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Integral of e to the 2x

Before going to find the integral of e to the 2x, let us recall a few facts about e2x. It is an exponential function because its base, which is ‘e’, is a constant (which is known as Euler’s number) and its exponent has a variable in it. It can be written as e^2x also. We apply the techniques of integrating exponential functions to find the integral of e to the 2x.

Let us find the integral of e to the 2x using multiple methods and also solve a few examples using this.

 1 What is the Integral of e to the 2x? 2 Integral of e to the 2x by Differentiation 3 Integral of e to the 2x by Substitution Method 4 Integral of e to the 2x Verification 5 Definite Integral of e to the 2x 6 FAQs on Integral of e to the 2x

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

The integral of e to the 2x is e2x/2 + C. This is mathematically written as ∫ e2x dx = e2x/2 + C. Here,

• ‘∫’ is the symbol of integration.
• e2x 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 Misconception about Integral of e to the 2x

Since ∫ ex dx = ex + C, do NOT think that ∫ e2x dx is e2x + C. We always have to divide the actual value of integral by the coefficient of x. Since the coefficient of x is 2, ∫ e2x dx = e2x/2 + C.

Let us prove the integral of e to the 2x to be e2x/2 + C using various methods and also we will verify the result using differentiation.

## Integral of e to the 2x by Differentiation

We know that differentiation and integration are the inverse operations of each other. Also, we know that the fundamental theorem of calculus is used to find the integral of a derivative. This theorem says, ∫ f'(x) dx = f(x) + C. Well, we will first find the derivative of e2x.

By chain rule,

(e2x)’ = 2e2x

Dividing both sides by 2,

(e2x)’ / 2 = e2x

By constant multiplication rule of derivatives,

(e2x / 2)’ = e2x

Taking integral on both sides,

∫ (e2x / 2)’ dx = ∫ e2x 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:

e2x / 2 + C = ∫ e2x dx

Hence proved.

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

We can find the integral of e to the 2x using the substitution method of integration. Consider the integral ∫ e2x dx. Here, we assume that 2x = u. Differentiating on both sides, we get 2 dx = du and it can be written as dx = du/2. Then the above integral becomes:

∫ eu (du/2) = (1/2) ∫ eu du

We know that the integral of ex is ex + C. Using this, the above integral becomes

= (1/2) (eu + C$$_1$$)

= (1/2) eu + C$$_1$$/2

= (1/2) e2x + C (where C$$_1$$/2 = C and u = 2x)

Hence, we have proved that ∫ e2x dx = (1/2) e2x + 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 e2x / 2 + C, we should prove that the derivative of e2x / 2 + C to be e2x. Let us find the derivative.

d/dx (e2x / 2 + C)
= d/dx(e2x / 2) + d/dx(C)
= (1/2) (2e2x) + 0 (by the chain rule)
= e2x

Hence, we have verified the integral of e2x.

## Definite Integral of e to the 2x

A definite integral is an integral with the bounds (lower and upper bounds). We will consider the definite integral of e to the 2x from a to b. i.e., ∫ₐb e2x dx. To evaluate this, we will first consider the fact that the integral of e2x is e2x/2 + C and then substitute the upper bound and lower bound one after the other in order and then subtract the results. i.e.,

∫ₐb e2x dx = (e2x/2 + C)ₐb

= (e2b/2 + C) – (e2a/2 + C)

= e2b/2 + C – e2a/2 – C

= e2b/2 – e2a/2

= (1/2) (e2b – e2a)

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

Important Notes on Integral of e to the 2x:

• The integral of e to the 2x is e2x/2 + C, where C is the integration constant.
• Hence, the integral of eax, in general, is eax/a + C.
• Extending this further, the integral of eax+b is eax+b/a + C.

Related Topics:

## Solved Examples on Integral of e to the 2x

1. Example 1: Find the integral of e to the 2x + 3.

Solution:

Let us find ∫ e2x+3 dx by substitution method of integration. For this, we assume that 2x+3 = u. Then 2 dx = du. From this, we have, dx = du/2.

The above integral becomes:

∫ eu (du/2)

= (1/2) ∫ eu du

=(1/2) eu + C

= (1/2) e2x+3+ C (u = 2x+3 is substituted back)

Answer: ∫ e2x+3 dx = (1/2) e2x+3+ C.

2. Example 2: Evaluate the integral of e to the 2x from -∞ to 0.

Solution:

We know that ∫ e2x dx = e2x/2 + C.

Applying the limits -∞ to 0,

∫$$_{-\infty}$$0 e2x dx

= e2(0)/2 – e2(-∞)/2

= e0/2 – 1/(2e∞) (By properties of exponents, a-m = 1/am)

= 1/2 – 1/∞ (Because e∞ = ∞)

= 1/2 – 0

= 1/2

Answer: ∫$$_{-\infty}$$0 e2x dx = 1/2.

3. Example 3: What is the integral of e to the power of -2x from 0 to ∞.

Solution:

First we will evaluate ∫ e-2x dx. For this, let -2x = u ⇒ -2 dx = du ⇒ dx = -du/2. Then the integral becomes:

∫ eu (-du/2) = (-1/2) ∫ eu du = (-1/2) eu = (-1/2) e-2x.

Now, we apply the limits.

∫₀∞ e-2x dx = (-1/2) (e-2∞- e0)

= (-1/2) (0 – 1) (Because e-2∞ = 1/(e2∞) = 1/∞ = 0)

= 1/2

Answer: ∫₀∞ e-2x dx = 1/2.

## 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 ∫ e2x dx = e2x/2 + C.

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

To find the ∫ e2x dx, assume that 2x = u. Then 2 dx = u (or) dx = du/2. Then the value of the integral is, (1/2) ∫ eu dx = (1/2) eu + C = (1/2) e2x + C.

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

No, the derivative of e2x is 2e2x whereas the integral of e2x is e2x/2 + C. i.e.,

• d/dx (e2x) = e2x
• ∫ e2x dx = e2x/2 + C

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

We know that the derivative of e2x is 2e2x. i.e.,

d/dx (e2x) = 2e2x

d/dx (e2x/2) = e2x

By taking integral on both sides,

∫ d/dx (e2x/2) dx = ∫ e2x dx

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

e2x/2 = ∫ e2x dx.

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

∫ e2x dx = e2x/2 + C.

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

To find the integral ∫ e2x+1 dx, assume that 2x+1 = u. Then 2 dx = du. From this, we have, dx = du/2. The integral becomes:

∫ eu du/2

=(1/2) eu + C

= (1/2) e2x+1 + C.

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