Integral of 1-y dx

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$$\int\limits_{0}^{1} \left(1 - y\right)\, dy$$

Integral(1 - y, (y, 0, 1))

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                      2
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 | (1 - y) dy = C + y - --
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$$\int \left(1 - y\right)\, dy = C - \frac{y^{2}}{2} + y$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1/2

$$\frac{1}{2}$$

1/2

$$\frac{1}{2}$$

1/2

Numerical answer [src]

0.5

0.5

The graph

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