Solve Special Integral Homework Equation

[Mechdude]

Homework Statement

im taking a course on methods including gamma beta ,bessel functions and related stuff, i met this question in the library on old exam papers for the course, $$\int^{3}_{0} \frac{t^3}{\sqrt{3-t}}dt$$

Homework Equations

im not sure how to do it but did think of the beta function
$$B(m,n)= \int^{1}_{0} t^{m-1}(1-t)^{n-1} dt$$

The Attempt at a Solution

$$\int^{3}_{0} t^{4-1}(1-t)^{\frac{1}{2} -1}} dt$$
but the limits are wrong and I am probably heading in the wrong direction

 

[vela]

Good try. You're going in the right direction, but you need to be a little more careful. What happened to the three in the radical?

 

[Mechdude]

Good try. You're going in the right direction, but you need to be a little more careful. What happened to the three in the radical?

thanks vela it should have remained since i did not do anything special to it, let me edit that.

 

[Mechdude]

Correction: this is what the attempt at a solution should be
$$\int^{3}_{0} t^{4-1}(3-t)^{\frac{1}{2} -1}} dt$$

 

[Mechdude]

Now if i make substitution $u= \frac{t}{3}$ at t=3, u=1
and t=0, u = 0. $$du=\frac {dt}{3}$$
now $$\int ^{1}_{0} 3^{4-1} u^{4-1} (1-u)^{\frac{1}{2}-1} 3 du$$
which is, $3^4 B(4, \frac{1}{2})$
im i ok?
And finaly $$= 3^4 \frac{gamma (4) gamma(\frac{1}{2} ) } {gamma(4+\frac{1}{2} ) }$$

 

[Mechdude]

Oops i fogot to take 1/sqrt(3) out when i divided throug the stuff in the radical by 3
correction
Now if i make substitution $u= \frac{t}{3}$ at t=3, u=1
and t=0, u = 0. $$du=\frac {dt}{3}$$
now $$\frac{1}{\sqrt{3}} \int ^{1}_{0} 3^{4-1} u^{4-1} (1-u)^{\frac{1}{2}-1} 3 du$$
which is, $\frac{1}{\sqrt{3}} 3^4 B(4, \frac{1}{2})$
im i ok?
And finaly $$= \frac{1}{\sqrt{3}} 3^4 \frac{gamma (4) gamma(\frac{1}{2} ) } {gamma(4+\frac{1}{2} ) }$$

 

[Dick]

Oops i fogot to take 1/sqrt(3) out when i divided throug the stuff in the radical by 3
correction
Now if i make substitution $u= \frac{t}{3}$ at t=3, u=1
and t=0, u = 0. $$du=\frac {dt}{3}$$
now $$\frac{1}{\sqrt{3}} \int ^{1}_{0} 3^{4-1} u^{4-1} (1-u)^{\frac{1}{2}-1} 3 du$$
which is, $\frac{1}{\sqrt{3}} 3^4 B(4, \frac{1}{2})$
im i ok?
And finaly $$= \frac{1}{\sqrt{3}} 3^4 \frac{gamma (4) gamma(\frac{1}{2} ) } {gamma(4+\frac{1}{2} ) }$$

You're ok. You can also evaluate it without the beta functions. Substitute u=3-t and expand t^3 in terms of u. You'll get the same thing.

 

[Mechdude]

You're ok. You can also evaluate it without the beta functions. Substitute u=3-t and expand t^3 in terms of u. You'll get the same thing.

thanks, it would work(i tried it) but since its a course on special methods, odinary integral calculus would not earn a mark.

 

[Dick]

thanks, it would work(i tried it) but since its a course on special methods, odinary integral calculus would not earn a mark.

Sure. But it's an easy way to check your solution.

 

[Mechdude]

Sure. But it's an easy way to check your solution.

thanks, il keep that in mind.