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

To evaluate the integral \(\int_{0}^{1} \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx\), we proceed as follows: 

We apply a conjugate multiplication trick to simplify the integrand. Consider multiplying the numerator and denominator by the conjugate of the denominator:

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

This simplifies using the identity \(a^2 - b^2 = (a-b)(a+b)\), giving:

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

Split this into two separate integrals:

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

To solve each part, use the substitution \(u = x + c\) where necessary and integrate:

1. For \(\int_{0}^{1} \sqrt{3+x} \, dx\), substitute \(u = 3 + x\), \(\frac{du}{dx} = 1\), \(dx = du\). The limits change from \(x=0\) to \(x=1\), resulting in \(u=3\) to \(u=4\):
2. For \(\int_{0}^{1} \sqrt{1+x} \, dx\), substitute \(v = 1 + x\), \(\frac{dv}{dx} = 1\), \(dx = dv\). The limits change from \(x=0\) to \(x=1\), resulting in \(v=1\) to \(v=2\):

Now, compute the values:

\(\frac{2}{3}[4^{3/2} - 3^{3/2}]\) gives \(\frac{2}{3}[8 - 3\sqrt{3}]\), a\)

\(\frac{2}{3}[2^{3/2} - 1] = \frac{2}{3}[2\sqrt{2} - 1]\)

Subtract the two results:

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

Simplifying:

\(= \frac{1}{2}\left(\frac{16}{3} - \frac{6\sqrt{3}}{3} - \frac{4\sqrt{2}}{3} + \frac{2}{3}\right)\)

\(= \frac{1}{2} \cdot \frac{18}{3} - \frac{6\sqrt{3} + 4\sqrt{2}}{3}\)\)

The answer is of the form \(a + b\sqrt{2}+ c\sqrt{3}\)

Here, \(a = 3\), \(b = -\frac{2}{3}\), and \(c = -1\)

Finally, compute \(2a + 3b - 4c = 2(3) + 3(-\frac{2}{3}) - 4(-1)\)

\(= 6 - 2 + 4 = 8\)

Therefore, the correct value is 8.

Answer 2

Given:

\(\int_0^1 \frac{1}{\sqrt[3]{x} + \sqrt[3]{x} + \sqrt[3]{1 + x}} \, dx\)
 

Step 1: Rationalizing the Denominator

Rationalize the denominator:
\(\int \frac{\sqrt[3]{x + \sqrt{x}} - \sqrt[3]{x - \sqrt{x}}}{\left(\sqrt[3]{x + \sqrt{x}} + \sqrt[3]{x - \sqrt{x}}\right)} \, dx = \int \frac{\sqrt[3]{x + \sqrt{x}}}{2} \, dx\)

Step 2: Separating the Integral

Separate the integral:

\(\frac{1}{2} \left( \int \sqrt[3]{1 + \sqrt{x}} \, dx - \int \sqrt[3]{1 - \sqrt{x}} \, dx \right)\)

Step 3: Evaluating the Integrals

1. For \(\int \sqrt[3]{1 + \sqrt{x}} \, dx\):
\(\int \sqrt[3]{1 + \sqrt{x}} \, dx = \frac{3}{2} \cdot \frac{3}{4} \cdot 2 + \frac{2}{5} \Rightarrow \frac{3}{2} \left( 2 + \sqrt[3]{3} - 2^{3/2} \right) = \frac{3}{2} (3 - 3\sqrt{3})\)

2. For \(\int \sqrt[3]{1 - \sqrt{x}} \, dx\):

  \(\int \sqrt[3]{1 - \sqrt{x}} \, dx = \frac{3}{2} (3 - \sqrt{3}) = \frac{3}{2} (2\sqrt{5} - 1)\)
 

Step 4: Combining the Results

Combine the results:

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

From this, we find:
\(a = 3, \quad b = -\frac{2}{3}, \quad c = -1\)
Calculate:
\(2a + 3b - c = 2\frac{4}{3} + 3\frac{-4}{3} - 4(-1)=8\)