g(x)=∫ (x on top, 0 on bottom) √(1+6t) dt, use fundamental theorem of calculus find g1(x) (derivative of g(x). )

Question

g(x)=∫ (x on top, 0 on bottom) √(1+6t) dt, use fundamental theorem of calculus find g1(x) (derivative of g(x). )

Answer 1

The Fundamental Theorem applies to a situation where an integral is used to define a new function. It says that if g(x) = ∫ f(t) dt from a to x, then g'(x) = f(t).

 

 ∫ f(t)dt = g(x) - g(0)  

 

So..

 

g'(x) = √(1+6x)

Answer 2

g(x)=∫₀^x √(1+6t) dt, find g′(x) = ?

 

Find solution of integral by method of substitution:

 

let 1+6t = u, 6dt =du, dt = (1/6)du substitute in the integral..

 

=(1/6)∫₀^x √u .du = (1/6)[ u^(1/2+1)/ (1/2+1)]^x₀ = (1/9)[u^(3/2)]^x₀ =

 

(1/9)[x^(3/2)-(1/9) = (1/9)[x^(3/2)-1], but x = 1+6t , substitute x in the

 

equation to get g(x) = (1/9)[(1+6t)^(3/2)-1]..find derivative g′(x)

 

g′(x) = (1/9)(3/2)[(1+6t)^(3/2-1)] (6), Use chain rule

 

g′(x) = (1/6)[(1+6t)(3/2-1)] (6) = (1+6t)^(1/2) or = √(1+6t)

 

Note that this is the value of the integrand in the integral

 

such that  g(x) = ∫ g′(x)