If _0^11√3+x+√1+xdx=a+b√2+c√3, where a,b,c are rational numbers, then 2a+3b-4c is equal to…

Question

If $\int _{0}^{1}\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx=a+b\sqrt{2}+c\sqrt{3}$, where $a,b,c$ are rational numbers, then $2a+3b-4c$ is equal to :

Accepted Answer

$\begin{gathered} \int _{0}^{1}\frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx=\int _{0}^{1}\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)}dx \ \frac{1}{2}[[|\int _{0}^{1}\sqrt{3+x}dx-\int _{0}^{1}(\sqrt{1+x})dx]]| \end{gathered}$

$\begin{gathered} \frac{1}{2}{[2\frac{(3+x{)}^{\frac{3}{2}}}{3}-\frac{2(1+x{)}^{\frac{3}{2}}}{3}]}_{1}^{0} \ \frac{1}{2}[\frac{2}{3}(8-3\sqrt{3})-\frac{2}{3}({2}^{\frac{3}{2}}-1)] \ \frac{1}{3}[8-3\sqrt{3}-2\sqrt{2}+1] \ =3-\sqrt{3}-\frac{2}{3}\sqrt{2}=a+b\sqrt{2}+c\sqrt{3} \ a=3,b=-\frac{2}{3},c=-1 \ 2a+3b-4c=6-2+4=8 \end{gathered}$

Solutions · 1