Integral of $e^{-2 x}$

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

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(-2 * x) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.43233236.

How to work it out

1
We want an antiderivative of f(x) = exp(-2 * x).
$\int e^{-2 x}\,dx$
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:
$\frac{d}{dx}\left[\frac{-e^{-2 x}}{2}\right] = \frac{-\left(\left(-4\right) e^{\left(-2\right) x}\right)}{4}$
5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:
$\int_{0}^{1} e^{-2 x}\,dx = \Big[\frac{-e^{-2 x}}{2}\Big]_{0}^{1} \approx 0.43233236$

LaTeX source of the formulas on this page

\[ \int e^{-2 x}\,dx = \frac{-e^{-2 x}}{2} + C \]
\[ \int e^{-2 x}\,dx \]
\[ \frac{d}{dx}\left[\frac{-e^{-2 x}}{2}\right] = \frac{-\left(\left(-4\right) e^{\left(-2\right) x}\right)}{4} \]
\[ \int_{0}^{1} e^{-2 x}\,dx = \Big[\frac{-e^{-2 x}}{2}\Big]_{0}^{1} \approx 0.43233236 \]

Definite integrals of the same function

Interval	Value
from 0 to 1	0.43233236
from 0 to 0.5	0.31606028

Questions

What is the integral of exp(-2 * x)?

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(-2 * x) from 0 to 1?

About 0.43233236.

Integral of e−2xe^{-2 x}e−2x

How to work it out

Definite integrals of the same function

Questions

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Integral of $e^{-2 x}$