integrate t^(1/2)sintdt. let lower bound=(x^1/2) upper bound=(x^3). use the chain rule and fundamental theorem calculas part1.

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- Nov 16th 2006, 07:23 PMmyoplex11integration using the fundamental theorem of calc part1
integrate t^(1/2)sintdt. let lower bound=(x^1/2) upper bound=(x^3). use the chain rule and fundamental theorem calculas part1.

- Nov 16th 2006, 08:00 PMThePerfectHacker
It seems to me that $\displaystyle x$ should be at least $\displaystyle x>0$

It seems you wish to integrate the function,

$\displaystyle f(x)=\int_{x^{1/2}}^{x^3}t^{1/2}\sin tdt$

Let use find whether $\displaystyle x^{1/2}=x^3$

The only possible solution is $\displaystyle x=0,1$ but $\displaystyle x>0$ thus, $\displaystyle x=1$.

In that case,

$\displaystyle f(1)=0$

If, $\displaystyle 0<x<1$ then, $\displaystyle x^{1/2}>x^3$

Thus,

$\displaystyle \int_{x^{1/2}}^{x^3}t^{1/2}\sin t dt=-\int_{x^3}^{x^{1/2}}t^{1/2}\sin tdt$

Therefore, there exists a point in the interval $\displaystyle [x^3,x^{1/2}]$

Using the integral subdivision rule,

$\displaystyle -\int_{x^3}^c t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt$

Thus,

$\displaystyle \int_c^{x^3} t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt$

Let,

$\displaystyle \mu (x)=\int_c^x t^{1/2}\sin tdt$

Therefore, those two functions can be expressed as composition functions,

$\displaystyle \mu (x^3)-\mu (x^{1/2})$

But by the Fundamental theorem,

$\displaystyle \mu ' (x)=x^{1/2}\sin x$

And since $\displaystyle x^3,x^{1/2}$ are both differenciable on $\displaystyle (0,1)$ we can use the chain rule.

$\displaystyle 3x^2 \mu' (x^3)+\frac{1}{2}x^{-1/2}\mu'(x^{1/2})=3 x^{7/2}\sin(x^3)+\frac{1}{2}x^{-1/2}\sin (x^{-1/2})$

Something interesting to note.

$\displaystyle \lim_{x\to 1}$ of this expression if not equal to $\displaystyle f(1)$ which means the function you mentioned is not everywhere differenciable.

The other part of the problem is to consider the case $\displaystyle x>1$. Since it is late now I cannot complete the solution but I think it will be the same as for $\displaystyle (0,1)$