Integral of e to the 2x
Before going to find the integral of e to the 2x, let us recall a few facts about e^{2x}. 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.
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 Misconception about Integral of e to the 2x
Since ∫ e^{x} dx = e^{x} + C, do NOT think that ∫ e^{2x} dx is e^{2x} + C. We always have to divide the actual value of integral by the coefficient of x. Since the coefficient of x is 2, ∫ e^{2x} dx = e^{2x}/2 + C.
Let us prove the integral of e to the 2x to be e^{2x}/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 e^{2x}.
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
We can find the integral of e to the 2x using the substitution method of integration. Consider the integral ∫ e^{2x }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:
∫ 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} + C\(_1\))
= (1/2) e^{u} + C\(_1\)/2
= (1/2) e^{2x} + C (where C\(_1\)/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
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., ∫ₐ^{b} e^{2x} dx. To evaluate this, we will first consider the fact that the integral of e^{2x} is e^{2x}/2 + C and then substitute the upper bound and lower bound one after the other in order and then subtract the results. 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).
Important Notes on Integral of e to the 2x:
 The integral of e to the 2x is e^{2x}/2 + C, where C is the integration constant.
 Hence, the integral of e^{ax, }in general, is e^{ax}/a + C.
 Extending this further, the integral of e^{ax}^{+b} is e^{ax+b}/a + C.
Related Topics:
Solved Examples on Integral of e to the 2x

Example 1: Find the integral of e to the 2x + 3.
Solution:
Let us find ∫ e^{2x+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:
∫ e^{u} (du/2)
= (1/2) ∫ e^{u} du
=(1/2) e^{u }+ C
= (1/2) e^{2x+3}+ C (u = 2x+3 is substituted back)
Answer: ∫ e^{2x+3} dx = (1/2) e^{2x+3}+ C.

Example 2: Evaluate the integral of e to the 2x from ∞ to 0.
Solution:
We know that ∫ e^{2x} dx = e^{2x}/2 + C.
Applying the limits ∞ to 0,
∫\(_{\infty}\)^{0} e^{2x} dx
= e^{2(0)}/2  e^{2(∞)}/2
= e^{0}/2  1/(2e^{∞) } (By properties of exponents, a^{m} = 1/a^{m})
= 1/2  1/∞ (Because e^{∞} = ∞)
= 1/2  0
= 1/2
Answer: ∫\(_{\infty}\)^{0} e^{2x} dx = 1/2.

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:
∫ e^{u} (du/2) = (1/2) ∫ e^{u} du = (1/2) e^{u} = (1/2) e^{2x}.
Now, we apply the limits.
∫₀^{∞} e^{2x} dx = (1/2) (e^{2∞} e^{0})
= (1/2) (0  1) (Because e^{2∞ }= 1/(e^{2∞}) = 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 ∫ 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.
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