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Integral of e3xe^{3 x}

The integral of e3xe^{3 x} is e3x3+C\frac{e^{3 x}}{3} + C.

Graph of f(x) = exp(3x) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.620406080100f(x)
Graph of f(x) = exp(3x) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 6.3618456.

How to work it out

  1. 1

    We want an antiderivative of f(x) = exp(3x).

    ∫e3x dx\int e^{3 x}\,dx
  2. 2

    Method: ∫ekxe^{kx} dx = ekxe^{kx}/k.

  3. 3

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

    ∫e3x dx=e3x3+C\int e^{3 x}\,dx = \frac{e^{3 x}}{3} + C
  4. 4

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

    ddx[e3x3]=9e3x9\frac{d}{dx}\left[\frac{e^{3 x}}{3}\right] = \frac{9 e^{3 x}}{9}
  5. 5

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

    ∫01e3x dx=[e3x3]01ā‰ˆ6.3618456\int_{0}^{1} e^{3 x}\,dx = \Big[\frac{e^{3 x}}{3}\Big]_{0}^{1} \approx 6.3618456
LaTeX source of the formulas on this page
\[ \int e^{3 x}\,dx = \frac{e^{3 x}}{3} + C \]
\[ \int e^{3 x}\,dx \]
\[ \frac{d}{dx}\left[\frac{e^{3 x}}{3}\right] = \frac{9 e^{3 x}}{9} \]
\[ \int_{0}^{1} e^{3 x}\,dx = \Big[\frac{e^{3 x}}{3}\Big]_{0}^{1} \approx 6.3618456 \]

Definite integrals of the same function

IntervalValue
from 0 to 16.3618456
from 0 to 0.51.160563

Questions

What is the integral of exp(3x)?

The integral of e3xe^{3 x} is e3x3+C\frac{e^{3 x}}{3} + 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(3x) from 0 to 1?

About 6.3618456.

More integrals

    Integral of exp(3x): exp(3x) / 3 + C (with steps) | WordGPT