# A few calc/integration questions.

• Nov 30th 2008, 09:33 PM
Kuma
A few calc/integration questions.
• Nov 30th 2008, 10:30 PM
akolman
For (c) use substitution, let $\displaystyle u=\ln(x)$ , so $\displaystyle du=\frac{1}{x} dx$

so the integral becomes $\displaystyle \int ^1 _0 u^{\frac{1}{n}} du$

$\displaystyle \int ^1_0 u^{\frac{1}{n}} du= \frac{u^{\frac{1}{n}+1}}{\frac{1}{n}+1} |^1 _0 =\frac{1}{\frac{1}{n}+1}$
• Nov 30th 2008, 10:44 PM
Kuma
Quote:

Originally Posted by akolman
For (c) use substitution, let $\displaystyle u=\ln(x)$ , so $\displaystyle du=\frac{1}{x} dx$

so the integral becomes $\displaystyle \int ^1 _0 u^{\frac{1}{n}} du$

$\displaystyle \int ^1_0 u^{\frac{1}{n}} du= \frac{u^{\frac{1}{n}+1}}{\frac{1}{n}+1} |^1 _0 =\frac{1}{\frac{1}{n}+1}$

great thx, i got that one but i just made it n/n+1 in the end :)
• Nov 30th 2008, 10:51 PM
akolman
For 3(a) expand$\displaystyle (t^2+3t)^2$

$\displaystyle (t^2+3t)^2=t^4+6t^3+9t^2$

So,

$\displaystyle \frac{(t^2+3t)^2}{t^3}=\frac{t^4+6t^3+9t^2}{t^3} = t + 6 + \frac{9}{t}$

that way the integral is easier

$\displaystyle \int \left( \frac{(t^2+3t)^2}{t^3} \right) dt= \int \left( t + 6 + \frac{9}{t} \right) dt$
• Dec 1st 2008, 01:10 AM
Kuma
Quote:

Originally Posted by akolman
For 3(a) expand$\displaystyle (t^2+3t)^2$

$\displaystyle (t^2+3t)^2=t^4+6t^3+9t^2$

So,

$\displaystyle \frac{(t^2+3t)^2}{t^3}=\frac{t^4+6t^3+9t^2}{t^3} = t + 6 + \frac{9}{t}$

that way the integral is easier

$\displaystyle \int \left( \frac{(t^2+3t)^2}{t^3} \right) dt= \int \left( t + 6 + \frac{9}{t} \right) dt$

Got this one as well; thx :)

I just need to know whether im right on the next 3.

For 10 i got c
For 5 i got c
For 3 i got b

are those correct?