The given integral is:

$I = \int_0^5 \frac{dx}{\sqrt[3]{2 - x}}$

Solution Approach:

Let's use substitution.

Step 1: Let

$t = 2 - x$ $dt = -dx$

Changing the limits:

• When $x = 0$, then $t = 2$.
• When $x = 5$, then $t = -3$.

Thus, the integral transforms as:

$I = \int_{2}^{-3} \frac{-dt}{\sqrt[3]{t}}$

Rewriting $\sqrt[3]{t}$ as $t^{1/3}$:

$I = - \int_{2}^{-3} t^{-1/3} dt$

Step 2: Apply Power Rule

The integral of $t^n$ is:

$\int t^n dt = \frac{t^{n+1}}{n+1}, \quad \text{for } n \neq -1.$

Applying this for $n = -1/3$:

$I = - \left[ \frac{t^{2/3}}{2/3} \right]_{2}^{-3}$

$I = - \left[ \frac{3}{2} t^{2/3} \right]_{2}^{-3}$

Step 3: Evaluate at Limits

$I = - \frac{3}{2} \left[ (-3)^{2/3} - (2)^{2/3} \right]$

Now, we need to evaluate $(-3)^{2/3}$ and $2^{2/3}$:

$(-3)^{2/3} = \left( 3^2 \right)^{1/3} = 9^{1/3}$

$2^{2/3} = 2^{2/3}$

Thus,

$I = -\frac{3}{2} \left( 9^{1/3} - 2^{2/3} \right)$

This is the final result.

Would you like a numerical approximation or more details? Let me know!

Related Questions:

1. How do you evaluate fractional exponents like $(-3)^{2/3}$?
2. What is the general formula for integrals of the form $\int x^n dx$?
3. How does substitution help in evaluating integrals?
4. Can this integral be solved using another method?
5. What happens if the integral had $\sqrt{2-x}$ instead of $\sqrt[3]{2-x}$?

Tip: When handling fractional exponents, always rewrite radicals as powers to simplify integration.