1. ## Definite Intergral

Hi, could someone tell me if this is the correct answer?

$\displaystyle \int_0^2 {3x^5e^{x^3}dx}$

I got:

$\displaystyle 7e^8 +1$

2. Originally Posted by millerst
Hi, could someone tell me if this is the correct answer?

$\displaystyle \int_0^2 {3x^5e^{x^3}dx}$

I got:

$\displaystyle 7e^8 +1$
yes that is correct

$\displaystyle \int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx$
where n is a natural number

(n^(1/n +1))/(1+n)

4. Originally Posted by millerst

$\displaystyle \int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx$
where n is a natural number

(n^(1/n +1))/(1+n)

Note quite the final answer should be

$\displaystyle -\frac{n}{n+1}$

5. Originally Posted by millerst

$\displaystyle \int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx$
where n is a natural number

(n^(1/n +1))/(1+n)
Here is what I did, can you point out where i am going wrong:
$\displaystyle \int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx$

let ln(x) = u
du = 1/x dx

change limits of integration
x=e then u=1
x=1, u=0

$\displaystyle =\int_{1}^{0} {u}^{1/n}du$

$\displaystyle = \frac {n{u}^{1/n +1}}{1+n}\$

6. Originally Posted by millerst
Here is what I did, can you point out where i am going wrong:
$\displaystyle \int_{e}^{1} \ \frac{(ln(x))^{1/n}}{x} \ dx$

let ln(x) = u
du = 1/x dx

change limits of integration
x=e then u=1
x=1, u=0

$\displaystyle =\int_{1}^{0} {u}^{1/n}du$

$\displaystyle = \frac {n{u}^{1/n +1}}{1+n}\$
$\displaystyle \int_{1}^{0}u^{\frac{1}{n}}du=\frac{n}{n+1}u^{\fra c{n+1}{n}} \bigg|_{1}^{0}=\frac{n}{n+1}(0^{\frac{n+1}{n}}-1^{\frac{n+1}{n}})=-\frac{n}{n+1}$