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Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the...

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yakar | Student | (Level 2) eNoter

Posted May 2, 2012 at 8:30 PM via web

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Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

y=  int((3t+(t)^(1/2))^(1/2)))dt, t=7...tanx))

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

y=  int((3t+(t)^(1/2))^(1/2)))dt, t=7...tanx))

y' = ________________

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thilina-g | College Teacher | (Level 1) Educator

Posted May 3, 2012 at 6:14 AM (Answer #1)

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`y=int_7^(tan(x))sqrt(3t+sqrt(t)))dt`

The first part of the theorem says,if,

`F(x) = int_a^xf(t)dt` then,

`F'(x) = f(x)`

In our case, we have tan x instead of x, so we can't apply this theorem straight away, so we need a substitution. we will use

`u =tan(x) `

so , `(du)/(dx) = sec^2(x)`

now our integral becomes,

`y=int_7^usqrt(3t+sqrt(t)))dt`

but we need '0' on the lower limit.

`y=int_0^usqrt(3t+sqrt(t)))dt - int_0^7sqrt(3t+sqrt(t))dt`

so,
 dy/du becomes,

`(dy)/(du) = f(u) - f(7)`

`(dy)/(du) = sqrt(3u+sqrt(u)) - sqrt(3*7+sqrt(7))`

`(dy)/(du) = sqrt(3u+sqrt(u)) - 4.862689`

 

But we know as per chain rule,

`(dy)/(dx) = (dy)/(du) * (du)/(dx)`

so we get,

`(dy)/(dx) = (sqrt(3u+sqrt(u)) - 4.862689) * sec^2(x)`

but u is tan(x)

`(dy)/(dx) = (sqrt(3tan(x)+sqrt(tan(x))) - 4.862689) * sec^2(x)`

 

 

 

 

 

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