Mister Exam

Lang:

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:
  
  $4^{x - 1} = \frac{4^{x}}{4}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{4^{x}}{4}\, dx = \frac{\int 4^{x}\, dx}{4}$
  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: $\frac{4^{x}}{4 \log{\left(4 \right)}}$
Now simplify:

$\frac{2^{2 x - 3}}{\log{\left(2 \right)}}$
Add the constant of integration:

$\frac{2^{2 x - 3}}{\log{\left(2 \right)}}+ \mathrm{constant}$

The answer is:

$\frac{2^{2 x - 3}}{\log{\left(2 \right)}}+ \mathrm{constant}$

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]

   3    
--------
8*log(2)

$$\frac{3}{8 \log{\left(2 \right)}}$$

   3    
--------
8*log(2)

$$\frac{3}{8 \log{\left(2 \right)}}$$

3/(8*log(2))

Numerical answer [src]

0.541010640333361

0.541010640333361

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

Method #1