Question

The value of the improper integral \( \int_0^1 \frac{1}{4\sqrt{1-x}} dx \) is equal to _______.

• A. 4
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( \infty \)

Hint:

When the denominator of a fraction contains a square root of a linear expression like \( \sqrt{a-x} \), substituting the entire expression under the root usually simplifies the integral to a basic power rule form.

Answer 1

The Correct Option is C

Concept:
An improper integral is an integral where either the limits are infinite or the integrand becomes infinite within the range of integration. Here, the integrand \( \frac{1}{4\sqrt{1-x}} \) approaches infinity as \( x \to 1 \). We solve this using limits and substitution.

Step 1: Apply substitution.
Let \( u = 1 - x \). Then \( du = -dx \), which means \( dx = -du \). Change the limits of integration:
• When \( x = 0 \), \( u = 1 - 0 = 1 \).
• When \( x = 1 \), \( u = 1 - 1 = 0 \).

Step 2: Rewrite and integrate.
The integral becomes: \[ I = \int_1^0 \frac{1}{4\sqrt{u}} (-du) = \int_0^1 \frac{1}{4} u^{-1/2} du \] Using the power rule for integration \( \int u^n du = \frac{u^{n+1}}{n+1} \): \[ I = \frac{1}{4} \left[ \frac{u^{1/2}}{1/2} \right]_0^1 = \frac{1}{4} \left[ 2\sqrt{u} \right]_0^1 \]

Step 3: Evaluate the limits.
\[ I = \frac{1}{4} \left[ 2\sqrt{1} - 2\sqrt{0} \right] \] \[ I = \frac{1}{4} \times 2 = \frac{2}{4} = \frac{1}{2} \]