Rearranging the differentiated equation gives (1 + x²) f'(x) = -4x f(x), which separates as (f'(x))/(f(x)) = -(4x)/(1 + x²). Integrating both sides gives ln |f(x)| = -2 ln(1 + x²) + C, so f(x) = (C)/((1 + x²)²). Using the initial condition from the original equation at x = 0, we have f(0) = 1, which gives C = 1. Thus, f(x) = (1)/((1 + x²)²).

Rearranging the differentiated equation gives (1 + x²) f'(x) = -4x f(x), which separates as (f'(x))/(f(x)) = -(4x)/(1 + x²). Integrating both sides gives ln |f(x)| = -2 ln(1 + x²) + C, so f(x) = (C)/((1 + x²)²). Using the initial condition from the original equation at x = 0, we have f(0) = 1, which gives C = 1. Thus, f(x) = (1)/((1 + x²)²).

Computing the first derivative of f(x) = (1 + x²)^(-2) gives f'(x) = -(4x)/((1 + x²)³). For all x > 0, f'(x) < 0, which means f(x) is strictly decreasing on [0, ∞). Therefore, the maximum value occurs at the boundary point x = 0, with maximum value f(0) = 1.

Computing the first derivative of f(x) = (1 + x²)^(-2) gives f'(x) = -(4x)/((1 + x²)³). For all x > 0, f'(x) < 0, which means f(x) is strictly decreasing on [0, ∞). Therefore, the maximum value occurs at the boundary point x = 0, with maximum value f(0) = 1.

Substitute x = tan θ, so dx = sec² θ dθ. As x ranges from 0 to ∞, θ ranges from 0 to (π)/(2). The integrand becomes (sec² θ)/((1 + tan² θ)²) = (sec² θ)/(sec^4 θ) = cos² θ. Evaluating the integral gives ∫_(0)^((π)/(2)) (1 + cos 2θ)/(2) dθ = [ (θ)/(2) + (sin 2θ)/(4) ]_0^((π)/(2)) = (π)/(4).

Substitute x = tan θ, so dx = sec² θ dθ. As x ranges from 0 to ∞, θ ranges from 0 to (π)/(2). The integrand becomes (sec² θ)/((1 + tan² θ)²) = (sec² θ)/(sec^4 θ) = cos² θ. Evaluating the integral gives ∫_(0)^((π)/(2)) (1 + cos 2θ)/(2) dθ = [ (θ)/(2) + (sin 2θ)/(4) ]_0^((π)/(2)) = (π)/(4).

Using the same substitution x = tan θ, when x = 1, θ = (π)/(4). The definite integral is ∫_(0)^((π)/(4)) cos² θ dθ = [ (θ)/(2) + (sin 2θ)/(4) ]_0^((π)/(4)) = ( (π)/(8) + (sin(π/2))/(4)) - 0 = (π)/(8) + (1)/(4).

Using the same substitution x = tan θ, when x = 1, θ = (π)/(4). The definite integral is ∫_(0)^((π)/(4)) cos² θ dθ = [ (θ)/(2) + (sin 2θ)/(4) ]_0^((π)/(4)) = ( (π)/(8) + (sin(π/2))/(4)) - 0 = (π)/(8) + (1)/(4).

Let $f \colon [0, \infty) \to \mathbb{R}$ be a continuous function satisfying the integral equation: $f(x) = \frac{1}{1 + x^2} - \frac{2}{1 + x^2} \int_{0}^{x} t \, f(t) \, dt \quad \text{for all } x \ge 0$ (i) 1. Find an explicit expression for f(x). 2. Determine the maximum value of f(x) on $[0, \infty)$ and find the value of $x$ at which this maximum is attained. 3. Evaluate the improper integral: $I = \int_{0}^{\infty} f(x) \, dx$ 4. Find the area of the region bounded by the curve $y = f(x)$ , the coordinate axes, and the vertical line $x = 1$ : $A = \int_{0}^{1} f(x) \, dx$ $

Answer: (i) $f(x) = \frac{1}{(1 + x^2)^2}$ (ii) Maximum value is $1$ attained at $x = 0$ (iii) $I = \frac{\pi}{4}$ (iv) $A = \frac{1}{4} + \frac{\pi}{8}$

Step-by-step solution

Step 1: Clear the denominator and differentiate using Leibniz's rule

Multiply both sides of the integral equation by $(1 + x^2)$ to obtain $(1 + x^2) f(x) = 1 - 2 \int_{0}^{x} t f(t) \, dt$ . Differentiating both sides with respect to $x$ using the product rule on the left-hand side and the Fundamental Theorem of Calculus (Leibniz's rule) on the right-hand side yields $2x f(x) + (1 + x^2) f'(x) = -2x f(x)$ .

Step 2: Solve the differential equation to find f(x)

Rearranging the differentiated equation gives $(1 + x^2) f'(x) = -4x f(x)$ , which separates as $\frac{f'(x)}{f(x)} = -\frac{4x}{1 + x^2}$ . Integrating both sides gives $\ln |f(x)| = -2 \ln(1 + x^2) + C$ , so $f(x) = \frac{C}{(1 + x^2)^2}$ . Using the initial condition from the original equation at $x = 0$ , we have $f(0) = 1$ , which gives $C = 1$ . Thus, $f(x) = \frac{1}{(1 + x^2)^2}$ .

Step 3: Find the maximum value of f(x) on [0, infinity)

Computing the first derivative of $f(x) = (1 + x^2)^{-2}$ gives $f'(x) = -\frac{4x}{(1 + x^2)^3}$ . For all $x > 0$ , f'(x) < 0, which means f(x) is strictly decreasing on $[0, \infty)$ . Therefore, the maximum value occurs at the boundary point $x = 0$ , with maximum value $f(0) = 1$ .

Step 4: Evaluate the improper integral from 0 to infinity

Substitute $x = \tan \theta$ , so $dx = \sec^2 \theta \, d\theta$ . As $x$ ranges from $0$ to $\infty$ , $\theta$ ranges from $0$ to $\frac{\pi}{2}$ . The integrand becomes $\frac{\sec^2 \theta}{(1 + \tan^2 \theta)^2} = \frac{\sec^2 \theta}{\sec^4 \theta} = \cos^2 \theta$ . Evaluating the integral gives $\int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \, d\theta = \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4}$ .

Step 5: Evaluate the area from $x = 0$ to $x = 1$

Using the same substitution $x = \tan \theta$ , when $x = 1$ , $\theta = \frac{\pi}{4}$ . The definite integral is $\int_{0}^{\frac{\pi}{4}} \cos^2 \theta \, d\theta = \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_{0}^{\frac{\pi}{4}} = \left( \frac{\pi}{8} + \frac{\sin(\pi/2)}{4} \right) - 0 = \frac{\pi}{8} + \frac{1}{4}$ .

Solve your own maths question free →