# Thread: ratio of two definite Integration

1. ## ratio of two definite Integration

$If I = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx and J = \int^{1}_{0}\frac{x^{\frac{3}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx . Then \displaystyle \frac{I}{J} is$

2. ## Re: ratio of two definite Integration

I think I know how to do this although my answer is pretty disgusting (could be wrong). I don't know how to use latex so I'll just tell you what I did.

So since the integrals have the same limits and variable, if you write the integral of I/J, its the same as finding the values of I and J and then dividing it. When you re-write the integral the (1-x)^7/2 terms cancel out and the x^5/2 is just reduced to x. So you're left with x(3+x)^8. Then it's just a sub for 3+x and bob's your uncle.

3. ## Re: ratio of two definite Integration

$\left(\int_a^b f(x)\, dx\right)/ \left(\int_a^b g(x)\, dx\right)\neq \int_a^b \frac{f(x)}{g(x)}\, dx$
in general.

Eg let a=0, f(x)=x^2, g(x)=x.

4. ## Re: ratio of two definite Integration

Oh. Man I just tried to type up my solution in LaTex for the 1st time but I didn't know how to put it as text in the document only as a picture. How do I put it as text like the posts above (even though ik it's wrong now -_- )

5. ## Re: ratio of two definite Integration

Sorry friends actually original question as

$If I = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx and J = \int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx . Then \displaystyle \frac{I}{J} is$

6. ## Re: ratio of two definite Integration

Originally Posted by juantheron
Sorry friends actually original question as

$If I = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx and J = \int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx . Then \displaystyle \frac{I}{J} is$
The answer is very simple.

$\noindent Let u = \frac{4x}{x+3} \Rightarrow \text{d}u = \frac{12 \, \text{d}x}{(x+3)^2} \\\\ x = 0 \Rightarrow u = 0 \\\\ x = 1 \Rightarrow u = 1 \\\\ \frac{1-x}{3+x} = \frac{1-u}{3} \\\\ \frac{x}{x+3} = \frac{u}{4} \\\\ J = \int_0^1 \frac{u^{\frac{5}{2}} (1-u)^{\frac{7}{2}}}{12 \times 4^{\frac{5}{2}} \times 3^{\frac{7}{2}}} = \frac{I}{10368 \sqrt{3}}$

The remainder of the problem is trivial.

7. ## Re: ratio of two definite Integration

Originally Posted by Paradoxica
The answer is very simple.

$\noindent Let u = \frac{4x}{x+3} \Rightarrow \text{d}u = \frac{12 \, \text{d}x}{(x+3)^2} \\\\ x = 0 \Rightarrow u = 0 \\\\ x = 1 \Rightarrow u = 1 \\\\ \frac{1-x}{3+x} = \frac{1-u}{3} \\\\ \frac{x}{x+3} = \frac{u}{4} \\\\ J = \int_0^1 \frac{u^{\frac{5}{2}} (1-u)^{\frac{7}{2}}}{12 \times 4^{\frac{5}{2}} \times 3^{\frac{7}{2}}} = \frac{I}{10368 \sqrt{3}}$

The remainder of the problem is trivial.
Can I ask how'd you get u = 4x/(x+3)?

8. ## Re: ratio of two definite Integration

Originally Posted by pikachu975
Can I ask how'd you get u = 4x/(x+3)?
A rational substitution of the form u = A + B/(x+3) will take out (x+3)^2 from the denominator, simply by existing. That is the primary purpose of the substitution. The remaining part of the denominator can be perfectly repartitioned across the two components of the denominator.

The secondary purpose is to pick such coefficients A and B such that the borders coincide, which can be done by substituting values.

The third purpose is to make it such that the integral is exactly transformed into a constant multiple of I, which just so happens to be possible because of the way this problem was designed. For arbitrary problems, such a transformation may not be possible.

It only took a few minutes of scribbling until I was done with the problem, at least intuitively, and the details were fleshed out while typesetting.

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