How do you use the second fundamental theorem of Calculus to find the derivative of given #int (2t-1)^3 dt# from #[x^2, x^7]#?

1 Answer
Feb 23, 2016

Answer:

The derivative is #(2x^7-1)^3 (7x^6) - (2x^2-1)^3 (2x)#

Explanation:

Suppose #f(x)# is continuous on #[a,b]# and both #p(x)# and #q(x)# are differentiable on #(a,b)#

Define the function

#F(x) = int_(p(x))^(q(x)) f(t) "d"t#

Combining the Second Fundamental Theorem of Calculus and the Chain Rule implies that #F(x)# is differentiable and

#F'(x) = f(q(x))*q'(x) - f(p(x))*p'(x)#

So, in this case we have

#f(t) = (2t-1)^3#
#p(x) = x^2#
#q(x) = x^7#

Plug it in to get

#frac{"d"}{"d"x}(int_(x^2)^(x^7) (2t-1)^3 "d"t)#

# = (2(x^7)-1)^3 frac{"d"}{"d"x}(x^7) - (2(x^2)-1)^3 frac{"d"}{"d"x}(x^2)#

# = (2x^7-1)^3 (7x^6) - (2x^2-1)^3 (2x)#