Mister Exam

Lang:

Solving integrals/ 4^(x+1)

Integral of 4^(x+1) dx

You have entered [src]

  1          
  /          
 |           
 |   x + 1   
 |  4      dx
 |           
/            
0

$$\int\limits_{0}^{1} 4^{x + 1}\, dx$$

Integral(4^(x + 1), (x, 0, 1))

Detail solution

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

$\frac{2^{2 x + 1}}{\log{\left(2 \right)}}$
Add the constant of integration:

$\frac{2^{2 x + 1}}{\log{\left(2 \right)}}+ \mathrm{constant}$

The answer is:

$\frac{2^{2 x + 1}}{\log{\left(2 \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                      
 |                  x + 1
 |  x + 1          4     
 | 4      dx = C + ------
 |                 log(4)
/

$$\int 4^{x + 1}\, dx = \frac{4^{x + 1}}{\log{\left(4 \right)}} + C$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  6   
------
log(2)

$$\frac{6}{\log{\left(2 \right)}}$$

  6   
------
log(2)

$$\frac{6}{\log{\left(2 \right)}}$$

6/log(2)

Numerical answer [src]

8.65617024533378

8.65617024533378

The graph

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

Method #1