Question

If \(\int _0^1\frac{dx}{\sqrt{x+1}-\sqrt{x}} = \sqrt{2}k\), then the value of \(k\) is...

• A. \(\frac{4}{\sqrt{3}}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{2}{\sqrt{3}}\)
• D. \(\frac{2}{3}\)

Hint:

Multiply by the conjugate: 1/(sqrt(x+1) - sqrt x) = sqrt(x+1) + sqrt x.

Answer 1

The Correct Option is B

Step 1: Idea:
The conjugate trick makes the denominator 1.

Step 2: Antiderivatives:
$\int\sqrt{x+1}dx = \frac23(x+1)^{3/2}$ and $\int\sqrt{x}dx = \frac23x^{3/2}$.

Step 3: Limits:
Upper: $\frac23(2\sqrt2 + 1)$. Lower: $\frac23(1)$. Difference: $\frac23\cdot2\sqrt2 = \frac{4\sqrt2}{3}$, so $\sqrt2k = \frac{4\sqrt2}{3}$ and $k = \frac43$.

Final Answer:
k equals 4/3, option (B). \[ \boxed{\frac{4}{3}} \]