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Identical expressions
11x+ forty (4x- sixteen)(x+ two)
11x plus 40(4x minus 16)(x plus 2)
11x plus forty (4x minus sixteen)(x plus two)
11x+404x-16x+2
11x+40(4x-16)(x+2)dx
Similar expressions
11x+40(4x+16)(x+2)
11x+40(4x-16)(x-2)
11x-40(4x-16)(x+2)

Integral of 11x+40(4x-16)(x+2) dx

The solution

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  1                                  
  /                                  
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 |  (11*x + 40*(4*x - 16)*(x + 2)) dx
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0

$$\int\limits_{0}^{1} \left(11 x + \left(x + 2\right) 40 \left(4 x - 16\right)\right)\, dx$$

Integral(11*x + (40*(4*x - 16))*(x + 2), (x, 0, 1))

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 11 x\, dx = 11 \int x\, dx$
  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: $\frac{11 x^{2}}{2}$
1. Rewrite the integrand:
  
  $\left(x + 2\right) 40 \left(4 x - 16\right) = 160 x^{2} - 320 x - 1280$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 160 x^{2}\, dx = 160 \int x^{2}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $\frac{160 x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 320 x\right)\, dx = - 320 \int x\, dx$
    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: $- 160 x^{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \left(-1280\right)\, dx = - 1280 x$
  The result is: $\frac{160 x^{3}}{3} - 160 x^{2} - 1280 x$
The result is: $\frac{160 x^{3}}{3} - \frac{309 x^{2}}{2} - 1280 x$
Now simplify:

$\frac{x \left(320 x^{2} - 927 x - 7680\right)}{6}$
Add the constant of integration:

$\frac{x \left(320 x^{2} - 927 x - 7680\right)}{6}+ \mathrm{constant}$

The answer is:

$\frac{x \left(320 x^{2} - 927 x - 7680\right)}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                      2        3
 |                                                  309*x    160*x 
 | (11*x + 40*(4*x - 16)*(x + 2)) dx = C - 1280*x - ------ + ------
 |                                                    2        3   
/

$$\int \left(11 x + \left(x + 2\right) 40 \left(4 x - 16\right)\right)\, dx = C + \frac{160 x^{3}}{3} - \frac{309 x^{2}}{2} - 1280 x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-8287/6

$$- \frac{8287}{6}$$

-8287/6

$$- \frac{8287}{6}$$

-8287/6

Numerical answer [src]

-1381.16666666667

-1381.16666666667

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

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Identical expressions

Similar expressions

Integral of 11x+40(4x-16)(x+2) dx

Limits of integration:

The graph:

Piecewise:

The solution