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Published on July 10, 2025
Verify Rolle's Theorem for two functions: f(x) = e^x(sin x - cos x) and f(x) = x^3 - 6x^2 + 11x - 6, by checking the conditions of the theorem and finding the value(s) of c where the derivative is zero.
This problem involves evaluating the integral of cos(ln x) using integration by parts. The solution involves applying integration by parts twice and solving for the integral.
Find the solution y(t) to the second-order linear non-homogeneous differential equation y" - 3y' - 10y = 1 with initial conditions y(0) = -1 and y'(0) = 2. The solution involves finding the complementary and particular solutions and applying initial conditions.
Find the solution to a differential equation given its Laplace transform. The solution involves inverse Laplace transforms and the time-shifting property.
This calculus problem involves evaluating the integral of (e^x + 1) / (e^x - 1) dx. The solution uses substitution and partial fraction decomposition to find the antiderivative.
Solve the second-order linear differential equation y'' - y = 0 using the Method of Undetermined Power Series Coefficients, resulting in a solution involving hyperbolic functions.
The provided text contains a detailed solution to various integration problems, including trigonometric substitutions, partial fraction decomposition, and integration by parts. It also demonstrates the use of properties of definite integrals to simplify calculations.
The Laplace transform of the piecewise function f(t) is calculated using Heaviside step functions and the time-shifting property of Laplace transforms. The result is expressed in terms of the exponential function and the reciprocal of s squared.
The function f(x,y) = -2x^2 + 3xy^2 - 3y^2 + 4 has a local maximum value of 4 at (0,0).
The double integral, using polar coordinates and integration by parts, evaluates to π(e^2 + 1).
The solution attempts to derive a reduction formula for the integral \(\int_0^\infty \frac{x}{\cosh^n(x)} dx\) using integration by parts and Feynman's trick, but ultimately finds that expressing the integral in terms of hypergeometric functions is a more suitable approach.