# Math Help - Integral

1. ## Integral

Hello, help me please with this integral :

prove that : $\left|\Large \int_{2x-1}^{x^{2}}t\sqrt{3+cost}\,dt \right| \leq x^{4}+ (2x-1)^{2}$

2. Originally Posted by linda2005
Hello, help me please with this integral :

prove that : $\left|\Large \int_{2x-1}^{x^{2}}t\sqrt{3+cost}\,dt \right| \leq x^{4}+ (2x-1)^{2}$
Note that $t \sqrt{3 + \cos t} \leq t \sqrt{3 + 1} = 2t$.

3. Originally Posted by linda2005
Hello, help me please with this integral :

prove that : $\left|\Large \int_{2x-1}^{x^{2}}t\sqrt{3+cost}\,dt \right| \leq x^{4}+ (2x-1)^{2}$
Well

$-1 \le \cos t \le 1$ so $2 \le 3 + \cos t \le 4$ so $\sqrt{2} \le \sqrt{3 + \cos t }\le 2$
$\int_{2x-1}^{x^2}\sqrt{2}t \,dt \le \int_{2x-1}^{x^2}t \sqrt{3 + \cos t }\, dt \le \int_{2x-1}^{x^{2}}2t\,dt$

$\left.\frac{\sqrt{2}}{2}t^2\right|_{2x-1}^{x^2} \le \int_{2x-1}^{x^2}t \sqrt{3 + \cos t }\, dt \le \left. t^2\right|_{2x-1}^{x^2}$

$\frac{\sqrt{2}}{2}\left( x^4 - (2x-1)^2\right)\le \int_{2x-1}^{x^2}t \sqrt{3 + \cos t }\, dt \le x^4 - (2x-1)^2\le x^4 + (2x-1)^2$