The integration of e^{3x} is e^{3x}/3. In this post, we will learn how to find the integral of e to the 3x. Let us recall the formula of the integral of e^{mx}:

$\int e^{mx} dx=\dfrac{e^{mx}}{m}+C$ where C is an integral constant. Thus, the integral of e^{3x} will be equal to $\int e^{3x} dx=\dfrac{e^{3x}}{3}+C$.

## What is the Integration of e^{3x}

**Question:** What is the integration of e^{3x}?

*Answer:* The integration of e^{3x} is $\dfrac{e^{3x}}{3}$.

**Explanation:**

**Step 1:** We will put $z=3x$.

Differentiating both sides of $z=3x$, we have

$\dfrac{dz}{dx}=3$

$\Rightarrow dx= \dfrac{dz}{3}$

**Step 2:** $\therefore \int e^{3x} dx = \int e^z \dfrac{dz}{3}$

$=\dfrac{1}{3} \int e^z dz$

$=\dfrac{1}{3} e^z +C$ as the integration of e^{x} is e^{x}

$=\dfrac{1}{3} e^{3x}+C$ as $z=3x$.

**Conclusion:** Thus, the integration of e^{3x}dx is $\dfrac{1}{3} e^{3x}+C$.

## Definite integral of e^{3x}

**Question:** Find the definite integral $\int_0^1 e^{3x} dx$.

*Answer: *

We have shown above that the integration of $e^{3x} dx$ is $\dfrac{1}{3} e^{3x}$. Thus, we have that

$\int_0^1 e^{3x} dx$

$=[\dfrac{1}{3} e^{3x}]_0^1$

$=\dfrac{1}{3}[e^{3x}]_0^1$

$=\dfrac{1}{3}(e^{3 \cdot 1} -e^0)$

$=\dfrac{1}{3}(e^3 -1)$

So the definite integration of e^{3x} from 0 to 1 is equal to (e^{3}-1)/3.

**Also Read:**

## FAQs

**Q1: What is the integration of e ^{3x}?**

**Answer:** The integration of e^{3x} is e^{3x}/3 +C where C is an integral constant.