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!
2 answers
Yes, you are correct :)
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
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
0 answers