Integration via Substitution

[blinlim]

I'm having a few problems on my current homework.

98) let g(x)= f(2x). show that the average value of f on the interval [0,2b] is the same as the average value of g on the interval [0,b].

In this problem I am not really sure what to do. I believe it might be something I should remember from Calculus 1, however I took that class a long time ago.

49) integrate x/[sqrt(1-x^4)] dx with the boundaries 0 to 1/sqrt(2).

Now I'm pretty comfortable with the methodology behind integration via substitution now, but this question seems to be a little tricky. I tried to do it using

w = x^4 as my substitution, and then again as x^2 as my substitution. This is my work so far.

Integral sign (0 to 1/sqrt(2)) : x/Sqrt[1-x^4] dx substitute w = x^2 so dw/2 = x

Integral sign (0 to 1/sqrt(2)) : 1/[2Sqrt(1-w^2)] dw

And then i have no way of integrating that equation

thanks for the help guys!

 

[srmichael]

I'm having a few problems on my current homework.

98) let g(x)= f(2x). show that the average value of f on the interval [0,2b] is the same as the average value of g on the interval [0,b].

In this problem I am not really sure what to do. I believe it might be something I should remember from Calculus 1, however I took that class a long time ago.

49) integrate x/[sqrt(1-x^4)] dx with the boundaries 0 to 1/sqrt(2).

Now I'm pretty comfortable with the methodology behind integration via substitution now, but this question seems to be a little tricky. I tried to do it using

w = x^4 as my substitution, and then again as x^2 as my substitution. This is my work so far.

Integral sign (0 to 1/sqrt(2)) : x/Sqrt[1-x^4] dx substitute w = x^2 so dw/2 = x

Integral sign (0 to 1/sqrt(2)) : 1/[2Sqrt(1-w^2)] dw

And then i have no way of integrating that equation

thanks for the help guys!

For #98, Average Value of f(x) on [a,b] = \(\displaystyle \dfrac{1}{b-a}\int^b_a f(x)dx\)

For #49, use the fact that \(\displaystyle \dfrac{d}{dx}sin^{-1}u=\dfrac{u'}{\sqrt{1-u^2}}\) where u is a function of x

 

[Deleted member 4993]

I'm having a few problems on my current homework.

98) let g(x)= f(2x). show that the average value of f on the interval [0,2b] is the same as the average value of g on the interval [0,b].

In this problem I am not really sure what to do. I believe it might be something I should remember from Calculus 1, however I took that class a long time ago.

49) integrate x/[sqrt(1-x^4)] dx with the boundaries 0 to 1/sqrt(2).

Now I'm pretty comfortable with the methodology behind integration via substitution now, but this question seems to be a little tricky. I tried to do it using

w = x^4 as my substitution, and then again as x^2 as my substitution. This is my work so far.

Integral sign (0 to 1/sqrt(2)) : x/Sqrt[1-x^4] dx substitute w = x^2 so dw/2 = x

Integral sign (0 to 1/sqrt(2)) : 1/[2Sqrt(1-w^2)] dw

And then i have no way of integrating that equation

thanks for the help guys!

\(\displaystyle \displaystyle{\int_0^{\frac{1}{\sqrt{2}}}\frac{x}{\sqrt{1-x^4}}dx}\)

let u = x²→ du = 2 dx

\(\displaystyle \displaystyle{\frac{1}{2}\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-u^2}}du}\)

This is standard integrand which results in sin^-1(u) ← Look it up Or

substitute again

sin(Θ) = u

du = cos(Θ) dΘ

then

\(\displaystyle \displaystyle{\frac{1}{2}\int_0^{\frac{\pi}{6}} \frac{cos(\theta)}{\sqrt{1-sin^2(\theta)}}d\theta}\)

\(\displaystyle \displaystyle{= \ \frac{1}{2}\int_0^{\frac{\pi}{6}} d\theta}\)

and finish it....