Find F'(x)
Original equation:
F(x)= definite integral of x+3 to 0 (10t+1) dt
Whoops, I meant:
f(x)=definite integral (10t+1)dt from x+3 to 0
int (10t+1) dt = 5 t^2 + t
at t = 0 that is 0
at t = x+3 that is 5(x+3)^2 +(x+3)
subtract
F(x) = - 5(x+3)^2 - x - 3
now do d/dx
-10(x+3) - 1
or
-10 x - 31
To find the derivative of the function F(x), we can use the Fundamental Theorem of Calculus. According to the theorem, if F(x) is the integral of a function f(t) with respect to t over a certain interval, then the derivative of F(x) with respect to x is equal to the integrand, f(t), evaluated at x.
Given that the original equation is:
F(x) = ∫[x+3 to 0] (10t+1) dt
To find F'(x), we need to differentiate with respect to x. However, since the limits of integration depend on the variable t, we need to write the limits in terms of x.
To do this, we can substitute the upper and lower limits of integration with their x-dependent counterparts:
When t = 0, we have x + 3 = 0, which gives x = -3.
When t = x + 3, we have x + 3 = x + 3, which is unchanged.
Therefore, we can rewrite the integral as:
F(x) = ∫[-3 to x+3] (10t+1) dt
Now we can differentiate F(x) with respect to x:
F'(x) = d/dx [ ∫[-3 to x+3] (10t+1) dt ]
According to the Fundamental Theorem of Calculus, we can treat the derivative of the integral as the integrand evaluated at the upper limit (x+3):
F'(x) = (10(x+3) + 1) = 10x + 31
So, the derivative of F(x) is F'(x) = 10x + 31.