Integral of e^x
Before finding the integral of e to the x, let us recall what is e^{x}. It is an exponential function as it has variable (x) in its exponent and constant (e) in its base. Here, 'e' is called Euler's number and its approximate value is 2.718. The integral of exponential function e^{x} is itself. Of course, we always add an integration constant to the value of every indefinite integral.
Let us find the integral of e^{x}, along with its verification using differentiation, and a few solved examples.
What is the Integral of e^x?
The integral of e^{x} is e^{x} itself. But we know that we add an integration constant after the value of every indefinite integral and hence the integral of e^{x} is e^{x} + C. We write it mathematically as ∫ e^{x} dx = e^{x} + C. Here,
 ∫ is the symbol of integration.
 e^{x} (which is followed by dx) is the integrand
 C is the integration constant
Integral of e^x Formula
The integral of e^{x} formula is
∫ e^{x} dx = e^{x} + C, where C is the integration constant.
Let us prove this by differentiation and the series expansion of e^{x}.
Integral of e to the x Proof by Differentiation
We are aware that integration and differentiation are the reverse processes of each other. So to find the integral of e^{x}, we have to see by differentiating what function will result in e^{x}. If we look into the formulas of differentiation, we can find that
d/dx (e^{x}) = e^{x}
Thus, we can directly say that the integral of e^{x} is e^{x} itself (or) we can prove this by the fundamental theorem of calculus.
Multiplying both sides by dx,
d/dx (e^{x}) dx = e^{x }dx
Taking the integral on both sides of the above equation,
∫ d/dx (e^{x}) dx = ∫ e^{x }dx
By the fundamental theorem of calculus, "d/dx" and the "∫ dx" symbol get cancelled with each other and we get,
e^{x} = ∫ e^{x }dx
Since we should add the integration constant C for every indefinite integral,
∫ e^{x }dx = e^{x} + C
Hence we proved the integral of e^x formula.
Integral of e^x Using Series Expansion
All standard functions have series expansions. The series expansion of e^{x} is, e^{x} = 1 + x + x^{2}/2! + x^{3}/3! + ... . By taking the integral on both sides, we get
∫ e^{x }dx = ∫ [1 + x + x^{2}/2! + x^{3}/3! + ...] dx
= x + x^{2}/2 + x^{3}/3(2!) + x^{4}/4(3!) + ... (by power rule of integration)
= x + x^{2}/2! + x^{3}/3! + x^{4}/4! + ...
Adding and subtracting 1,
∫ e^{x }dx = 1 + x + x^{2}/2! + x^{3}/3! + x^{4}/4! + ...  1
We know that 1 + x + x^{2}/2! + x^{3}/3! + x^{4}/4! + ... = e^{x}. Also, we can replace the constant 1 by the integration constant C. Thus,
∫ e^{x }dx = e^{x} + C
Hence proved.
Verification of Integral e^x Formula
We know that the value of any indefinite integral can be verified by using the process of differentiation. To prove the integral of e to the x to be e^{x} + C, we just need to prove that the derivative of e^{x} + C to be e^{x}. Let us find the derivative now.
d/dx (e^{x} + C) = d/dx (e^{x}) + d/dx (C)
= e^{x} + 0
= e^{x}
Thus, we have verified the formula of the integral of exponential function e^{x}.
Definite Integral of e^x
To evaluate the definite integral of e^{x}, we can ignore the integration constant (C) and substitute the bounds in just e^{x}. Let us see how to evaluate the definite integral of e^{x} by looking at a few examples below.
Integral of e^x From 0 to 1
∫₀^{1} e^{x} dx = [e^{x}]₀^{1}
= e^{1}  e^{0}
= e  1
Thus, the integral of e^x from 0 to 1 is e  1.
Integral of e to the x From 0 to 1
∫₀^{∞} e^{x} dx = [e^{x}]₀^{∞}
= e^{∞}  e^{0}
= ∞  1
= ∞
Thus, the integral of e^x from 0 to ∞ is diverges.
Integral of Exponential Function
We have already proved that the integral of the exponential function e^{x} is itself. But in general, the integral of any exponential function a^{x }is NOT itself. Then what is ∫ a^{x} dx? Let us see. We know that the derivative of a^{x} is a^{x} ln a. i.e.,
d/dx (a^{x}) = a^{x} ln a
Dividing both sides by ln a,
d/dx [ a^{x}/ln a] = a^{x}
Since the derivative of a^{x}/ln a is a^{x}, and since integral is the reverse operation of differentiation, we can say that the integral of a^{x} is a^{x}/ln a. i.e.,
∫ a^{x} dx = a^{x }/ ln a + C
Here, C is the integration constant.
Thus, the integral of an exponential function a^{x} is a^{x }/ ln a.
Important Notes on Integral of e^x:
 The integral of e^{x} is itself. i.e., ∫ e^{x }dx = e^{x} + C.
 ∫ e^{ax }dx = e^{ax} / a + C by using integration by substitution.
 The integral of e^(x^{2}) involves the error function. i.e., ∫ e^{}ˣ² dx = √π/2 erf (x) + C.
 The integral of exponential function a^{x} is NOT itself, instead, ∫ a^{x} dx = a^{x }/ ln a + C.
 ∫ a^{kx} dx = a^{kx }/ (k ln a) + C by using integration by substitution.
☛ Related Topics:
Examples Using Integral of e^x

Example 1: Evaluate the integral ∫ e^{x }sin (e^{x}) dx.
Solution:
Let us solve this by using integration by substitution.
Let e^{x} = u. Then e^{x} dx = du.
By substituting these, the given integral becomes
∫ e^{x} sin (e^{x}) dx = ∫ sin u du =  cos u + C
(This is because the integral of sin x is cos x + C)
Substituting u = e^{x} back here,
∫ e^{x} sin (e^{x}) dx =  cos (e^{x}) + C.
Answer: The integral of e^x sin (e^x) is  cos (e^x) + C.

Example 2: What is the value of the integral ∫ e^{x }sin x dx?
Solution:
We will solve this using the integration by parts.
Let u = sin x and dv = e^{x} dx.
Then du = cos x dx and v = e^{x}.
By using the integration by parts formula,
∫ u dv = uv  ∫ v du
∫ e^{x} sin x dx = (sin x) (e^{x})  ∫ e^{x }cos x dx
If we again apply integration by parts (by assuming u = cos x and dv = e^{x} dx this time), we get
∫ e^{x} sin x dx = (sin x) (e^{x})  [(cos x) (e^{x})  ∫ e^{x} (sin x) dx]
∫ e^{x} sin x dx = e^{x }sin x  e^{x} cos x  ∫ e^{x }sin x dx
Adding ∫ e^{x} sin x dx on both sides,
2∫ e^{x} sin x dx = e^{x} (sin x  cos x)
Dividing both sides by 2,
∫ e^{x} sin x dx = (e^{x}/2) (sin x  cos x) + C
Answer: The integral of e^x sin x is (e^x/2) (sin x  cos x) + C.

Example 3: What is the value of the definite integral ∫₋∞^{0} e^{x} dx?
Solution:
We know that ∫ e^{x }dx = e^{x }+ C.
Thus,
∫₋∞^{0} e^{x} dx = [e^{x}]₋∞^{0}
= e^{0}  e^{∞}
= 1  1/e^{∞}
= 1  1/∞
= 1  0
= 1
Answer: The integral of e^{x} from ∞ to 0 is 1.
FAQs on Integral of e to the x
What is the Integral of e to the x?
The integral of e^{x} is e^{x} + C. Symbolically it is written as ∫ e^{x }dx = e^{x }+ C, where C is the integration constant.
How to Find the Integral of e^x?
We know that the derivative of e^{x} is e^{x}. Since the integral is the inverse operation of differentiation, the integral of e^{x} is also e^{x}. i.e., ∫ e^{x }dx = e^{x }+ C. Here, C is the integration constant.
Why is the Integral of e^x is Itself?
Since the derivative of e^{x} is itself, the integral of e^{x} is also itself. This is because differentiation and integration are the inverse operations of each other.
What is the Integral of e?
"e" is a constant. Thus, we can write the integral of e as: ∫ e dx = e ∫ 1 dx. Here, ∫ 1 dx = x and so, ∫ e dx = ex + C.
What is the Value of the Integral of e^x^2?
We know that ∫ e^{}ˣ² dx = √π/2 erf (x) + C. Thus, ∫ eˣ² dx = ∫ e^{(}ˣ²^{)} dx= ∫ e^{(ix}^{)² }dx. Now let ix = u then i dx = du (or) dx = du/i = i du. Then the above integral becomes i ∫ e^{u}^{² }du = (i√π)/2 erf (u) + C = (i√π)/2 erf (ix) + C.
How to Find the Integral of x e^x?
To find the ∫ x e^{x} dx, assume that u = x and dv = e^{x} dx. Then du = dx and v = e^{x}. Then by integration by parts, ∫ u dv = uv  ∫ v du = x e^{x}  ∫ e^{x} dx = xe^{x}  e^{x} + C.
How to Solve the Integral of e^{x} cos x?
To find the ∫ e^{x} cos x dx, assume that u = cos x and dv = e^{x} dx. Then du =  sin x dx and v = e^{x}. By using the integration by parts formula, ∫ u dv = uv  ∫ v du. From this, ∫ e^{x }cos x dx = (cos x) (e^{x})  ∫ e^{x }( sin x) dx = cos x e^{x} + ∫ e^{x }sin x dx. We again apply integration by parts (with u = sin x and dv = e^{x} dx this time). Then we get ∫ e^{x }cos x dx = (cos x) (e^{x}) + [(sin x) (e^{x})  ∫ e^{x }(cos x) dx] = e^{x} cos x + e^{x} sin x  ∫ e^{x }cos x dx. From this, 2∫ e^{x }cos x dx = e^{x} (cos x + sin x). Then ∫ e^{x }cos x dx = (e^{x}/2) (cos x + sin x) + C
What is the Integral of 1/e^x dx?
We can write ∫ 1/e^{x} dx as ∫ e^{x} dx. Now assume that x = u. Then dx = du (or) dx = du. Then the above integral becomes ∫ e^{u} (du) =  e^{u} + C = e^{x} + C.
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