ratio of two definite Integration (1 Viewer)

BenHowe · Feb 14, 2017

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.

seanieg89 · Feb 14, 2017

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

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

BenHowe · Feb 14, 2017

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 -_- )

juantheron · Mar 30, 2017

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$

Paradoxica · Apr 2, 2017

juantheron said:
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.

pikachu975 · Apr 2, 2017

Paradoxica said:
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)?

Paradoxica · Apr 2, 2017

pikachu975 said:
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.

ratio of two definite Integration (1 Viewer)

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