problem check

[Ryan Rigdon]

my answer has to be in exact terms of pi. if someone could check my answer to problem number 3 below would be greatly appreciated.

 

[Ryan Rigdon]

could the final answer be written as ln (72/pi +1)^(1/32)

 

[mmm4444bot]

I did not check your work; I calculated the integral myself, instead.

Your answer looks good, to me.

Yes, you could write it either way.

I wrote mine like this:

(1/32) * ln(72 + ?) - (1/32) * ln(?)

PS: In future posts, please reduce your image sizes.

 

[Ryan Rigdon]

thank you and i will try to reduce future posts. peace

 

[BigGlenntheHeavy]

\(\displaystyle \int_{0}^{3}\frac{x^3dx}{8x^4+9\pi}\)

\(\displaystyle Let \ u \ + \ 8x^4+9\pi, \ then \ \frac{du}{32} \ = \ x^3dx\)

\(\displaystyle Hence, \ we \ have \ \frac{1}{32}\int_{9\pi}^{648+9\pi}\frac{du}{u} \ = \ \frac{1}{32}\bigg[ln|u|\bigg]_{9\pi}^{648+9\pi}\)

\(\displaystyle = \ \frac{1}{32}[ln(648+9\pi)-ln(9\pi)] \ = \ \frac{1}{32}\bigg[ln\frac{648+9\pi}{9\pi}\bigg]\)

\(\displaystyle =\frac{1}{32}\bigg[ln\frac{9(72+\pi)}{9\pi}\bigg] \ = \ ln\bigg[\frac{72+\pi}{\pi}\bigg]^{1/32} \ \dot= \ .099207635934\)

 

[Ryan Rigdon]

it didnt ask for an approximation but leave it in terms of pi so would u agree bigglenntheheavy that my final is ok.

 

[BigGlenntheHeavy]

\(\displaystyle \frac{72}{\pi}+1 \ = \ \frac{72+\pi}{\pi}, \ either \ way \ is \ fine.\)

 

[mmm4444bot]

You've been told twice that your answer is okay.

You already know that (72 + ?)/? is the same as 72/? + 1.

Are you looking for some type of self-validation, here ? :?