By 2nd fundamental theorem of calculus, please evaluate..... d/dx integral (from 0 to x) [dt/(2+3√t)]?

Question

Integration.
2nd fundamental theorem of calculus.

Answer 1

replace the expression with "f(t)" and you are literally restating the theorem w/a=0 thus i suspect that answer is merely 1/(2+3*root(x)) if the integral were evaluated first the 0 lower limit would generate a constant which would vanish upon differentiation with respect to x

Answer 2

The fundamental theorem states that for a function f(t) continuous on an interval [c,x], the integral from c to x of f(t) dt is equal to F(x) - F(c), where F is a function that satisfies F'(x) = f(x). If we allow the upper limit x to be variable and c remain constant, we see that differentiating with respect to x yields (in Mathematica notation)

D[Integrate[f[t], {t,c,x}],x] = f[x].

Therefore, your answer is simply 1/(2+3 Sqrt(x)).

Answer 3

1/(2+3√x)

Viola!

Answer 4

Quite a strange wording...
Probably, you mean:
[I = Integral symbol]

I(2+3*sqrt(t)).dt
=2t + 3*t^(1+1/2) / (1+1/2)
=2t + 3*t^(3/2)*2/3
=2t + 2*t^(3/2)
=2t + 2*sqrt(t^3)

Answer 5

Easy, next

Answer 6

yeah what they said

Answer 7

dont know