# Thread: Prove that ... Help!!!

1. ## Prove that ... Help!!!

Prove that
9 < integral (0,3) (4x + 1)^(1/4)dx + integral (1,3) (4x − 1)^(1/4)dx < 9.0001

2. Originally Posted by mathdanny
Prove that
9 < integral (0,3) (4x + 1)^(1/4)dx + integral (1,3) (4x − 1)^(1/4)dx < 9.0001
$\displaystyle \int_0^3{(4x + 1)^{\frac{1}{4}}\,dx} + \int_1^3{(4x - 1)^{\frac{1}{4}}\,dx} = \frac{1}{4}\int_0^3{4(4x + 1)^{\frac{1}{4}}\,dx} + \frac{1}{4}\int_1^3{(4x - 1)^{\frac{1}{4}}\,dx}$

$\displaystyle = \frac{1}{4}\left[\frac{4}{5}(4x + 1)^{\frac{5}{4}}\right]_0^3 +\frac{1}{4}\left[\frac{4}{5}(4x - 1)^{\frac{5}{4}}\right]_1^3$

$\displaystyle = \frac{1}{5}\left[(4x + 1)^{\frac{5}{4}}\right]_0^3 + \frac{1}{5}\left[(4x - 1)^{\frac{5}{4}}\right]_1^3$

$\displaystyle = \frac{1}{5}\left[13^{\frac{5}{4}} - 1^{\frac{5}{4}}\right] + \frac{1}{5}\left[11^{\frac{5}{4}} - 3^{\frac{5}{4}}\right]$

What does the calculator say this is?

3. Originally Posted by Prove It
$\displaystyle \int_0^3{(4x + 1)^{\frac{1}{4}}\,dx} + \int_1^3{(4x - 1)^{\frac{1}{4}}\,dx} = \frac{1}{4}\int_0^3{4(4x + 1)^{\frac{1}{4}}\,dx} + \frac{1}{4}\int_1^3{(4x - 1)^{\frac{1}{4}}\,dx}$

$\displaystyle = \frac{1}{4}\left[\frac{4}{5}(4x + 1)^{\frac{5}{4}}\right]_0^3 +\frac{1}{4}\left[\frac{4}{5}(4x - 1)^{\frac{5}{4}}\right]_1^3$

$\displaystyle = \frac{1}{5}\left[(4x + 1)^{\frac{5}{4}}\right]_0^3 + \frac{1}{5}\left[(4x - 1)^{\frac{5}{4}}\right]_1^3$

$\displaystyle = \frac{1}{5}\left[13^{\frac{5}{4}} - 1^{\frac{5}{4}}\right] + \frac{1}{5}\left[11^{\frac{5}{4}} - 3^{\frac{5}{4}}\right]$

What does the calculator say this is?
I calculated this value and I got 7.9538... This value is too small! So, there is should the other the way to prove it! I tried many ways, but all gave me so small value!