Question

Evaluate the integral \[ I = \int_0^1 \frac{x}{(1 - x)^{3/4}} \, dx \]

• A. \( \frac{4}{5} \)
• B. \( \frac{8}{15} \)
• C. \( \frac{14}{5} \)
• D. \( \frac{16}{5} \)

Hint:

For integrals with simple substitutions, break them into manageable parts and apply known integration rules.

Answer 1

The Correct Option is D

Step 1: Substitution Let: \[ u = 1 - x \] Then \( du = -dx \), and when \( x = 0 \), \( u = 1 \), and when \( x = 1 \), \( u = 0 \). The integral becomes: \[ I = \int_1^0 \frac{(1 - u)}{u^{3/4}} (-du) = \int_0^1 \frac{(1 - u)}{u^{3/4}} \, du \] We can break the integral into two parts: \[ I = \int_0^1 \frac{1}{u^{3/4}} \, du - \int_0^1 \frac{u}{u^{3/4}} \, du \] Step 2: Evaluate the first integral \[ I_1 = \int_0^1 u^{-3/4} \, du = \left[ 4u^{1/4} \right]_0^1 = 4 \] Step 3: Evaluate the second integral \[ I_2 = \int_0^1 u^{1/4} \, du = \left[ \frac{4u^{5/4}}{5} \right]_0^1 = \frac{4}{5} \] Step 4: Combine the results \[ I = I_1 - I_2 = 4 - \frac{4}{5} = \frac{16}{5} \] Thus, the value of the integral is: \[ \boxed{\frac{16}{5}} \]

Answer 2

Evaluate the integral:
\[ I = \int_0^1 \frac{x}{(1 - x)^{3/4}} \, dx \]

Step 1: Substitute \( t = 1 - x \), so \( x = 1 - t \) and \( dx = -dt \). When \( x = 0 \), \( t = 1 \); when \( x = 1 \), \( t = 0 \).

Step 2: Rewrite the integral:
\[ I = \int_{t=1}^{0} \frac{1 - t}{t^{3/4}} (-dt) = \int_0^1 \frac{1 - t}{t^{3/4}} dt = \int_0^1 (1 - t) t^{-3/4} dt \]

Step 3: Split the integral:
\[ I = \int_0^1 t^{-3/4} dt - \int_0^1 t^{1 - 3/4} dt = \int_0^1 t^{-3/4} dt - \int_0^1 t^{1/4} dt \]

Step 4: Evaluate each integral:
\[ \int_0^1 t^{-3/4} dt = \left[ \frac{t^{1/4}}{1/4} \right]_0^1 = 4 (1^{1/4} - 0) = 4 \]
\[ \int_0^1 t^{1/4} dt = \left[ \frac{t^{5/4}}{5/4} \right]_0^1 = \frac{4}{5} (1^{5/4} - 0) = \frac{4}{5} \]

Step 5: Calculate \( I \):
\[ I = 4 - \frac{4}{5} = \frac{20}{5} - \frac{4}{5} = \frac{16}{5} \]

Therefore,
\[ \boxed{\frac{16}{5}} \]