration, which in this case is 2x:
(1/2) * (4(√2)^2 + 1)^(-1/2) * 4(√2)
Simplifying further:
(1/2) * (8 + 1)^(-1/2) * 4(√2)
5. Finally, simplify the expression:
(1/2) * 9^(-1/2) * 4(√2) = 2/√9 * 4(√2) = 2/3 * 4(√2) = 8/3(√2)
So, the derivative of the integral ∫[1 to 2x] √(t^2 + 1) dt at x = √2 is 8/3(√2).
It seems that the answer you provided, 6, is not correct.