Question

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?

Answer 1

`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`

Answer 2

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