Integral of (2-x)dx dx

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\int\limits_{0}^{1} \left(- x + 2\right) 1\, dx

Integral(2 - x*1, (x, 0, 1))

Detail solution

Let $u = 2 - x$ .

Then let $du = - dx$ and substitute $- du$ :

$\int u\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- u\right)\, du = - \int u\, du$
  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int u\, du = \frac{u^{2}}{2}$
  So, the result is: $- \frac{u^{2}}{2}$
Now substitute $u$ back in:

$- \frac{\left(2 - x\right)^{2}}{2}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                          2
 |                    (2 - x) 
 | (2 - x)*1 dx = C - --------
 |                       2    
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2\,x-{{x^2}\over{2}}

Simplify

The graph

Plot the graph f(x)

The answer [src]

3/2

{{3}\over{2}}

3/2

\frac{3}{2}

Simplify

Numerical answer [src]

1.5

1.5

The graph

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