# integration using the fundamental theorem of calc part1

• Nov 16th 2006, 07:23 PM
myoplex11
integration 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 PM
ThePerfectHacker
Quote:

Originally Posted by myoplex11
integrate t^(1/2)sintdt. let lower bound=(x^1/2) upper bound=(x^3). use the chain rule and fundamental theorem calculas part1.

It seems to me that $x$ should be at least $x>0$
It seems you wish to integrate the function,
$f(x)=\int_{x^{1/2}}^{x^3}t^{1/2}\sin tdt$

Let use find whether $x^{1/2}=x^3$
The only possible solution is $x=0,1$ but $x>0$ thus, $x=1$.
In that case,
$f(1)=0$

If, $0 then, $x^{1/2}>x^3$
Thus,
$\int_{x^{1/2}}^{x^3}t^{1/2}\sin t dt=-\int_{x^3}^{x^{1/2}}t^{1/2}\sin tdt$
Therefore, there exists a point in the interval $[x^3,x^{1/2}]$
Using the integral subdivision rule,
$-\int_{x^3}^c t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt$
Thus,
$\int_c^{x^3} t^{1/2}\sin tdt-\int_c^{x^{1/2}}t^{1/2}\sin tdt$
Let,
$\mu (x)=\int_c^x t^{1/2}\sin tdt$
Therefore, those two functions can be expressed as composition functions,
$\mu (x^3)-\mu (x^{1/2})$
But by the Fundamental theorem,
$\mu ' (x)=x^{1/2}\sin x$
And since $x^3,x^{1/2}$ are both differenciable on $(0,1)$ we can use the chain rule.
$3x^2 \mu' (x^3)+\frac{1}{2}x^{-1/2}\mu'(x^{1/2})=3 x^{7/2}\sin(x^3)+\frac{1}{2}x^{-1/2}\sin (x^{-1/2})$
Something interesting to note.
$\lim_{x\to 1}$ of this expression if not equal to $f(1)$ which means the function you mentioned is not everywhere differenciable.

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