Question

You can help me?


`\sf \: find \: the\:answer \: of : \\ \bf \to \int^(1)_( - 1) \: (2x + 1)dx = . \: . \: . \\`
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Answer 1

~Hello! I will try to help you!

`\: \Rightarrow\sf \int^(1)_( - 1) \: (2x + 1)dx = . \: . \: . \\`

»Our first step is to find the integration (2x + 1), then:-

`\Rightarrow \sf ( (2)/(1 + 1) {x}^(1 + 1) ) + x \: ]^(1)_( - 1) \\ \Rightarrow \sf( \cancel{(2)/(2) } {x}^(2) ) + x \: ]^(1)_( - 1) = \rm {x}^(2) + x \: ]^(1)_( - 1)`

»The next step is to just substitute for the upper limit of 1 and the lower limit of -1 to find the result.

`\Rightarrow \sf( {1}^(2) + 1) - (( - 1) {}^(2) + ( - 1)) \\ \Rightarrow \sf(1 + 1) - (1 + ( - 1)){}{} \\ \Rightarrow \sf2 - 0 \\ \Rightarrow \bf2`

With this it's done!

Answer 2

Definite Integration

`\rightarrow \sf \int\limits^1_(-1) {(2x+1)} \, dx`

integrate the following

`\rightarrow \sf { [ (2x^2)/(2) +x ] }_(-1)^1`

simplify

`\rightarrow \sf [x^2+x]_(-1)^1`

apply limits

`\rightarrow \sf (1)^2+(1) - ( (-1)^2 +(-1))`

simplify

`\rightarrow \sf 2 - 0`

`\rightarrow \sf 2`