Show that the integral of x(sqrt(1 + sinx)) dx is less than

[maeveoneill]

Show that the integral of x(sqrt(1 + sinx)) dx is less than or equal to 1/sqrt2 on the interval from 0 - 1.

Could someone show me how to do thisss!! THANKS...i'm almost done the studying!

 

[galactus]

Less than \(\displaystyle \frac{1}{\sqrt{2}}\) over what interval?. For instance,

\(\displaystyle \L\\\int_{0}^{\frac{3{\pi}}{2}}x\sqrt{1+sin(x)}dx\approx{9.33}\)

That's certainly greater than \(\displaystyle \frac{1}{\sqrt{2}}\)

 

[pka]

\(\displaystyle \L
\begin{array}{l}
0 \le x \le 1\quad \Rightarrow \quad x\sqrt {1 + \sin (x)} \le \left( x \right)\sqrt {1 + 1} = x\sqrt 2 \\
\int\limits_0^1 {\sqrt 2 xdx} = \left. {\frac{{\sqrt 2 x^2 }}{2}} \right|_0^1 = \frac{{\sqrt 2 }}{2} = \frac{1}{{\sqrt 2 }} \\
\end{array}\)

 

[maeveoneill]

\(\displaystyle \L
\begin{array}{l}
0 \le x \le 1\quad \Rightarrow \quad x\sqrt {1 + \sin (x)} \le \left( x \right)\sqrt {1 + 1} = x\sqrt 2 \\
\int\limits_0^1 {\sqrt 2 xdx} = \left. {\frac{{\sqrt 2 x^2 }}{2}} \right|_0^1 = \frac{{\sqrt 2 }}{2} = \frac{1}{{\sqrt 2 }} \\
\end{array}\)

how do you get the part that incldues the sqrt of 1+1??? how do we know that it is greater than/equal to the function?