Antiderivative of e Antiderivative of e Antiderivative is the term used in the calculus mathematics and especially in the topic of the differential equations. The anti derivatives are the type of the integral equations in which we don’t have limits on the integration symbol. It is the reverse process of the derivatives or we can say it as the process of reverse differentiation in the calculus mathematics. Let's take an example of the derivative of the expression 3 x4 + 5 is 12 x3 . The anti derivative of the 12 x3 will be 3 x4 + 5. Another example can be 5 x 3 – 2 x 2 – 3 x + 5, its anti derivative will be (x 4/ 4) – 5 (x 3/ 3) – 7 (x 2 /2) + 5 x = x 4/ 4 – 5 x 3/ 3 – 7 x 2 /2 + 5 x + k. Here the term k is constant and called as the constant of integration in the calculus. Finding the anti derivative is the process of solving the differential equations F' = f for any function F. . Know More About How to solve Second order Differential Equations

These differential equations describe the state of any particle. For example any moving particle can be given by its velocity and its acceleration at a particular time, and all the equation related to the velocity is the first order differential equation and the equation related to the acceleration is in the form of second order equation. In initial value problems some additional conditionals are given that allows us to pick a particular solution form the general solution set. For example we have a question as the initial value problem is dx /dt = 2t +sin t and the initial condition of the problem is value of X at initial point (x0 = 0). Now talking about the solving initial value problems in antiderivatives, we have to go through by some of the example and solutions of them. Say for example: Suppose F(0) = 1 , and F'(x) = 8 x3 - 10 sec2 (5x), Then we have to find the value of the function F(x). So here we are given in the example, the derivative of the function and some of the conditions with it, and we are asked to find the original function with the help of all given values. Thus we will go through the anti derivative of the given derivative. Now applying the concept for finding the anti derivative we have: Learn More About Odd Integrands

F(x) = ∫ (8 x3 - 10 sec2 (5x)) dx, F(x) = 8 ∫ x3 dx - 10 ∫ sec2 (5x) dx, F(x) = 8 x4/4 - 10 tan (5x) / 5 + k, F(x) = 2 x4 - 2 tan (5x) + k (k is integration constant) Now we can use all the given initial values of the conditions. So putting the values in functions for k: F(x) = 2 x4 - 2 tan (5x) + k, As F(0) = 1 so, 1 = 2 x 0 -2 tan (0) + k, That’s why k = 1; Hence we will have solution of the question: F(x) = 2 x4 - 2 tan (5x) + 1

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