What is the derivative of # int_x^(x^2) t^3 \ dt # wrt #x#?

3 Answers
May 26, 2017

Answer:

#d/(dx) int_x^(x^2) t^3 dt =x^6-x^3#

Explanation:

According to the #2^(nd)# Fundamental Theorem of Calculus,
#d/(dx) int_a^b f(t) dt = f(b) - f(a)#.

Using your example, #d/(dx) int_x^(x^2) t^3 dt =(x^2)^3 - x^3 = x^6 - x^3#.

May 26, 2017

Answer:

# d/dx \ int_x^(x^2) t^3 \ dt =2x^7 - x^3 #

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

# d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a#

(ie the derivative of an integral gives us the original function back).

We are asked to find:

# d/dx \ int_x^(x^2) t^3 \ dt #

Initially we can manipulate the integral as follows (although we have chosen #0# as the lower limit we could in fact choose any constant:

# int_0^(x^2) t^3 \ dt = int_0^x t^3 \ dt + int_x^(x^2) t^3 \ dt#

And so:

# int_x^(x^2) t^3 \ dt = int_0^(x^2) t^3 \ dt - int_0^x t^3 \ dt #

And therefore differentiating we get:

# d/dx \ int_x^(x^2) t^3 \ dt = d/dx \ int_0^(x^2) t^3 \ dt - d/dx \ int_0^x t^3 \ dt #

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

# u=x^2 => (du)/dx = 2x #

The substituting into the first integral we get:

# d/dx \ int_x^(x^2) t^3 \ dt = d/dx \ int_0^u t^3 \ dt - d/dx \ int_0^x t^3 \ dt #

# " " = (du)/dx*d/(du) \ int_0^u t^3 \ dt - d/dx \ int_0^x t^3 \ dt #

# " " = 2xd/(du) \ int_0^u t^3 \ dt - d/dx \ int_0^x t^3 \ dt #

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

# d/dx \ int_x^(x^2) t^3 \ dt = 2x \ u^3-x^3 #
# " " = 2x \ (x^2)^3-x^3 #
# " " = 2x \ x^6 - x^3 #
# " " = 2x^7 - x^3 #

May 27, 2017

Another way to do this is to apply Leibniz's integral rule:

#d/(dx)[int_(a(x))^(b(x)) f(x,t)dt] = int_(a(x))^(b(x)) del/(delx)[f(x,t)]dt + f(x,b(x))(db)/(dx) - f(x,a(x))(da)/(dx)#

In this case, we have:

#d/(dx)[int_(x)^(x^2) t^3 dt]#,

where #a(x) = x#, #b(x) = x^2#, and #f(x,t) = t^3#.

We obtain, by plugging in #a(x)# and #b(x)# in place of #t# for the #f(x,a(x))# and #f(x,b(x))# terms:

#= cancel(int_(x)^(x^2) 0dt)^(0) + stackrel(|[t^3]|_(t = x^2))overbrace((x^2)^3)cdotcolor(red)((db)/(dx)) - stackrel(|[t^3]|_(t = x))overbrace((x)^3)cdotcolor(red)((da)/(dx))#

since the derivative of #t^3# with respect to #x# is #0# (#t# does not vary when #x# varies), and the integral of #0# is #0# (a flat horizontal line along the #x# axis has no area).

We then take the derivative of #b(x) = x# with respect to #x# and #a(x) = x^2# with respect to #x# to get:

#=> color(blue)(d/(dx)[int_(x)^(x^2) t^3 dt]) = x^6 cdot color(red)(2x) - x^3 cdot color(red)(1)#

#= color(blue)(2x^7 - x^3)#