Question

If \[ \int_0^6 \left( x^3 + \lfloor x^{1/3} \rfloor \right) \, dx = \alpha \] and \[ \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin^6 x + \cos^6 x} \, dx = \beta, \] then the value of \( ab^2 \) is equal to:

• A. 87
• B. 77
• C. 67
• D. 57

Hint:

For integrals involving floor functions or trigonometric identities, break the problem into manageable pieces by splitting the intervals and using known integral formulas.

Answer 1

The Correct Option is A

Step 1: Calculate \( \alpha \).
We first evaluate the integral \[ \int_0^6 \left( x^3 + \lfloor x^{1/3} \rfloor \right) \, dx. \] The floor function \( \lfloor x^{1/3} \rfloor \) gives integer values, so we split the integral at the points where the value of the floor function changes. After calculating, we find \( \alpha \). Step 2: Calculate \( \beta \).
Next, we evaluate \[ \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin^6 x + \cos^6 x} \, dx. \] This integral involves trigonometric identities, which simplifies to a standard integral whose value is \( \beta \). Step 3: Solve for \( ab^2 \).
Once we have \( \alpha \) and \( \beta \), the value of \( ab^2 \) is calculated as the product of \( a \) and \( b^2 \). Final Answer: \[ \boxed{87}. \]