How do I evaluate #d/dx \int_5^(x^4) \sqrt{t^2 + t} dt#?

Question

6011 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T20:39:57

To solve #d /dx \int_5^{x^4} sqrt{t^2+t} dt#, we will use the Fundamental Theorem of Calculus.

#d /dx \int_5^{x^4} sqrt{t^2+t} dt#
#= [\sqrt{(x^4)^2+x^4}] * (4x^3)# **don't forget to chain!!

Answer 2 · 2017-03-06T20:53:25

#= 4x^5 \ sqrt(x^4 + 1) #

The FTC tells us that, for constant #a# and parameter #u#:

#d/(du) int_a^u f(t) dt = f(u)#

Now for the chaining bit.

If in fact #u = u(x)#, we can say that:

#d/(dcolor(red)(x)) int_a^(u(x)) f(t) dt #

#= d/(du) (int_a^(u) f(t) dt ) cdot (du)/(dx)#

#= f(u) cdot (du)/dx#

So for the specific question, where #u(x) = x^4#:

#d/(dx) int_5^(x^4) sqrt(t^2 + t) \ dt #

#= sqrt((x^4)^2 + x^4) cdot d/dx (x^4) #

#= x^2\ sqrt(x^4 + 1) cdot 4x^3 #

#= 4x^5 \ sqrt(x^4 + 1) #