The integration of e^{2x} is equal to e^{2x}/2. In this post, we will learn how to integrate e to the 2x. The integration of e^{2x} can be derived directly from the formula of the integral of e^{mx} which is given below:

∫e^{mx} dx = $\dfrac{e^{mx}}{m}$ +C where C is an integration constant.

Putting m=2 in the above formula, the integral of e to the 2x is equal to

∫e^{2x} dx = $\dfrac{e^{2x}}{2}$ +C.

## What is the Integration of e^{2x}

**Question:** What is the integration of e^{2x}?

*Answer:* The integration of e^{2x} is $\dfrac{e^{2x}}{2}$+C.

**Explanation:**

We will find the integration of e^{2x} by the substitution method, we will follow the below steps.

**Step 1:** Let us put $t=2x$.

Differentiating both sides with respect to x, we have

$\dfrac{dt}{dx}=2$

⇒ $dx= \dfrac{dt}{2}$

**Step 2:** Let us now put the values of 2x and dx in the original integration. By doing so, we will get that

∴ ∫e^{2x} dx = ∫ $e^t \dfrac{dt}{2}$

= $\dfrac{1}{2}$ ∫ e^{t} dt as 1/2 is a constant, so we can take it out of the integration.

= $\dfrac{1}{2}$ e^{t} +C as the integration of e^{x} is e^{x}

= $\dfrac{1}{2}$ e^{2x} + C as t=2x.

So the integration of e^{2x}dx is $\dfrac{1}{2}$ e^{2x} +C where C is an integral constant.

**Verification:** To verify the integration of e^{2x} is e^{2x}/2 +C, we need to show that the derivative of e^{2x}/2 +C is equal to e^{2x}.

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

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

= 2e^{2x}/2 + 0 as the derivative of a constant is zero and d/dx (e^{ax}) = ae^{ax}.

= e^{2x}.

So the integration of e^{2x} is equal to e^{2x}/2 +C is verified.

**More Integrals:** Integral of e^{3x}

Integration of tan x | Integration of cot x

Integration of secx | Integration of cosecx

## Definite Integral of e^{2x}

**Question:** Find the definite integral $\int_0^1 e^{2x} dx$.

*Answer: *

We have shown above that the integration of $e^{2x} dx$ is $\dfrac{1}{2} e^{2x}$. Thus, we have that

$\int_0^1 e^{2x} dx$

$=[\dfrac{1}{2} e^{2x}]_0^1$

$=\dfrac{1}{2}[e^{2x}]_0^1$

$=\dfrac{1}{2}(e^{2 \cdot 1} -e^0)$

$=\dfrac{1}{2}(e^2 -1)$

So the definite integration of e^{2x} from 0 to 1 is equal to (e^{2}-1)/2.

Also Read:

Derivative of $\sqrt{x}+\dfrac{1}{\sqrt{x}}$

## FAQs

**Q1: What is the integration of e ^{2x}?**

**Answer:** The integration of e^{2x} is e^{2x}/2 +C where C is an integral constant.