# Integral of e to the 2x

Let’s review some information about e2x before attempting to determine the integral of e to the 2x. It is an exponential function since its exponent contains a variable and its base, “e,” which is a constant (sometimes referred to as Euler’s number). It can alternatively be written as e2x. To get the integral of e to the 2x, we use methods for integrating exponential functions.

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Let’s use various techniques to determine the integral of e to the 2x and use this information to solve a few cases.

## What is the Integral of e to the 2x?

The **integral of e to the 2x** is e^{2x}/2 + C. This is mathematically written as **∫ e ^{2x} dx = e^{2x}/2 + C**. Here,

- ‘∫’ is the symbol of integration.
- e
^{2x}that is next to dx is the integrand. - C is the integration constant which is written along with the indefinite integral value of any function.

**Common Illusion Regarding Integral of e to the 2x**

Do not assume that e2x dx is e2x + C since ex dx = ex + C. The real integral value must always be divided by the x-coefficient. Because x has a coefficient of 2, e2x dx = e2x/2 + C.

Let’s use several techniques to demonstrate that the integral of e to the 2x is e2x/2 + C and then use differentiation to confirm the outcome.

### Integral of e to the 2x by Differentiation

We are aware that differentiation and integration operate in the opposite directions from one another. Additionally, we are aware that the integral of a derivative can be calculated using the calculus fundamental theorem. According to this theory, f'(x) dx = f(x) + C. So let’s start by determining the derivative of e2x.

By chain rule,

(e^{2x})’ = 2e^{2x}

Dividing both sides by 2,

(e^{2x})’ / 2 = e^{2x}

By constant multiplication rule of derivatives,

(e^{2x} / 2)’ = e^{2x}

Taking integral on both sides,

∫ (e^{2x} / 2)’ dx = **∫** e^{2x }dx

Now, by the fundamental theorem of calculus, the integral and derivative symbols get canceled with each other on the left side and we will be left with:

e^{2x} / 2 + C = **∫** e^{2x }dx

Hence proved.

**Integral of e to the 2x by Substitution Method**

The substitution method of integration allows us to get the integral of e to the 2x. Think about the integral e2x dx. In this case, we’ll suppose 2x = u. We obtain 2 dx = du by differentiating on both sides, which is equivalent to writing dx = du/2. The above integral then changes to:

∫ e^{u} (du/2) = (1/2) ∫ e^{u} du

We know that the integral of e^{x} is e^{x} + C. Using this, the above integral becomes

= (1/2) (e^{u} + C1)

= (1/2) e^{u} + C1/2

= (1/2) e^{2x} + C (where C1/2 = C and u = 2x)

Hence, we have proved that **∫** e^{2x }dx = (1/2) e^{2x} + C by using the substitution method.

## Integral of e to the 2x Verification

Since the integrals and derivatives of inverses of each other, to verify the integral of e to the x to be e^{2x} / 2 + C, we should prove that the derivative of e^{2x} / 2 + C to be e^{2x}. Let us find the derivative.

d/dx (e^{2x} / 2 + C)

= d/dx(e^{2x} / 2) + d/dx(C)

= (1/2) (2e^{2x}) + 0 (by the chain rule)

= e^{2x}

Hence, we have verified the integral of e^{2x}.

### Definite Integral of e to the 2x

An integral with boundaries is a definite integral (lower and upper bounds). The definite integral of e to the 2x from a to b will be taken into consideration. Hence, ab e2x dx. To assess this, we will first take into account the fact that the integral of e2x is equal to e2x/2 + C, following which we will substitute the upper bound and lower bound in turn, one after the other, before subtracting the outcomes. i.e.,

**∫**ₐ^{b} e^{2x} dx = (e^{2x}/2 + C)ₐ^{b}

= (e^{2b}/2 + C) – (e^{2a}/2 + C)

= e^{2b}/2 + C – e^{2a}/2 – C

= e^{2b}/2 – e^{2a}/2

= (1/2) (e^{2b} – e^{2a})

Thus, the integration constant doesn’t play any role while calculating the definite integral (because it got canceled).

### FAQs on Integral of e to the 2x

### What is the Value of the Integral of e to the 2x?

The integral of e^2x is e^2x/2 + C. We can write this mathematically using the integration symbol as ∫ e^{2x} dx = e^{2x}/2 + C.

### How to Find the Integral of e to the power of 2x?

To find the ∫ e^{2x} dx, assume that 2x = u. Then 2 dx = u (or) dx = du/2. Then the value of the integral is, (1/2) ∫ e^{u} dx = (1/2) e^{u} + C = (1/2) e^{2x} + C.

### Is the Derivative of Integral of e to the 2x the Same?

No, the derivative of e^{2x} is 2e^{2x} whereas the integral of e^{2x} is e^{2x}/2 + C. i.e.,

- d/dx (e
^{2x}) = e^{2x} - ∫ e
^{2x}dx = e^{2x}/2 + C

### How to Find the Integral of e to the 2x by Differentiation?

We know that the derivative of e^{2x} is 2e^{2x}. i.e.,

d/dx (e^{2x}) = 2e^{2x}

d/dx (e^{2x}/2) = e^{2x}

By taking integral on both sides,

∫ d/dx (e^{2x}/2) dx = ∫ e^{2x }dx

The integral and d/dx get canceled with each other on the left. So we will be left with

e^{2x}/2 = ∫ e^{2x }dx.

Since we usually add an integration constant C for every indefinite integral,

∫ e^{2x} dx = e^{2x}/2 + C.

### What is the Integral of e to the 2x + 1?

To find the integral ∫ e^{2x+1} dx, assume that 2x+1 = u. Then 2 dx = du. From this, we have, dx = du/2. The integral becomes:

∫ e^{u} du/2

=(1/2) e^{u }+ C

= (1/2) e^{2x+1} + C.