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Integral of (1/2)^x dx

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$$\int\limits_{4}^{-2} \left(\frac{1}{2}\right)^{x}\, dx$$

Integral((1/2)^x, (x, 4, -2))

Detail solution

Rewrite the integrand:

$\left(\frac{1}{2}\right)^{x} = 2^{- x}$
Let $u = - x$ .

Then let $du = - dx$ and substitute $- du$ :

$\int \left(- 2^{u}\right)\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2^{u}\, du = - \int 2^{u}\, du$
  1. The integral of an exponential function is itself divided by the natural logarithm of the base.
    
    $\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$
  So, the result is: $- \frac{2^{u}}{\log{\left(2 \right)}}$
Now substitute $u$ back in:

$- \frac{2^{- x}}{\log{\left(2 \right)}}$
Add the constant of integration:

$- \frac{2^{- x}}{\log{\left(2 \right)}}+ \mathrm{constant}$

The answer is:

$- \frac{2^{- x}}{\log{\left(2 \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                   
 |                -x  
 |  -x           2    
 | 2   dx = C - ------
 |              log(2)
/

$$\int \left(\frac{1}{2}\right)^{x}\, dx = C - \frac{2^{- x}}{\log{\left(2 \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   -63   
---------
16*log(2)

$$- \frac{63}{16 \log{\left(2 \right)}}$$

   -63   
---------
16*log(2)

$$- \frac{63}{16 \log{\left(2 \right)}}$$

-63/(16*log(2))

Numerical answer [src]

-5.68061172350029

-5.68061172350029

Use the examples entering the upper and lower limits of integration.