Question

The value of the integral \[ \int_{0}^{9} \frac{x - 1}{1 + \sqrt{x}} \, dx \] is _____________. (in integer) 
 

Hint:

When a radical like $\sqrt{x}$ appears, substitute $t = \sqrt{x}$ to simplify the integral into a polynomial form.

Answer 1

Correct Answer: 9

Step 1: Substitution.
Let $\sqrt{x} = t \Rightarrow x = t^2, \, dx = 2t \, dt$. Limits: when $x=0, t=0$; when $x=9, t=3$.
Step 2: Substitute into the integral. \[ \int_{0}^{9} \frac{x - 1}{1 + \sqrt{x}} \, dx = \int_{0}^{3} \frac{t^2 - 1}{1 + t} \times 2t \, dt \] \[ = 2 \int_{0}^{3} \frac{t(t^2 - 1)}{1 + t} \, dt = 2 \int_{0}^{3} t(t - 1) \, dt \quad (\text{since } t^2 - 1 = (t - 1)(t + 1)) \]
Step 3: Simplify and integrate. \[ 2 \int_{0}^{3} (t^2 - t) \, dt = 2 \left[ \frac{t^3}{3} - \frac{t^2}{2} \right]_0^3 = 2 \left( 9 - \frac{9}{2} \right) = 2 \times \frac{9}{2} = 9. \] Correction check: Wait — algebra simplification step rechecked. Actually, \[ \frac{t^2 - 1}{1 + t} = t - 1 + \frac{t}{1+t} - \text{No, expand properly.} \] Let’s properly divide $(t^2 - 1)$ by $(1 + t)$: \[ \frac{t^2 - 1}{1 + t} = t - 1. \] Then, \[ I = 2\int_0^3 t(t-1)\,dt = 2\int_0^3 (t^2 - t)\,dt = 2\left[\frac{t^3}{3} - \frac{t^2}{2}\right]_0^3 = 2(9 - 4.5) = 9. \]
Step 4: Conclusion.
\[ \boxed{I = 9} \] (round to integer).