How do you calculate the derivative of #int(2t-1)^3 dt # from #t=x^2# to #t=x^7#?

2 Answers
Feb 15, 2017

Answer:

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

Explanation:

Let #f(t) = (2t - 1)^3#. Then, if #F(t)# an antiderivative of #f(t)#, it holds that:

#F'(t) = f(t)#.

We also know that #int f(t)dt = F(t) + c#. But also,

#(F(t) + c)' = f(t)# because #(c)' = 0# for any #cinRR#.

Therefore,

#[int f(t)dt]' = f(t)#,

so evaluating the derivative of that integral is exactly the same as evaluating #f(t)# at the two given #t# points, then taking their difference (as the question says to evaluate from #x^2# to #x^7#)

Finally,

#{[int f(t)dt]'}_(x^2)^(x^7) = f(x^7) - f(x^2)#

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

Note that further simplification is applicable (but not required) since this is a difference of cubes.

Mar 6, 2017

Answer:

# = x( 7x^5 (2x^7-1)^3 - 2 (2x^2-1)^3 )#

Explanation:

#d/dx ( int_(x^2)^(x^7) (2t-1)^3 dt )#

We can use the Fundamental Theorem of Calculus , Part 1:

#d/dx ( int_(a)^(x) f(t) \ dt ) = f(x)#

But if it is:

#d/dx ( int_(a)^(u) f(t) \ dt ) # where #u = u(x)#, then we must also use the chain rule:

#d/dx ( int_(a)^(u(x)) f(t) \ dt ) = d/(du) ( int_(a)^(u) f(t) \ dt ) (du)/dx = f(u(x)) cdot (du)/dx#

To make this particular integral fit within the FTC, we need also do some interval manipulation:

#d/dx ( int_(x^2)^(x^7) (2t-1)^3 dt )#

Splitting out the integration intervals:

#= d/dx ( int_(x^2)^(0) (2t-1)^3 dt + int_(0)^(x^7) (2t-1)^3 dt )#

Switching the first interval limits:

# = - d/dx ( int_0^(x^2) (2t-1)^3 dt ) + d/dx ( int_(0)^(x^7) (2t-1)^3 dt )#

Applying the FTC:

# = - (2(x^2)-1)^3 d/dx(x^2) + (2(x^7)-1)^3 d/dx (x^7)#

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

# = x( 7x^5 (2x^7-1)^3 - 2 (2x^2-1)^3 )#

Maybe that can be simplified a bit more.