1. ## integral substitution

evaluate integral for interval between x=0 and x=1

INT: (x^2)(1+x)^0.5 dx

thanks

2. Originally Posted by daaavo
evaluate integral for interval between x=0 and x=1

INT: (x^2)(1+x)^0.5 dx

thanks
Try a u sub

let $\displaystyle u=x+1$ then $\displaystyle du=dx$ and $\displaystyle u-1=x$

using the above and substition
$\displaystyle \int x^2(x+1)^{\frac{1}{2}}dx=\int(u-1)^2 u^{\frac{1}{2}}du$

expanding the binomial we get...

$\displaystyle \int (u^2-2u+1)u^{\frac{1}{2}}du=\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}du$

From here it is off to the races...

3. Originally Posted by daaavo

evaluate integral for interval between x=0 and x=1

INT: (x^2)(1+x)^0.5 dx
After substitution $\displaystyle u^2 = 1 + x$ the integral becomes $\displaystyle 2\int_1^{\sqrt 2 } {u^2 \left( {u^2 - 1} \right)^2 \,du}.$

Now this is routine. After some straightforward calculations your integral is equal to $\displaystyle \frac{4} {{105}}\left( {11\sqrt 2 - 4} \right).$