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Identical expressions
t^ two -t- two
t squared minus t minus 2
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t²-t-2
t to the power of 2-t-2
Similar expressions
t^2+t-2
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Integral of t^2-t-2 dt

The solution

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  1                
  /                
 |                 
 |  / 2        \   
 |  \t  - t - 2/ dt
 |                 
/                  
0

$$\int\limits_{0}^{1} \left(\left(t^{2} - t\right) - 2\right)\, dt$$

Integral(t^2 - t - 2, (t, 0, 1))

Detail solution

Integrate term-by-term:
1. Integrate term-by-term:
  1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int t^{2}\, dt = \frac{t^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- t\right)\, dt = - \int t\, dt$
    1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int t\, dt = \frac{t^{2}}{2}$
    So, the result is: $- \frac{t^{2}}{2}$
  The result is: $\frac{t^{3}}{3} - \frac{t^{2}}{2}$
1. The integral of a constant is the constant times the variable of integration:
  
  $\int \left(-2\right)\, dt = - 2 t$
The result is: $\frac{t^{3}}{3} - \frac{t^{2}}{2} - 2 t$
Now simplify:

$\frac{t \left(2 t^{2} - 3 t - 12\right)}{6}$
Add the constant of integration:

$\frac{t \left(2 t^{2} - 3 t - 12\right)}{6}+ \mathrm{constant}$

The answer is:

$\frac{t \left(2 t^{2} - 3 t - 12\right)}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                   
 |                              2    3
 | / 2        \                t    t 
 | \t  - t - 2/ dt = C - 2*t - -- + --
 |                             2    3 
/

$$\int \left(\left(t^{2} - t\right) - 2\right)\, dt = C + \frac{t^{3}}{3} - \frac{t^{2}}{2} - 2 t$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-13/6

$$- \frac{13}{6}$$

-13/6

$$- \frac{13}{6}$$

-13/6

Numerical answer [src]

-2.16666666666667

-2.16666666666667

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

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Identical expressions

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Integral of t^2-t-2 dt

Limits of integration:

The graph:

Piecewise:

The solution