Antiderivative Calculator is a free online tool that displays antiderivative (integration) functions. STUDYQUERIES’S online antiderivative calculator tool speeds up calculations, and it displays the integrated value in a fraction of a second.

**How to Use the Antiderivative Calculator With Steps? **

To use the antiderivative calculator, follow these steps:

**Step 1:**Enter the function into the input field**Step 2:**Click the “Solve” button to get the antiderivative**Step 3:**In the new window, the antiderivative of the given function will be displayed

Antiderivative Calculator

**What Is Antiderivative And Antiderivative Calculator?**

The antiderivative is the name we sometimes, (rarely) give to the operation that goes backward from the derivative of a function to the function itself. Since the derivative does not determine the function completely (you can add any constant to your function and the derivative will be the same), you have to add additional information to go back to an explicit function as anti-derivative.

Thus we sometimes say that the antiderivative of a function is a function plus an arbitrary constant. Thus the antiderivative of \(cos(x)\) is \(sin(x)+c\).

The more common name for the antiderivative is the indefinite integral. This is the identical notion, merely a different name for it.

A wavy line is used as a symbol for it. Thus the sentence “the antiderivative of \(cos(x)\) is \(sin(x)+c\)” is usually stated as: the indefinite integral of \(cos(x)\) is \(sin(x)+c\), and this is generally written as

$$\int_{}^{}cos(x)dx=sin(x)+c$$

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Actually, this is bad notation. The variable x that occurs on the right is a variable and represents the argument of the sine function. The symbols on the left merely say that the function whose antiderivative we are looking for is the cosine function. You will avoid confusion if you express this using an entirely different symbol (say y) on the left to denote this. The proper way to write this is then

$$\int_{}^{}cos(y)dy=sin(y)+c$$

**What Is +C?**

It is a reminder that the derivative of a constant is 0 so an anti-derivative as an inverse operation to a derivative is not completely determined. You can add any constant to an anti-derivative and get another one. Some believe that it was invented by pedants to torture students by penalizing them for occasionally ignoring this boring fact.

We can apply this to each term in a polynomial, and find its anti-derivative.

Thus, the anti-derivative of

$$3x^3-4x^2-x+7$$

is $$\frac{3x^4}{4} – \frac{4x^3}{3} – \frac{x^2}{2} + 7x + c$$

Students typically find this so easy that when they are forced to find such an anti-derivative on a test, often their minds are already focused on the next question, and they absent-mindedly forget and differentiate instead of anti-differentiating one or perhaps all terms. Please avoid this error.

**Calculating Areas With Antiderivatives**

A general method to find the area under a graph \(y=f(x)\) between \(x=a\)\ and \(x=b\)\ is given by the following important theorem.

**Theorem**

Let \(f(x)\) be a continuous real-valued function on the interval \([a,b]\). Then

$$\int_{a}^{b}f(x)dx=[F(x)]_b^a=F(b)−F(a)$$

where \(F(x)\) is any antiderivative of \(f(x)\).

The notation \([F(x)]_b^a\) is just a shorthand to substitute \(x=a\)\ and \(x=b\) into F(x) and subtract; it is synonymous with F(b)−F(a).

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**Note** that any antiderivative of \(f(x)\) will work in the above theorem. Indeed, if we have two different antiderivatives \(F(x)\)\ and \(G(x)\)\ of \(f(x)\), then they must differ by a constant, so \(G(x)=F(x)+c\) for some constant \(c\). Then we have to get the same answer whether we use \(F\)\ or \(G\), since

\(G(b)−G(a)=(F(b)+c)−(F(a)+c)=F(b)−F(a)\).

We give a proof of the theorem, assuming our previous statements:

- The derivative of the area function is \(f(x)\(
- Any two antiderivatives of \(f(x)\( differ by a constant.

\(\color{red}{Find \int_{0}^{5}(x^2+1)dx.}\)

\(\int_{0}^{5}(x^2+1)dx\)

\(=[\frac{1}{3}(x^3)+x]_0^5\)

\(=[\frac{1}{3}(5^3)+5]-[\frac{1}{3}(0^3)+0]\)

\(=\frac{125}{3}+5\)

\(=\frac{140}{3}\)

**How Can We Calculate Velocity And Acceleration With Antiderivative?**

For now, let’s look at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. We examine various techniques for finding antiderivatives of more complicated functions later in the text (Introduction to Techniques of Integration).

This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration, we introduced the kinematic functions of velocity and acceleration using the derivative.

By taking the derivative of the position function we found the velocity function, and likewise, by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.

**Kinematic Equations from Integral Calculus**

Let’s begin with a particle with an acceleration \(a(t)\) is a known function of time. Since the time derivative of the velocity function is acceleration,

$$\frac{d}{dt}v(t)=a(t)$$,

we can take the indefinite integral of both sides, finding

$$\int_{}^{}\frac{d}{dt}v(t)dt=\int_{}^{}a(t)dt+C_1$$

where \(C_1\) is a constant of integration. Since \(\int_{}^{}\frac{d}{dt}v(t)dt=v(t)\), the velocity is given by

$$v(t)=\int_{}^{}a(t)dt+C_1$$

Similarly, the time derivative of the position function is the velocity function,

$$\frac{d}{dt}x(t)=v(t)$$

Thus, we can use the same mathematical manipulations we just used and find

$$x(t)=\int_{}^{}v(t)dt+C_2$$

where \(C_2\) is the second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With \(a(t) = a\)\ constant, and doing the integration in, we find

$$v(t)=\int_{}^{}a(t)dt+C_1=at+C_1$$

If the initial velocity is \(v(0) = v_0\), then

\(v_0=0+C_1\)

Then, \(C_1 = v_0\) and

\(v(t)=v_0+at\),

which is (Equation). Substituting this expression gives

\(x(t)=\int_{}^{}(v_0+at)dt+C_2\)

Doing the integration, we find

\(x(t)=v_0t+\frac{1}{2}at^2+C_2\)

If \(x(0) = x_0), we have

\(x_0=0+0+C_2\);

so, \(C_2 = x_0\). Substituting back into the equation for \(x(t)\), we finally have

\(x(t)=x_0+v_0t+\frac{1}{2}at^2\).

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\(\color{red}{A\ motorboat\ is\ traveling\ at\ a\ constant\ velocity\ of\ 5.0 \frac{m}{s}}\),

\(\color{red}{when\ it\ starts\ to\ decelerate\ to\ arrive\ at\ the\ dock.\ Its\ acceleration\ is\ a(t)=−\frac{1}{4}t(\frac{m}{s^2})}\).

- What is the velocity function of the motorboat?
- At what time does the velocity reach zero?
- What is the position function of the motorboat?
- What is the displacement of the motorboat from the time it begins to decelerate to when the velocity is zero?
- Graph the velocity and position functions.

**Strategy**

- To get the velocity function we must integrate and use initial conditions to find the constant of integration.
- We set the velocity function equal to zero and solve for t.
- Similarly, we must integrate to find the position function and use initial conditions to find the constant of integration.
- Since the initial position is taken to be zero, we only have to evaluate the position function at \(t=0\).

We take \(t = 0\) to be the time when the boat starts to decelerate.

1: From the functional form of the acceleration we can solve to get \(v(t)\):

$$v(t)=\int_{}^{}a(t)dt+C_1$$

$$=\int_{}^{}(−\frac{1}{4}t)dt+C_1$$

$$=−\frac{1}{8}t^2+C_1$$

At \(t = 0\) we have \(v(0) = 5.0 \frac{m}{s} = 0 + C_1, so C_1 = 5.0 \frac{m}{s}\) or

$$v(t)=5.0\frac{m}{s}−\frac{1}{8}t^2$$

**2:** $$v(t)=0$$

$$0=5.0\frac{m}{s}−\frac{1}{8}t^2\Longrightarrow t=6.3s$$

**3:** $$x(t)=\int_{}^{}v(t)dt+C_2$$

$$=\int_{}^{}(5.0−\frac{1}{8}t^2)dt+C_2$$

$$=5.0t−\frac{1}{24}t^3+C_2$$

At \(t = 0\), we set \(x(0) = 0 = x_0\), since we are only interested in the displacement from when the boat starts to decelerate. We have

\(x(0)=0=C_2\). Therefore, the equation for the position is

$$x(t)=5.0t−\frac{1}{24}t^3$$

**4:** Graph A is a plot of velocity in meters per second as a function of time in seconds. Velocity is five meters per second in the beginning and decreases to zero. Graph B is a plot of position in meters as a function of time in seconds. The position is zero at the beginning, increases reaching a maximum between six and seven seconds, and then starts to decrease.

**Initial Value Problems**

We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.

A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation

$$\frac{dy}{dx}=f(x)$$

is a simple example of a differential equation. Solving this equation means finding a function y with a derivative \(f\). Therefore, the solutions of Equation are the antiderivatives of \(f\). If \(F\) is one antiderivative of \(f \), every function of the form \(y=F(x)+C\) is a solution of that differential equation. For example, the solutions of

$$\frac{dy}{dx}=6x^2$$

are given by

$$y=\int_{}^{}6x^2dx=2x^3+C$$

Sometimes we are interested in determining whether a particular solution curve passes through a certain point \((x_0,y_0)\) —that is, \(y(x_0)=y_0\). The problem of finding a function y that satisfies a differential equation

$$\frac{dy}{dx}=f(x)$$

with the additional condition \(y(x_0)=y_0\) is an example of an initial-value problem. The condition \(y(x_0)=y_0\) is known as an initial condition. For example, looking for a function \(y\) that satisfies the differential equation

$$\frac{dy}{dx}=6x^2$$

and the initial condition \(y(1)=5\) is an example of an initial-value problem. Since the solutions of the differential equation are \(y=2x^3+C\), to find a function y that also satisfies the initial condition, we need to find \(C\) such that \(y(1)=2(1)^3+C=5\). From this equation, we see that \(C=3\), and we conclude that \(y=2x^3+3\) is the solution of this initial-value problem as shown in the following graph.

Some of the solution curves of the differential equation \(\frac{dy}{dx}=6x^2\) are displayed. The function \(y=2x^3+3\) satisfies the differential equation and the initial condition \(y(1)=5\).

**FAQs**

**What is an antiderivative calculator?**

Anti-derivative Calculator is an online tool used to calculate the value of a given indefinite integral. Integration can be used to find the area under a curve. It can also be used to determine the volume of a three-dimensional solid shape.

**Is integral the same as antiderivative?**

Antiderivatives are related to definite integrals through the fundamental theorem of calculus: the definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.

**What are Antiderivatives used for?**

An antiderivative is a function that reverses what the derivative does. One function has many antiderivatives, but they all take the form of a function plus an arbitrary constant. Antiderivatives are a key part of indefinite integrals.

**What is a general antiderivative?**

General Antiderivative The function F(x) + C is the General Antiderivative of the function f(x) on an interval I if F (x) = f(x) for all x in I and C is an arbitrary constant. Indefinite Integral The Indefinite Integral of f(x) is the General Antiderivative of f(x).

**What is antiderivative sin?**

The general antiderivative of sin(x) is −cos(x)+C . With an integral sign, this is written: ∫sin(x) dx=−cos(x)+C .