How do you use the second fundamental theorem of Calculus to find the derivative of given #int cos(sqrt( t))dt# from #[6, x]#?

Question

3064 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-15T14:05:33

See the explanation below.

The second Fundamental Theorem of Calculus (a.k.a The Fundamental Theorem of Calculus, Part 2) tells us that

#int_6^x f(t) dt = F(x) - F(6)#

where #F# is an antiderivative of #f#.

That is, where #F'(x) = f(x)#

Now, #F(6)# is a constant, so its derivative is #0#.

Therefore,

#d/dx(F(x) - F(6)) = d/dx(F(x))-d/dx(F(6))#

# = d/dx(F(x)) - 0#

# = F'(x) = f(x) = cos(sqrt(x))#

Note that I did not find an expression for #F(x)# because I was not asked to. If you are expected to do that, use integration by parts for

#2 int underbrace(sqrtx)_u underbrace(cos(sqrtx) 1/(2sqrtx) dx)_(dv)#