Integral of e^x

• Introduction
• Formula of Integral `e^x`
• Integral of `e^x`  Proof by Differentiation 
• Integral of `e` to the `x` Using Series Expansion
• Integral `e^x` Formula Verification
• Integral of `e^x` Important Notes
• Solved Examples on Integral of `e^x`
• Practice problems on Integral of `e^x`
• Frequently Asked Questions on Integral of `e^x`

Introduction

The integral of \(e^x\) with respect to \(x\) is a fundamental concept in calculus. The notation for the integral of `e` to the `x` is given by:

\( \int e^x \,dx \)

This expression represents the antiderivative of the function \(e^x\).

The exponential function \(e^x\) is special in that its derivative and integral are essentially the same, with a constant factor. This property makes it a key function in calculus and is often encountered in various mathematical and scientific applications.

Formula of Integral `e^x`

The formula for the integration of `e` with respect to \(x\) is:

Here, \(C\) is the constant of integration, which accounts for any constant that may have been present in the original function but was lost when taking the derivative.

This formula represents the antiderivative of the function \(e^x\). In other words, if you were to find the derivative of \(e^x + C\) with respect to \(x\), you would get \(e^x\).

Proof of Integral of `e^x` by Differentiation

Here is is the Proof of Integral of `e^x` by Differentiation step by step:

Step `1`: Differentiation of \(e^x\)

First, you correctly identified that the differentiation of \(e^x\) with respect to \(x\) yields \(e^x\) itself:

\( \frac{d}{dx} e^x = e^x \)

Step `2`: Multiplying by \(dx\)

Multiplying both sides by \(dx\) doesn't change the equation's essence but sets the stage for integrating both sides, leading to:

\( \frac{d}{dx} e^x dx = e^x dx \)

Step `3`: Integrating Both Sides

You then integrated both sides:

\( \int \frac{d}{dx} e^x dx = \int e^x dx \)

Step `4`: Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus allows us to cancel out the differentiation and integration on the left side because they are inverse operations. This results in:

\( e^x = \int e^x dx \)

Step `5`: Adding the Constant of Integration

Finally, recognizing that the integration process can include an arbitrary constant (since the differentiation of a constant is zero), you correctly conclude:

\( \int e^x dx = e^x + C \)

Integral of `e` to the `x` Using Series Expansion

Proving the integral of \(e^x\) through its series expansion involves first expressing \(e^x\) as its series expansion, then integrating term by term, and finally showing that the resulting series is indeed the series expansion of \(e^x\) plus a constant. This method not only illustrates the integral of \(e^x\) but also demonstrates the power of series in calculus.

Series Expansion of \(e^x\)

The exponential function \(e^x\) can be expanded into an infinite series as follows:

\(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)

Now, we integrate the series with respect to \(x\):

\( \int e^x \, dx = \int \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) dx = \sum_{n=0}^{\infty} \int \frac{x^n}{n!} \, dx\)

\( \sum_{n=0}^{\infty} \int \frac{x^n}{n!} \, dx = \sum_{n=0}^{\infty} \frac{1}{n!} \int x^n \, dx \)

The integral of \(x^n\) with respect to \(x\) is \(\frac{x^{n+1}}{n+1}\), so we get:

\( \sum_{n=0}^{\infty} \frac{1}{n!} \int x^n \, dx = \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)n!} + C\)

Notice that \((n+1)n! = (n+1)!\), so we can rewrite the series as:

\( \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} + C = \sum_{n=0}^{\infty} \frac{x^n}{n!} -1 + C\)

We can combine `-1 + C` to write it as `C`.

\( \sum_{n=0}^{\infty} \frac{x^n}{n!} + C = e^x + C\)

Therefore \( \int e^x \,dx = e^x + C \).

Integral `e^x` Formula Verification

To verify the integral of \(e^x\) using the standard formula \(\int e^x \,dx = e^x + C\), where \(C\) is the constant of integration, we can simply differentiate the expression \(e^x + C\) with respect to \(x\) and confirm that it indeed results in \(e^x\).

Let's consider the function:

\( F(x) = e^x + C \)

Now, differentiate \(F(x)\) with respect to \(x\):

\( \frac{d}{dx} (F(x)) = \frac{d}{dx} (e^x + C) \)

Using the rules of differentiation, the derivative of \(e^x\) is \(e^x\), and the derivative of a constant \(C\) is zero:

\( \frac{d}{dx} (F(x)) = e^x + 0 = e^x \)

So, the derivative of \(F(x)\) with respect to \(x\) is indeed \(e^x\). This confirms that \(F(x) = e^x + C\) is a valid antiderivative of \(e^x\), and the standard integral formula \(\int e^x \,dx = e^x + C\) holds true.

Important Notes on Integral of `e^x`

A few important notes on the integral of \(e^x\):

`1`. Integral Formula:

\( \int e^x \,dx = e^x + C \)

where \(C\) is the constant of integration.

`2`. Constant of Integration (\(C\)):

• The constant of integration (\(C\)) is included because the derivative of a constant is zero.
   
• It accounts for the family of functions that differ by a constant.

`3`. Uniqueness of \(e^x\):

• The exponential function \(e^x\) is unique in that its derivative and integral are essentially the same, up to a constant factor.
   
• This property is not shared by other functions.

`4`. Role of \(e^x\) in Calculus:

• The function \(e^x\) frequently appears in calculus and mathematical modeling due to its special properties.
   
• It is the base of the natural logarithm and has applications in various fields.

`5`. Application in Differential Equations:

• The exponential function \(e^x\) often arises in the solutions of differential equations, particularly those involving exponential growth or decay.

Solved Examples

Example `1`: Find the integral of \( e^{2x} \) with respect to \( x \).

Solution:

To solve this, you can use substitution. Let \( u = 2x \), then \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \). Substituting this into the integral:

\( \int e^{2x} \, dx = \frac{1}{2} \int e^u \, du \)

Now, integrate \( e^u \) with respect to \( u \), remembering to account for the constant factor:

\( \frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C \)

Substituting back \( u = 2x \):

\( \frac{1}{2} e^{2x} + C \)

Example `2`: Evaluate \( \int e^{-3x} \, dx \).

Solution:

Similar to the previous example, we use substitution. Let \( u = -3x \), then \( du = -3 \, dx \) or \( dx = -\frac{1}{3} \, du \). Substituting this into the integral:

\( \int e^{-3x} \, dx = -\frac{1}{3} \int e^u \, du \)

Integrate \( e^u \) with respect to \( u \):

\( -\frac{1}{3} \int e^u \, du = -\frac{1}{3} e^u + C \)

Substituting back \( u = -3x \):

\( -\frac{1}{3} e^{-3x} + C \)

Example `3`:

Evaluate the integral of \( xe^x \) with respect to \( x \).

\( \int xe^x \, dx \)

Solution:

To solve this integral, we can use integration by parts. Let \( u = x \) and \( dv = e^x \, dx \). Then, \( du = dx \) and \( v = e^x \). Applying the integration by parts formula:

\( \int u \, dv = uv - \int v \, du \)

\( = xe^x - \int e^x \, dx \)

The integral of \( e^x \) with respect to \( x \) is simply \( e^x \), so:

\( = xe^x - e^x + C \)

Example `4`:

Find the integral :   \( \int e^x \cos(x) \, dx \)

Solution:

The integral \(\int e^x \cos(x) \, dx\) using integration by parts.

For integration by parts, the formula is \(\int u \, dv = uv - \int v \, du\). We apply this formula twice, using a clever choice for \(u\) and \(dv\) to create a solvable equation.

First Application of Integration by Parts

Let's choose:

\(u = e^x\), meaning \(du = e^x \, dx\)

\(dv = \cos(x) \, dx\), meaning \(v = \sin(x)\)

Applying these choices:

\(\int e^x \cos(x) \, dx = e^x \sin(x) - \int e^x \sin(x) \, dx\)

Second Application of Integration by Parts

For the second integral, we apply integration by parts again, but this time to \(\int e^x \sin(x) \, dx\):

Let's choose:

\(u = e^x\), meaning \(du = e^x \, dx\)

\(dv = \sin(x) \, dx\), meaning \(v = -\cos(x)\)

Applying these choices:

\(\int e^x \sin(x) \, dx = -e^x \cos(x) - \int -e^x \cos(x) \, dx\)

\(\int e^x \sin(x) \, dx = -e^x \cos(x) + \int e^x \cos(x) \, dx\)

Combining the Two

Substituting the second result back into the first:

\(\int e^x \cos(x) \, dx = e^x \sin(x) - (-e^x \cos(x) + \int e^x \cos(x) \, dx)\)

Simplifying:

\(\int e^x \cos(x) \, dx = e^x \sin(x) + e^x \cos(x) - \int e^x \cos(x) \, dx\)

Now, we have \(\int e^x \cos(x) \, dx\) on both sides. Moving the integral term to one side gives us:

\(2\int e^x \cos(x) \, dx = e^x \sin(x) + e^x \cos(x)\)

Dividing by `2`:

\(\int e^x \cos(x) \, dx = \frac{1}{2} e^x (\sin(x) + \cos(x)) + C\)

Where \(C\) is the constant of integration. This is the correct solution to the integral.

Practice Problems 

Q`1`. Find the integral of \( e^{-x} \) with respect to \( x \). 

1. \( -e^{-x} + C \)
2. \( e^x + C \)
3. \( e^{-x} + C \)
4. \( -e^{-x} - C \)

Answer: a

Q`2`. Calculate the integral: \( \int e^{3x} \, dx \)

1. \( -e^{-x} + C \)
2. \( \frac{1}{3} e^{3x} + C \)
3. \( \frac{1}{3} e^{2x} + C \)
4. \( \frac{1}{2} e^{3x} + C \)

Answer: b

Q`3`. Calculate the integral : \( \int 2e^{-x/2} \, dx \)

1. \( \frac{1}{3} e^{3x} + C \)
2. \( 4e^{x/2} + C \)
3. \( -4e^{-x/2} + C \)
4. \( -4e^{-x/4} + C \)

Answer: c

Q`4`. Find the integral of \( e^x \sin(x) \) with respect to \( x \).

1. \( \frac{1}{2}e^x (\sin(x) +\cos(x) + C\)
2. \( \frac{1}{3}e^x \cos(x) + C \)
3. \( \frac{1}{2}e^x (\sin(x) -\cos(x) + C\)
4. \( \frac{1}{2}e^x \cos(x) + C \)

Answer: c

Q`5`. Find the integral of \( xe^x \) with respect to \( x \).

1. \( \frac{1}{3} e^{3x} + C \)
2. \( xe^x - e^x + C \)
3. \( xe^x + e^x + C \)
4. \( xe^2x - e^x + C \)

Answer: b

Frequently Asked Questions

Q`1`. What is the integral of \(e^x\) with respect to \(x\)?

Answer: The integral of \(e^x\) with respect to \(x\) is given by \(\int e^x \,dx = e^x + C\), where \(C\) is the constant of integration.

Q`2`. Why does the integral of \(e^x\) include a constant (\(C\))?

Answer: The constant of integration (\(C\)) is added because the derivative of a constant is zero. It accounts for the family of functions that differ by a constant.

Q`3`. What is the significance of \(e^x\) in calculus?

Answer: The exponential function \(e^x\) is significant in calculus due to its unique properties. It is the base of the natural logarithm and plays a key role in various mathematical applications.

Q`4`. How is the integral of \(e^x\) used in differential equations?

Answer: The integral of \(e^x\) often appears in the solutions of differential equations, especially those involving exponential growth or decay.

Q`5`. Can the integral of \(e^x\) be applied to real-world problems?

Answer: Yes, the integral of \(e^x\) is used in various real-world problems, such as modeling population growth, radioactive decay, and other exponential processes.