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Integral of eāˆ’4xe^{-4 x}

The integral of eāˆ’4xe^{-4 x} is āˆ’eāˆ’4x4+C\frac{-e^{-4 x}}{4} + C.

Graph of f(x) = exp(-4 * x) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.62468f(x)
Graph of f(x) = exp(-4 * x) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.24542109.

How to work it out

  1. 1

    We want an antiderivative of f(x) = exp(-4 * x).

    ∫eāˆ’4x dx\int e^{-4 x}\,dx
  2. 2

    Method: ∫eāˆ’kxe^{āˆ’kx} dx = āˆ’eāˆ’kxe^{āˆ’kx}/k.

  3. 3

    That gives the antiderivative; add the constant C because any constant disappears when you differentiate:

    ∫eāˆ’4x dx=āˆ’eāˆ’4x4+C\int e^{-4 x}\,dx = \frac{-e^{-4 x}}{4} + C
  4. 4

    Check by differentiating the answer; you must get the original function back:

    ddx[āˆ’eāˆ’4x4]=āˆ’((āˆ’16)e(āˆ’4)x)16\frac{d}{dx}\left[\frac{-e^{-4 x}}{4}\right] = \frac{-\left(\left(-16\right) e^{\left(-4\right) x}\right)}{16}
  5. 5

    As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:

    ∫01eāˆ’4x dx=[āˆ’eāˆ’4x4]01ā‰ˆ0.24542109\int_{0}^{1} e^{-4 x}\,dx = \Big[\frac{-e^{-4 x}}{4}\Big]_{0}^{1} \approx 0.24542109
LaTeX source of the formulas on this page
\[ \int e^{-4 x}\,dx = \frac{-e^{-4 x}}{4} + C \]
\[ \int e^{-4 x}\,dx \]
\[ \frac{d}{dx}\left[\frac{-e^{-4 x}}{4}\right] = \frac{-\left(\left(-16\right) e^{\left(-4\right) x}\right)}{16} \]
\[ \int_{0}^{1} e^{-4 x}\,dx = \Big[\frac{-e^{-4 x}}{4}\Big]_{0}^{1} \approx 0.24542109 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.24542109
from 0 to 0.50.21616618

Questions

What is the integral of exp(-4 * x)?

The integral of eāˆ’4xe^{-4 x} is āˆ’eāˆ’4x4+C\frac{-e^{-4 x}}{4} + 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(-4 * x) from 0 to 1?

About 0.24542109.

More integrals

    Integral of exp(-4 * x): -exp(-4 * x) / 4 + C (with steps) | WordGPT