Find d/dx integral(with upper bound of 4x and lower bound of 1) of square root of t^2 +1 dt.
A. sqrt(4x+1)
B. sqrt(16x^2+1) + sqrt(2)
C. 4 sqrt(16x^2+1)
D. sqrt(16x^2+1) <------ My solution/choice
Am I right? Thannk you!
This is just the chain rule in reverse.
If dF(t)/dt = f(t)
then ∫f(t) dt = F(t)
Now, you know that
∫[a..b] f(t) dt = F(b) - F(a)
So now, plug in your limits.
∫[1..4x] f(t) dt = F(4x) - F(1)
Now just take the derivative with respect to x, and you have
d/dx F(4x) = f(4x) * d(4x)/dx = 4f(4x) = 4√((4x)^2+1) = 4√(16x^2+1)
d/dx F(1) = 0 since F(1) is just a constant
You had the right idea, but you forgot the chain rule.
d/dx ∫[u..v] f(t) dt = f(v) dv/dx - f(u) du/dx
where u and v are functions of x
To find the derivative of an integral, we can use the Fundamental Theorem of Calculus. The theorem states that if we have a function f(x) defined as the integral of another function F(t) with respect to t, then the derivative of f(x) with respect to x is equal to F(x).
In this case, the integral of sqrt(t^2+1) with respect to t is denoted as F(t). So, we need to find F(x), the antiderivative of sqrt(t^2+1), and then take its derivative with respect to x.
To find the antiderivative, we can use the substitution method. Let u = t^2+1, then du = 2t dt. Rearranging, we have t dt = (1/2) du. Therefore, our integral becomes:
∫(sqrt(t^2+1)) dt = ∫(sqrt(u)) (1/2) du
Now, we can solve the integral with respect to u:
(1/2) ∫(sqrt(u)) du
Using the power rule for integration, the integral becomes:
(1/2) * (2/3) * u^(3/2) + C
= (1/3) * u^(3/2) + C
Substituting back u = t^2+1, we have:
(1/3) * (t^2+1)^(3/2) + C
Finally, we differentiate with respect to x:
d/dx [(1/3) * (t^2+1)^(3/2)] = (2/3) * t * (t^2+1)^(1/2)
Substituting t = 4x, we get:
(2/3) * 4x * (16x^2+1)^(1/2)
= (8/3) * x * (16x^2+1)^(1/2)
Comparing this with the given options, we see that the correct answer is option C: 4 sqrt(16x^2+1).