
A few calc/integration questions.

For (c) use substitution, let $\displaystyle u=\ln(x)$ , so $\displaystyle du=\frac{1}{x} dx$
so the integral becomes $\displaystyle \int ^1 _0 u^{\frac{1}{n}} du$
$\displaystyle \int ^1_0 u^{\frac{1}{n}} du= \frac{u^{\frac{1}{n}+1}}{\frac{1}{n}+1} ^1 _0 =\frac{1}{\frac{1}{n}+1}$

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 :)

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

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?