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: \[ \boxed{\frac{16}{5}} \]