Question

alguem consegue resolver?$\int\limits^4_{-1} {x} \sqrt{x+5} \, dx$

Answer 1

Vamos fazer uma substituição. Seja $u=x+5$. Então:

$u=x+5\Longrightarrow du=dx$

Redefinindo os limites de integração:

Para x=-1: $u=-1+5\Longrightarrow u=4$
Para x=4: $u=4+5\Longrightarrow u=9$

Agora vamos desenvolver a integral dada:

$\displaystyle I=\int_{-1}^{4} x\sqrt{x+5}\,dx\\\\ I=\int_{4}^{9} (u-5)\sqrt{u}\,dx\\\\ I=\int_{4}^{9}(u\sqrt u-5\sqrt u)\,dx\\\\ I=\int_{4}^{9}(u^{\frac 3 2}-5 u^{\frac1 2})\,dx\\\\ I=\int_{4}^{9} u^{\frac 3 2}\,dx-5\int_{4}^{9} u^{\frac1 2}\,dx\\\\ I=\left[\dfrac{u^{\frac 3 2 +1}}{\frac 3 2 +1}\right]_{4}^{9}-5\left[\dfrac{u^{\frac 1 2 +1}}{\frac 1 2 +1}\right]_{4}^{9}$

$I=\left[\dfrac{u^{\frac 5 2}}{\frac 5 2}\right]_{4}^{9}-5\left[\dfrac{u^{\frac 3 2}}{\frac 3 2}\right]_{4}^{9}\\\\ I=\dfrac{2}{5}\left[u^{\frac 5 2}\right]_{4}^{9}-\dfrac{10}{3}\left[u^{\frac 3 2}\right]_{4}^{9}\\\\ I=\dfrac{2}{5}\left[9^{\frac 5 2}-4^{\frac 5 2}\right]-\dfrac{10}{3}\left[9^{\frac 3 2}-4^{\frac 3 2}\right]\\\\ I=\dfrac{2}{5}\left[3^5-2^5\right]-\dfrac{10}{3}\left[3^3-2^3\right]\\\\ I=\dfrac{2}{5}\left[243-32\right]-\dfrac{10}{3}\left[27-8\right]$

$I=\dfrac{2}{5}[211]-\dfrac{10}{3}[19]\\\\ I=\dfrac{422}{5}-\dfrac{190}{3}\\\\ \boxed{I=\dfrac{316}{15}}$