Question

The value of $\int_{-1}^{1} (\sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}) dx$ is

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Hint:

Whenever you see an integral from $-a$ to $a$, always check if the function is odd first!

Answer 1

The Correct Option is C

Step 1: Concept
Use the property of definite integrals for even and odd functions: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function.

Step 2: Meaning
A function is odd if $f(-x) = -f(x)$.

Step 3: Analysis
Let $f(x) = \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}$. Substitute $-x$ for $x$: $f(-x) = \sqrt{1 + (-x) + (-x)^2} - \sqrt{1 - (-x) + (-x)^2}$. $f(-x) = \sqrt{1 - x + x^2} - \sqrt{1 + x + x^2}$. $f(-x) = -(\sqrt{1 + x + x^2} - \sqrt{1 - x + x^2}) = -f(x)$. Since $f(x)$ is odd, the integral over the symmetric interval $[-1, 1]$ is zero.

Step 4: Conclusion
The value of the integral is 0. Final Answer: (C)