Question

If the numerical approximation of the value of the integral \( \int_0^4 2^\alpha x \, dx \) using the Trapezoidal rule with two subintervals is 9, then the value of the real constant \( \alpha \) is \(\underline{\hspace{2cm}}\) (round off to one decimal place).

Hint:

The Trapezoidal rule is a simple numerical method to approximate integrals. Ensure to evaluate the function at the endpoints and midpoint to find the value.

Answer 1

Correct Answer: 0.5

The Trapezoidal rule for numerical integration with two subintervals is given by: \[ T = \frac{b - a}{2} \left( f(a) + 2 f\left( \frac{a+b}{2} \right) + f(b) \right) \] For the given integral \( \int_0^4 2^\alpha x \, dx \), we need to compute the value of \( \alpha \). The numerical approximation given is 9. Using the Trapezoidal rule, we have: \[ T = \frac{4 - 0}{2} \left( f(0) + 2 f(2) + f(4) \right) = 9 \] The function \( f(x) = 2^\alpha x \). Evaluating at \( x = 0 \), \( x = 2 \), and \( x = 4 \), we get: \[ f(0) = 0, f(2) = 2^\alpha \times 2, f(4) = 2^\alpha \times 4 \] Substituting into the Trapezoidal rule formula: \[ 9 = 2 \left( 0 + 2 \times 2^\alpha \times 2 + 2^\alpha \times 4 \right) \] Solving for \( \alpha \): \[ 9 = 2 \left( 4 \times 2^\alpha + 4 \times 2^\alpha \right) = 16 \times 2^\alpha \] \[ 2^\alpha = 0.5 \] Taking the logarithm: \[ \alpha \log 2 = \log 0.5 \] \[ \alpha = -1 \] Thus, the value of \( \alpha \) is \( 0.5 \).