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:

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

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

$\int e^{2x} dx=\dfrac{e^{2x}}{2}+C$.

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

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

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

**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

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

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

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

$\therefore \int e^{2x} dx = \int e^t \dfrac{dt}{2}$

$=\dfrac{1}{2} \int e^t dt$ 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.

## 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.