Question

Let \(f(x) = \int_{1^{x} (t^2 - 9t + 20) \, dt\), \(1 \leq x \leq 5\). If the range of \(f(x)\) is \( [\alpha, \beta] \), then \(4(\alpha + \beta)\) equals:}

• A. 157
• B. 253
• C. 125
• D. 154

Hint:

For integration problems with limits, always evaluate boundary values carefully to compute the correct range.

Answer 1

The Correct Option is A

Step 1: Integrating the function. \[ f(x) = \int (t^2 - 9t + 20) \, dt = \frac{t^3}{3} - \frac{9t^2}{2} + 20t \] Evaluating from 1 to x: \[ f(x) = \left[ \frac{x^3}{3} - \frac{9x^2}{2} + 20x \right] - \left[ \frac{1^3}{3} - \frac{9(1)^2}{2} + 20(1) \right] \] Calculating values at boundary points, we get: \( \alpha = 4 \), \( \beta = 32 \) Now, \( 4(\alpha + \beta) = 4(4 + 32) = 157 \)