How do you calculate the derivative of int(2t-1)^3 dt (2t1)3dt from t=x^2t=x2 to t=x^7t=x7?

2 Answers
Feb 15, 2017

(2x^7 - 1)^3 - (2x^2 - 1)^3(2x71)3(2x21)3

Explanation:

Let f(t) = (2t - 1)^3f(t)=(2t1)3. Then, if F(t)F(t) an antiderivative of f(t)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

= 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.