Evaluate the two improper integrals: a) \( \int_{0}^{\infty} e^{-4 x} d x \) (4 Marks) b) \( \int_{0}^{\infty} \frac{d x}{x^{2}+9} \) (5 Marks)

To evaluate the given improper integrals, we'll solve them one at a time. a) For the integral \( \int_{0}^{\infty} e^{-4 x} dx \): This is an exponential integral. We can compute it as follows: \[ \int e^{-4 x} dx = -\frac{1}{4} e^{-4x} + C \] Now, we need to evaluate the definite integral from 0 to \( \infty \): \[ \int_{0}^{\infty} e^{-4 x} dx = \lim_{b \to \infty} \left( -\frac{1}{4} e^{-4x} \right) \Big|_{0}^{b} \] Evaluating the boundary terms: At \( b \to \infty \), \( e^{-4b} \to 0 \) and at \( x = 0 \), we have: \[ \left. -\frac{1}{4} e^{-4x} \right|_{0}^{b} = 0 - \left( -\frac{1}{4} e^{-0} \right) = 0 + \frac{1}{4} = \frac{1}{4} \] Therefore, \[ \int_{0}^{\infty} e^{-4 x} dx = \frac{1}{4}. \] b) For the integral \( \int_{0}^{\infty} \frac{dx}{x^{2}+9} \): This integral can also be evaluated using the formula for the arctangent function. Recall the integral: \[ \int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \] In our case, \( a = 3 \) (since \( 9 = 3^2 \)). Thus, we have: \[ \int \frac{dx}{x^{2}+9} = \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + C \] Now evaluate the definite integral: \[ \int_{0}^{\infty} \frac{dx}{x^{2}+9} = \lim_{b \to \infty} \left( \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) \right) \Big|_{0}^{b} \] Evaluating the limits: At \( b \to \infty \), \( \tan^{-1} \left( \frac{b}{3} \right) \to \frac{\pi}{2} \) and at \( x = 0 \), we have: \[ = \frac{1}{3} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{6} \] Thus, \[ \int_{0}^{\infty} \frac{dx}{x^{2}+9} = \frac{\pi}{6}. \] In summary, the values of the integrals are: a) \( \int_{0}^{\infty} e^{-4 x} dx = \frac{1}{4} \) b) \( \int_{0}^{\infty} \frac{dx}{x^{2}+9} = \frac{\pi}{6} \)