Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of e^x/x^2
Integral of (1+x)/x
Integral of coshx
Integral of a^x*e^x
Identical expressions
(seven -2x)*lg(x)
(7 minus 2x) multiply by lg(x)
(seven minus 2x) multiply by lg(x)
(7-2x)lg(x)
7-2xlgx
(7-2x)*lg(x)dx
Similar expressions
(7+2x)*lg(x)

Integral of (7-2x)*lg(x) dx

The solution

You have entered [src]

 1000                   
   /                    
  |                     
  |  (7 - 2*x)*log(x) dx
  |                     
 /                      
 100

$$\int\limits_{100}^{1000} \left(7 - 2 x\right) \log{\left(x \right)}\, dx$$

Integral((7 - 2*x)*log(x), (x, 100, 1000))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \log{\left(x \right)}$ .
  
  Then let $du = \frac{dx}{x}$ and substitute $du$ :
  
  $\int \left(- 2 u e^{2 u} + 7 u e^{u}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u e^{2 u}\right)\, du = - 2 \int u e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      So, the result is: $- u e^{2 u} + \frac{e^{2 u}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 7 u e^{u}\, du = 7 \int u e^{u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      Now evaluate the sub-integral.
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $7 u e^{u} - 7 e^{u}$
    The result is: $- u e^{2 u} + 7 u e^{u} + \frac{e^{2 u}}{2} - 7 e^{u}$
  Now substitute $u$ back in:
  
  $- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x$
Method #2
1. Rewrite the integrand:
  
  $\left(7 - 2 x\right) \log{\left(x \right)} = - 2 x \log{\left(x \right)} + 7 \log{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x \log{\left(x \right)}\right)\, dx = - 2 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 \log{\left(x \right)}\, dx = 7 \int \log{\left(x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      Now evaluate the sub-integral.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $7 x \log{\left(x \right)} - 7 x$
  The result is: $- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x$
Method #3
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 7 - 2 x$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
  
  To find $v{\left(x \right)}$ :
  1. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 7\, dx = 7 x$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 x\right)\, dx = - 2 \int x\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $- x^{2}$
    The result is: $- x^{2} + 7 x$
  Now evaluate the sub-integral.
2. Rewrite the integrand:
  
  $\frac{- x^{2} + 7 x}{x} = 7 - x$
3. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 7\, dx = 7 x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x\right)\, dx = - \int x\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $- \frac{x^{2}}{2}$
  The result is: $- \frac{x^{2}}{2} + 7 x$
Method #4
1. Rewrite the integrand:
  
  $\left(7 - 2 x\right) \log{\left(x \right)} = - 2 x \log{\left(x \right)} + 7 \log{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x \log{\left(x \right)}\right)\, dx = - 2 \int x \log{\left(x \right)}\, dx$
    1. Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
    So, the result is: $- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 \log{\left(x \right)}\, dx = 7 \int \log{\left(x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
      
      Now evaluate the sub-integral.
    2. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    So, the result is: $7 x \log{\left(x \right)} - 7 x$
  The result is: $- x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x$
Now simplify:

$\frac{x \left(- 2 x \log{\left(x \right)} + x + 14 \log{\left(x \right)} - 14\right)}{2}$
Add the constant of integration:

$\frac{x \left(- 2 x \log{\left(x \right)} + x + 14 \log{\left(x \right)} - 14\right)}{2}+ \mathrm{constant}$

The answer is:

$\frac{x \left(- 2 x \log{\left(x \right)} + x + 14 \log{\left(x \right)} - 14\right)}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                           2                               
 |                           x           2                    
 | (7 - 2*x)*log(x) dx = C + -- - 7*x - x *log(x) + 7*x*log(x)
 |                           2                                
/

$$\int \left(7 - 2 x\right) \log{\left(x \right)}\, dx = C - x^{2} \log{\left(x \right)} + \frac{x^{2}}{2} + 7 x \log{\left(x \right)} - 7 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

488700 - 993000*log(1000) + 9300*log(100)

$$- 993000 \log{\left(1000 \right)} + 9300 \log{\left(100 \right)} + 488700$$

488700 - 993000*log(1000) + 9300*log(100)

$$- 993000 \log{\left(1000 \right)} + 9300 \log{\left(100 \right)} + 488700$$

488700 - 993000*log(1000) + 9300*log(100)

Numerical answer [src]

-6327872.90929957

-6327872.90929957

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (7-2x)*lg(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4