How do you use Part 1 of the fundamental theorem of calculus to find the derivative of the function #y= int (1+v^2)^10 dv# from sinx to cosx? Please help have been looking at problem for a hour can't figure out how to work it?

2 Answers
May 28, 2018

Answer:

#d/(dx) [int_(sinx)^(cosx) (1+v^2)^10dv] = sinx +cosx -(1+cos^2x)^10 sinx -(1+sin^2x)^10cosx#

Explanation:

Let:

#F(u) = int_0^u (1+v^2)^10dv#

based on the fundamental theorem of calculus:

#(dF)/(du) = (1+u^2)^10-1#

Now:

# int_(sinx)^(cosx) (1+v^2)^10dv = F(cosx)-F(sinx)#

and using the linearity of the derivative and the chain rule:

#d/(dx) [int_(sinx)^(cosx) (1+v^2)^10dv] = F'(cosx) d/dx cosx - F'(sinx)d/dx sinx#

#d/(dx) [int_(sinx)^(cosx) (1+v^2)^10dv] = -((1+cos^2x)^10 -1)sinx -((1+sin^2x)^10 -1)cosx#

#d/(dx) [int_(sinx)^(cosx) (1+v^2)^10dv] = sinx +cosx -(1+cos^2x)^10 sinx -(1+sin^2x)^10cosx#

May 28, 2018

Answer:

See below

Explanation:

FTC, Part 1 states, broadly, that if:

  • #F(x)= int _a^x f(t)\ dt#

then:

  • #F'(x)=d/(dx) int _a^x f(t)\ dt = f(x) #

This can then be modified for an interval that includes a function #u = u(x)#, by using the chain rule:

  • #d/dx = d/(du) (du)/(dx) implies d/(dx) int _a^(u(x)) f(t)\ dt = f(u(x)) * u'(x)#

Then by manipulating the interval, you can get a complete result:

  • # int_(v(x))^(u(x)) f(t)\ dt = int_(0)^(u(x)) f(t)\ dt - int_(0)^(v(x)) f(t)\ dt #

  • #implies d/(dx) int _(v(x))^(u(x)) f(t)\ dt = f(u(x)) * u'(x) - f(v(x)) * v'(x) #

In this case, #v # is the dummy variable instead of #t#. Matters not.

#d/dx int_(sin x)^(cos x) (1+v^2)^10 dv#

# = (1+cos^2x)^10 * (- sin x) - (1+sin^2x)^10 * cos x#

....which might simplify a little