answer:To find the Fourier transform of the function f(x) = 2 + 3sin(2x) - 4cos(5x), we will use the following formula for the continuous Fourier transform: F(ω) = ∫[f(x) * e^(-jωx)] dx where F(ω) is the Fourier transform of f(x), j is the imaginary unit (j^2 = -1), and ω is the angular frequency. First, let's rewrite the function f(x) using Euler's formula: f(x) = 2 + 3sin(2x) - 4cos(5x) = 2 + 3[Im(e^(j2x))] - 4[Re(e^(j5x))] Now, we can substitute this into the Fourier transform formula: F(ω) = ∫[2 + 3[Im(e^(j2x))] - 4[Re(e^(j5x))] * e^(-jωx)] dx We can split this integral into three parts: F(ω) = ∫[2 * e^(-jωx)] dx + ∫[3[Im(e^(j2x))] * e^(-jωx)] dx - ∫[4[Re(e^(j5x))] * e^(-jωx)] dx Now, we can solve each integral separately: 1) ∫[2 * e^(-jωx)] dx = 2 * ∫[e^(-jωx)] dx = 2 * [-j/(ω) * e^(-jωx)] + C₁ 2) ∫[3[Im(e^(j2x))] * e^(-jωx)] dx = 3 * ∫[Im(e^(j2x)) * e^(-jωx)] dx = 3 * ∫[Im(e^(j(2-ω)x))] dx = 3 * [-j/(2-ω) * e^(j(2-ω)x)] + C₂ (if ω ≠ 2) 3) ∫[4[Re(e^(j5x))] * e^(-jωx)] dx = 4 * ∫[Re(e^(j5x)) * e^(-jωx)] dx = 4 * ∫[Re(e^(j(5-ω)x))] dx = 4 * [1/(5-ω) * e^(j(5-ω)x)] + C₃ (if ω ≠ 5) Now, we can combine these results to find the Fourier transform F(ω): F(ω) = 2 * [-j/(ω) * e^(-jωx)] + 3 * [-j/(2-ω) * e^(j(2-ω)x)] - 4 * [1/(5-ω) * e^(j(5-ω)x)] + C where C is the constant of integration. Since we are considering values of x between 0 and 2π, we can evaluate the Fourier transform at these limits: F(ω) = 2 * [-j/(ω) * (e^(-jω(2π)) - e^(0))] + 3 * [-j/(2-ω) * (e^(j(2-ω)(2π)) - e^(0))] - 4 * [1/(5-ω) * (e^(j(5-ω)(2π)) - e^(0))] + C Since e^(j2πn) = 1 for any integer n, we have: F(ω) = 2 * [-j/(ω) * (1 - 1)] + 3 * [-j/(2-ω) * (1 - 1)] - 4 * [1/(5-ω) * (1 - 1)] + C F(ω) = C Thus, the Fourier transform of the function f(x) = 2 + 3sin(2x) - 4cos(5x) for values of x between 0 and 2π is a constant, which depends on the initial conditions of the problem.

answer:To find the Fourier transform of the function f(x) = e^(-2|x|), we need to break it into two parts, one for x ≥ 0 and another for x < 0, since the function is defined using the absolute value of x. For x ≥ 0, f(x) = e^(-2x), and for x < 0, f(x) = e^(2x). The Fourier transform F(k) of a function f(x) is defined as: F(k) = ∫[f(x) * e^(-ikx)] dx, where the integral is taken from -∞ to ∞. Now, we can write the Fourier transform of f(x) as the sum of two integrals: F(k) = ∫[e^(-2x) * e^(-ikx) dx] + ∫[e^(2x) * e^(-ikx) dx], where the first integral is taken from 0 to ∞, and the second integral is taken from -∞ to 0. Let's evaluate the first integral: ∫[e^(-2x) * e^(-ikx) dx] = ∫[e^(-2x - ikx) dx] = ∫[e^(-x(2 + ik)) dx] from 0 to ∞. Using the formula for the integral of an exponential function, we get: = [-1/(2 + ik)] * e^(-x(2 + ik)) evaluated from 0 to ∞. = [-1/(2 + ik)] * [e^(-∞) - e^(0)] = (-1/(2 + ik)) * (0 - 1) = 1/(2 + ik). Now, let's evaluate the second integral: ∫[e^(2x) * e^(-ikx) dx] = ∫[e^(x(2 - ik)) dx] from -∞ to 0. = [1/(2 - ik)] * e^(x(2 - ik)) evaluated from -∞ to 0. = [1/(2 - ik)] * [e^(0) - e^(-∞)] = (1/(2 - ik)) * (1 - 0) = 1/(2 - ik). Finally, we can write the Fourier transform F(k) as the sum of these two results: F(k) = 1/(2 + ik) + 1/(2 - ik). To combine these two fractions, we can find a common denominator: F(k) = [(2 - ik) + (2 + ik)] / [(2 + ik)(2 - ik)]. Simplifying the numerator: F(k) = [4] / [4 + k^2]. So, the Fourier transform of the function f(x) = e^(-2|x|) is: F(k) = 4 / (4 + k^2).

answer:The Fourier transform of a function f(x) is defined as: F(k) = ∫[f(x) * e^(-2πikx)] dx, where the integral is taken over the range (-∞, +∞). For the given function f(x) = x^2, we have: F(k) = ∫[(x^2) * e^(-2πikx)] dx, where the integral is taken over the range (-∞, +∞). To solve this integral, we can use integration by parts twice. Let u = x^2 and dv = e^(-2πikx) dx. Then, du = 2x dx and v = (-1/(2πik)) * e^(-2πikx). Applying integration by parts: F(k) = uv | (-∞, +∞) - ∫[v * du] = (-1/(2πik)) * x^2 * e^(-2πikx) | (-∞, +∞) - ∫[(-1/(2πik)) * e^(-2πikx) * 2x dx]. The first term, (-1/(2πik)) * x^2 * e^(-2πikx) | (-∞, +∞), goes to zero as x approaches ±∞. So, we are left with: F(k) = (1/(πik)) * ∫[x * e^(-2πikx) dx]. Now, we apply integration by parts again. Let u = x and dv = e^(-2πikx) dx. Then, du = dx and v = (-1/(2πik)) * e^(-2πikx). F(k) = (1/(πik)) * [uv | (-∞, +∞) - ∫[v * du]] = (1/(πik)) * [(-1/(2πik)) * x * e^(-2πikx) | (-∞, +∞) - ∫[(-1/(2πik)) * e^(-2πikx) dx]]. The first term, (-1/(2πik)) * x * e^(-2πikx) | (-∞, +∞), goes to zero as x approaches ±∞. So, we are left with: F(k) = (1/(πik)) * [(-1/(2πik)) * ∫[e^(-2πikx) dx]] = (1/(πik)) * [(-1/(2πik)) * (-1/(2πik)) * e^(-2πikx) | (-∞, +∞)]. The term e^(-2πikx) | (-∞, +∞) goes to zero as x approaches ±∞. Therefore, the Fourier transform of the function f(x) = x^2 is: F(k) = 0.

answer:To find the Fourier Transform of the given signal f(t) = 2cos(3πt) + 4sin(6πt), we will use the definition of the continuous-time Fourier Transform: F(ω) = ∫[f(t) * e^(-jωt)] dt, where the integral is taken over all time t. Let's break the signal into two parts and find the Fourier Transform of each part separately: 1) g(t) = 2cos(3πt) 2) h(t) = 4sin(6πt) Now, let's find the Fourier Transform of each part: 1) G(ω) = ∫[2cos(3πt) * e^(-jωt)] dt 2) H(ω) = ∫[4sin(6πt) * e^(-jωt)] dt For G(ω), we can use the Euler's formula: cos(x) = (e^(jx) + e^(-jx))/2 G(ω) = ∫[2 * (e^(j3πt) + e^(-j3πt))/2 * e^(-jωt)] dt G(ω) = ∫[e^(j(3π - ω)t) + e^(-j(3π + ω)t)] dt Now, we can integrate G(ω) with respect to t: G(ω) = [1/(j(3π - ω)) * e^(j(3π - ω)t) - 1/(j(3π + ω)) * e^(-j(3π + ω)t)] evaluated from -∞ to ∞ Since the signal is given for time t between 0 and 1, we can evaluate the integral from 0 to 1: G(ω) = [1/(j(3π - ω)) * (e^(j(3π - ω)) - 1) - 1/(j(3π + ω)) * (e^(-j(3π + ω)) - 1)] Now, let's find the Fourier Transform of h(t): H(ω) = ∫[4sin(6πt) * e^(-jωt)] dt Using the Euler's formula for sine: sin(x) = (e^(jx) - e^(-jx))/(2j) H(ω) = ∫[4 * (e^(j6πt) - e^(-j6πt))/(2j) * e^(-jωt)] dt H(ω) = ∫[(e^(j(6π - ω)t) - e^(-j(6π + ω)t))/j] dt Now, we can integrate H(ω) with respect to t: H(ω) = [1/(j(6π - ω)) * e^(j(6π - ω)t) - 1/(j(6π + ω)) * e^(-j(6π + ω)t)] evaluated from -∞ to ∞ Again, since the signal is given for time t between 0 and 1, we can evaluate the integral from 0 to 1: H(ω) = [1/(j(6π - ω)) * (e^(j(6π - ω)) - 1) - 1/(j(6π + ω)) * (e^(-j(6π + ω)) - 1)] Now, we can find the Fourier Transform of the original signal f(t) by adding the Fourier Transforms of g(t) and h(t): F(ω) = G(ω) + H(ω) F(ω) = [1/(j(3π - ω)) * (e^(j(3π - ω)) - 1) - 1/(j(3π + ω)) * (e^(-j(3π + ω)) - 1)] + [1/(j(6π - ω)) * (e^(j(6π - ω)) - 1) - 1/(j(6π + ω)) * (e^(-j(6π + ω)) - 1)] This is the Fourier Transform of the given signal f(t) = 2cos(3πt) + 4sin(6πt).