How do you evaluate the definite integral $\int 4 - x$ $int 4-x$ from $[0, 2]$ $[0,2]$ ?

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How do you evaluate the definite integral $\int 4 - x$ $int 4-x$ from $[0, 2]$ $[0,2]$ ?

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Answer 1 · 2016-10-28T15:24:32

The integral $\int_{0}^{2} (4 - x) d x = 6$ $int_0^2(4-x)dx=6$

Explanation:

You have to use $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$ $(n \neq - 1)$ $(n!=-1)$
$\int (4 - x) d x = \int 4 d x - \int x d x = 4 x - \frac{x^{2}}{2}$ $int(4-x)dx=int4dx-intxdx=4x-x^2/2$

then you have to put the limits

$\int_{0}^{2} (4 - x) d x = {(4 x - \frac{x^{2}}{2})}_{0}^{2} = (4 \cdot 2 - 2 \cdot \frac{2}{2}) = 6$ $int_0^2(4-x)dx=(4x-x^2/2)_0^2=(4*2-2*2/2)=6$

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How do you evaluate the definite integral $\int 4 - x$ $int 4-x$ from $[0, 2]$ $[0,2]$ ?

1 Answer

Explanation:

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How do you evaluate the definite integral $\int 4 - x$ $int 4-x$ from $[0, 2]$ $[0,2]$ ?