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Solving integrals/ 2^(x+2)

Integral of 2^(x+2) dx

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  1          
  /          
 |           
 |   x + 2   
 |  2      dx
 |           
/            
0

$$\int\limits_{0}^{1} 2^{x + 2}\, dx$$

Integral(2^(x + 2), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = x + 2$ .
  
  Then let $du = dx$ and substitute $du$ :
  
  $\int 2^{u}\, du$
  1. The integral of an exponential function is itself divided by the natural logarithm of the base.
    
    $\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$
  Now substitute $u$ back in:
  
  $\frac{2^{x + 2}}{\log{\left(2 \right)}}$
Method #2
1. Rewrite the integrand:
  
  $2^{x + 2} = 4 \cdot 2^{x}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 4 \cdot 2^{x}\, dx = 4 \int 2^{x}\, dx$
  1. The integral of an exponential function is itself divided by the natural logarithm of the base.
    
    $\int 2^{x}\, dx = \frac{2^{x}}{\log{\left(2 \right)}}$
  So, the result is: $\frac{4 \cdot 2^{x}}{\log{\left(2 \right)}}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

  4   
------
log(2)

$$\frac{4}{\log{\left(2 \right)}}$$

  4   
------
log(2)

$$\frac{4}{\log{\left(2 \right)}}$$

Упростить

Numerical answer [src]

5.77078016355585

5.77078016355585

The graph

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

Method #1