Question

$\int\limits^1_0 {t}\sqrt{t^{2}+1 } \, dx$

Answer 1

Resposta:

$\frac{1}{3} (2\sqrt{2} -1)$

Explicação passo a passo:

$\displaystyle\int_{0}^{1}t\sqrt{t^2+1}~dt \\\\t^2+1=u \implies 2tdt=du \implies tdt\impliestdt= \frac{1}{2}du\\\\p/t=0 \implies u=1\\\\p/t=1 \implies u=2 \\\\ \displaystyle\int_{0}^{1}t\sqrt{t^2+1}~dt = \displaystyle\int_{1}^{2}\sqrt{u}~\frac{1}{2}du=\frac{1}{2}$

$\frac{1}{2} \displaystyle\int_{1}^2}\sqrt{u}~ du=\frac{1}{2} u^{\frac{1}{2} } du=\frac{1}{2} *\frac{u^{\frac{1}{2}+1 } }{\frac{1}{2}+1\left \\] {{2} \atop {1}} \right. } =\frac{1}{2}*\frac{u^{\frac{3}{2} } }{\frac{3}{2} } \left ] {{2} \atop {1}} \right. =\frac{1}{2}*\frac{2}{3} \sqrt{u^{\frac{3}{2} } } \left ] {{2} \atop {1}} \right. =\frac{1}{3}*\sqrt{u^3} \left ] {{2} \atop {1}} \right. =\frac{1}{3} }( \sqrt{2^3}-\sqrt{1^3})=\frac{1}{3} (2\sqrt{2}-1)$