Question

Evaluate the integral  \(\int_{0}^{1} \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx\) Given that the integral can be expressed in the form \(a + b\sqrt{2} + c\sqrt{3}\), where \(a, b, c\) are rational numbers, find the value of \(2a + 3b - 4c\).

• A. \( 4 \)
• B. \( 10 \)
• C. \( 7 \)
• D. \( 8 \)

Answer 1

The Correct Option is D

The integral \(\int_{0}^{1} \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx\) is evaluated as follows:

Multiply the numerator and denominator by the conjugate of the denominator to simplify the integrand:

\(\int_{0}^{1} \frac{\sqrt{3+x} - \sqrt{1+x}}{\left(\sqrt{3+x} + \sqrt{1+x}\right)\left(\sqrt{3+x} - \sqrt{1+x}\right)} \, dx\)

Using the difference of squares identity \((a-b)(a+b) = a^2 - b^2\), this becomes:

\(\int_{0}^{1} \frac{\sqrt{3+x} - \sqrt{1+x}}{(3+x) - (1+x)} \, dx = \int_{0}^{1} \frac{\sqrt{3+x} - \sqrt{1+x}}{2} \, dx\)

Separate into two integrals:

\(\frac{1}{2}\left(\int_{0}^{1} \sqrt{3+x} \, dx - \int_{0}^{1} \sqrt{1+x} \, dx\right)\)

Integrate each part using a substitution of the form \(u = x + c\):

1. For \(\int_{0}^{1} \sqrt{3+x} \, dx\), let \(u = 3 + x\), so \(du = dx\). The limits change from \(x=0\) to \(u=3\) and \(x=1\) to \(u=4\).
2. For \(\int_{0}^{1} \sqrt{1+x} \, dx\), let \(v = 1 + x\), so \(dv = dx\). The limits change from \(x=0\) to \(v=1\) and \(x=1\) to \(v=2\).

Compute the definite integrals:

\(\int_{3}^{4} \sqrt{u} \, du = \left[\frac{2}{3}u^{3/2}\right]_{3}^{4} = \frac{2}{3}(4^{3/2} - 3^{3/2}) = \frac{2}{3}(8 - 3\sqrt{3})\)

\(\int_{1}^{2} \sqrt{v} \, dv = \left[\frac{2}{3}v^{3/2}\right]_{1}^{2} = \frac{2}{3}(2^{3/2} - 1^{3/2}) = \frac{2}{3}(2\sqrt{2} - 1)\)

Subtract the results:

\(\frac{1}{2}\left(\frac{2}{3}(8 - 3\sqrt{3}) - \frac{2}{3}(2\sqrt{2} - 1)\right)\)

Simplify:

\(= \frac{1}{3}\left((8 - 3\sqrt{3}) - (2\sqrt{2} - 1)\right)\)

\(= \frac{1}{3}(8 - 3\sqrt{3} - 2\sqrt{2} + 1)\)

\(= \frac{1}{3}(9 - 2\sqrt{2} - 3\sqrt{3})\)

\(= 3 - \frac{2}{3}\sqrt{2} - \sqrt{3}\)

This is in the form \(a + b\sqrt{2} + c\sqrt{3}\), where \(a = 3\), \(b = -\frac{2}{3}\), and \(c = -1\).

Calculate \(2a + 3b - 4c\):

\(2(3) + 3(-\frac{2}{3}) - 4(-1) = 6 - 2 + 4 = 8\)

The final value is 8.