How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #h(x) = int sqrt(6 + r^3) dx# from 7 to #x^2#?

1 Answer
Aug 22, 2015

Answer:

We need FTC 1 and the chain rule.

Explanation:

#h(x) = int_7^(x^2) sqrt(6 + r^3) dx# does not quite fit the form:

#g(x) = int_a^b f(t) dt# whose derivative is #f(x)#

But we can use the chain rule here;

The #h(x)# can be written as

#h(u) = int_7^u sqrt(6 + r^3) dx# where #u = x^2#

So

#d/dx(h) = (dh)/(du)* (du)/dx#

# = sqrt(6 + u^3) (du)/dx#

# = sqrt(6 + (x^2)^3)*(2x)#

# = 2xsqrt(6 + x^6)#

Summary

For
#h(x) = int_a^(g(x)) f(t) dt# #" "#(With the right kind of #f# and domain of #h#)

We have

#h'(x) = f(g(x))*g'(x)# #" "# (Which is one way of writing the chain rule.)