Integral of $e^{2 x}$

Q: What is the integral of exp(2x)?

The integral of e^2 x is e^2 x2 + C.

The integral of $e^{2 x}$ is $\frac{e^{2 x}}{2} + C$ .

\int e^{2 x}\,dx = \frac{e^{2 x}}{2} + C

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Graph of f(x) = exp(2x) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 3.194528.

How to work it out

1
We want an antiderivative of f(x) = exp(2x).
$\int e^{2 x}\,dx$
2
Method: ∫ $e^{kx}$ dx = $e^{kx}$ /k.
3
That gives the antiderivative; add the constant C because any constant disappears when you differentiate:
$\int e^{2 x}\,dx = \frac{e^{2 x}}{2} + C$
4
Check by differentiating the answer; you must get the original function back:
$\frac{d}{dx}\left[\frac{e^{2 x}}{2}\right] = \frac{4 e^{2 x}}{4}$
5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:
$\int_{0}^{1} e^{2 x}\,dx = \Big[\frac{e^{2 x}}{2}\Big]_{0}^{1} \approx 3.194528$

LaTeX source of the formulas on this page

\[ \int e^{2 x}\,dx = \frac{e^{2 x}}{2} + C \]
\[ \int e^{2 x}\,dx \]
\[ \frac{d}{dx}\left[\frac{e^{2 x}}{2}\right] = \frac{4 e^{2 x}}{4} \]
\[ \int_{0}^{1} e^{2 x}\,dx = \Big[\frac{e^{2 x}}{2}\Big]_{0}^{1} \approx 3.194528 \]

Definite integrals of the same function

Interval	Value
from 0 to 1	3.194528
from 0 to 0.5	0.85914091

Questions

What is the integral of exp(2x)?

The integral of $e^{2 x}$ is $\frac{e^{2 x}}{2} + C$ .

Why do we add + C?

Every function F(x) + C has the same derivative, so the antiderivative is only defined up to a constant.

What is the integral of exp(2x) from 0 to 1?

About 3.194528.

Integral of e2xe^{2 x}e2x

How to work it out

Definite integrals of the same function

Questions

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Integral of $e^{2 x}$