Answer
$- \dfrac{8}{15} \approx 0.5333$
Work Step by Step
We are given the integral $I=\int_{-1}^0 2x\sqrt {x+1} \ dx$
We will solve the given integral by using u-substitution method.
Let us consider that $u=x+1 \implies dx=du$
So, we can write as: $\int_{-1}^0 2x\sqrt {x+1} \ dx =2 \int_{-1}^0 (u-1) u^{1/2} \ du \\ =2 \int_{-1}^0 (u^{3/2}-u^{1/2}) \ du$
Now, use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Therefore, we have: $ 2 \int_{-1}^0 (u^{3/2}-u^{1/2}) \ du =2 [\dfrac{u^{5/2}}{5/2}-\dfrac{u^{3/2}}{3/2}]_{-1}^0+C$
or, $=[\dfrac{4(x+1)^{5/2}}{5}-\dfrac{4(x+1)^{3/2}}{3}]_{-1}^0+C$
or, $=[\dfrac{4(0+1)^{5/2}}{5}-\dfrac{4(0+1)^{3/2}}{3}]-[\dfrac{4((-1)+1)^{5/2}}{5}-\dfrac{4((-1)+1)^{3/2}}{3}]$
or, $= - \dfrac{8}{15} \approx 0.5333$
