# Math Help - Definite Integral - no clue.....

1. ## Definite Integral - no clue.....

Hi,

my math lector gave me the following "homework"

calculate the definite integral:

I tried using substitution where U=
√(x^2+2x) and dx=(3*du)/2*(x^2+x^3/3)^3/2 but I don't really know where to go with this problem

∫[0,2] √(x^2+2x)*(2x+2) dx

the intgegral is between 0 to 2..... ∫[0,2]

2. ## Re: Definite Integral - no clue.....

Try the substitution $u=x^2+2x$ instead, and don't forget the rewrite the limits of integration in terms of the new variable...

3. ## Re: Definite Integral - no clue.....

Originally Posted by FireTheft
Hi,

my math lector gave me the following "homework"

calculate the definite integral:

I tried using substitution where U=
√(x^2+2x) and dx=(3*du)/2*(x^2+x^3/3)^3/2 but I don't really know where to go with this problem

∫[0,2] √(x^2+2x)*(2x+2) dx

the intgegral is between 0 to 2..... ∫[0,2]

You are making life difficult for yourself with your particular substitution. When dealing with integration by substitution, in your integrand, you need an "inner" function, and you also need the derivative of this inner function as a factor. Here, it's pretty easy to see that an easy inner function is \displaystyle \begin{align*} x^2 + 2x \end{align*}, because its derivative, \displaystyle \begin{align*} 2x + 2 \end{align*} is a factor.

Anyway, making the substitution \displaystyle \begin{align*} u = x^2 + 2x \implies du = (2x + 2)\,dx \end{align*}, and noting that \displaystyle \begin{align*} u(0) = 0 \end{align*} and \displaystyle \begin{align*} u(2) = 8 \end{align*} gives

\displaystyle \begin{align*} \int_0^2{\sqrt{x^2 + 2x}\left(2x + 2\right)dx} &= \int_0^8{\sqrt{u}\,du} \\ &= \int_0^8{u^{\frac{1}{2}}\,du} \end{align*}

which is an easy integral to solve.

4. ## Re: Definite Integral - no clue.....

thx a lot guys.....

the substitution u=x^2+2x makes it lot easier ;-)......