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Integral of xāˆ’5x^{-5}

The integral of xāˆ’5x^{-5} is xāˆ’4āˆ’4+C\frac{x^{-4}}{-4} + C.

Graph of f(x) = x^(-5) with the area from 1 to 2 shaded0.40.60.811.21.41.61.822.22.42.61020304050f(x)
Graph of f(x) = x^(-5) with the area from 1 to 2 shaded. The shaded area equals the definite integral, about 0.234375.

How to work it out

  1. 1

    We want an antiderivative of f(x) = xāˆ’5x^{-5}.

    ∫xāˆ’5 dx\int x^{-5}\,dx
  2. 2

    Method: power rule with a negative exponent.

  3. 3

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

    ∫xāˆ’5 dx=xāˆ’4āˆ’4+C\int x^{-5}\,dx = \frac{x^{-4}}{-4} + C
  4. 4

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

    ddx[xāˆ’4āˆ’4]=16xāˆ’516\frac{d}{dx}\left[\frac{x^{-4}}{-4}\right] = \frac{16 x^{-5}}{16}
  5. 5

    As a definite integral, evaluate the antiderivative at both ends and subtract. Between 1 and 2:

    ∫12xāˆ’5 dx=[xāˆ’4āˆ’4]12ā‰ˆ0.234375\int_{1}^{2} x^{-5}\,dx = \Big[\frac{x^{-4}}{-4}\Big]_{1}^{2} \approx 0.234375
LaTeX source of the formulas on this page
\[ \int x^{-5}\,dx = \frac{x^{-4}}{-4} + C \]
\[ \int x^{-5}\,dx \]
\[ \frac{d}{dx}\left[\frac{x^{-4}}{-4}\right] = \frac{16 x^{-5}}{16} \]
\[ \int_{1}^{2} x^{-5}\,dx = \Big[\frac{x^{-4}}{-4}\Big]_{1}^{2} \approx 0.234375 \]

Definite integrals of the same function

IntervalValue
from 1 to 20.234375
from 1 to 1.50.20061728

Questions

What is the integral of xāˆ’5x^{-5}?

The integral of xāˆ’5x^{-5} is xāˆ’4āˆ’4+C\frac{x^{-4}}{-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 xāˆ’5x^{-5} from 1 to 2?

About 0.234375.

More integrals

    Integral of x^(-5): x^(-4) / -4 + C (with steps) | WordGPT