Convergence: Test for Convergence of ∫ sin∅/sqrt(pi-∅) d∅

[whatlifeforme]

Homework Statement

Test for convergence or divergence.

Homework Equations

\displaystyle\int_0^∏ {\frac{sin∅}{\sqrt{pi-∅}} d∅}

The Attempt at a Solution

the solution manual does the following:

pi -∅ = x

\displaystyle\int_0^pi {\frac{sinx}{\sqrt{x}} d∅}


0 <= \frac{sinx}{\sqrt{x}} <= \frac{1}{\sqrt{x}}

but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

 

[SammyS]

Homework Statement

Test for convergence or divergence.

Homework Equations

\displaystyle\int_0^\pi {\frac{sin∅}{\sqrt{pi-∅}} d∅}

The Attempt at a Solution

the solution manual does the following:

pi -∅ = x

\displaystyle\int_0^\pi {\frac{sinx}{\sqrt{x}} d∅}0 &lt;= \frac{sinx}{\sqrt{x}} &lt;= \frac{1}{\sqrt{x}}

but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

(Use "\theta" in LaTeX for θ and "\pi" for pi.)

Yes, but dx = -dθ , and \displaystyle \ \ -\int_{a}^{b}f(x)\,dx=\int_{b}^{a}f(x)\,dx

 

[whatlifeforme]

(Use "\theta" in LaTeX for θ and "\pi" for pi.)

Yes, but dx = -dθ , and \displaystyle \ \ -\int_{a}^{b}f(x)\,dx=\int_{b}^{a}f(x)\,dx

i'm sorry what do you mean?

 

[SammyS]

i'm sorry what do you mean?

I thought you were having difficulty understanding why the limits of integration for the integral in x don't go from π to 0 .

OK, then what did you mean by the following?

the solution manual does the following:

pi -∅ = x

\displaystyle\int_0^pi {\frac{sinx}{\sqrt{x}} d∅}


0 &lt;= \frac{sinx}{\sqrt{x}} &lt;= \frac{1}{\sqrt{x}}

but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏​

Especially the statement following the word "but" .

 

[whatlifeforme]

they are replacing (∏ - ∅) by x (which is greater than which is essentially a replacement for ∅.

in other words:

sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

but when (∅> ∏): sqrt( ∏ - ∅) > sqrt(∅)

how does this apply on the integral of the problem. *checked* oopps.

the interval is from 0 to pi, so sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏ applies.

 

[SammyS]

they are replacing (∏ - ∅) by x (which is greater than which is essentially a replacement for ∅.

in other words:

sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

but when (∅> ∏): sqrt( ∏ - ∅) > sqrt(∅)

how does this apply on the integral of the problem. *checked* oopps.

the interval is from 0 to pi, so sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏ applies.

Oops is right, but you continue writing incorrect statements. (Also, please use θ, not the symbol for the empty set, ∅ .)

If θ = π/2 , then \displaystyle \ \sqrt{\pi-\theta\,}=\sqrt{\theta\,}\ .

If 0 < θ < π/2, then \displaystyle \ \sqrt{\pi-\theta\,}&gt;\sqrt{\theta\,}\ .

If π/2 < θ < π, then \displaystyle \ \sqrt{\pi-\theta\,}&lt;\sqrt{\theta\,}\ .

However, I don't see what any of this has to do with your problem.