Question

Evaluate the integral \[ I = \int_{-1}^{1} \left( \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2} \right) \, dx \]

• A. \( 2 \)
• B. \( 4 \)
• C. \( 0 \)
• D. \( 8 \)

Hint:

When integrating functions over symmetric intervals, check if the function is odd or even. If it’s odd, the integral over a symmetric range will be zero.

Answer 1

The Correct Option is C

Step 1: Simplifying the integrand The given integral is: \[ I = \int_{-1}^{1} \left( \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2} \right) \, dx \] Notice that the integrand involves symmetric limits of integration, i.e., from \( -1 \) to \( 1 \). Let's check if the two terms \( \sqrt{1 + x + x^2} \) and \( \sqrt{1 - x + x^2} \) are symmetric.
Since the integrand is an odd function (the subtraction of two functions that are symmetric about the origin), the integral of an odd function over a symmetric interval from \( -a \) to \( a \) is zero. Thus, the value of the integral is: \[ I = 0 \]

Answer 2

Evaluate the integral:
\[ I = \int_{-1}^{1} \left( \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2} \right) \, dx \]

Step 1: Define the integrand as:
\[ f(x) = \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2} \]

Step 2: Check the symmetry of \( f(x) \):
Calculate \( f(-x) \):
\[ f(-x) = \sqrt{1 - x + x^2} - \sqrt{1 + x + x^2} = -f(x) \]

Step 3: Since \( f(-x) = -f(x) \), the function \( f(x) \) is odd.

Step 4: The integral of an odd function over symmetric limits \( [-a, a] \) is zero:
\[ \int_{-a}^a f(x) \, dx = 0 \]

Therefore,
\[ \boxed{0} \]